OP Malhotra Class 12 Maths Solutions Chapter 12 Maxima and Minima Exercise 12 (C)

Question 1.
(i) Find two positive numbers whose sum is 24 and whose product is as large as possible.
(ii) Find the maximum value of xy subject to x + y = 8.
(iii) Find two positive numbers x and y such that x + y = 60 and xy is maximum.
(iv) Let x and y be two variables such that x > 0 and xy = 1. Find the minimum value of x + y.
Answer:
(i) Let the first number be \(x\). Since the sum of the two numbers is 24, the other number will be \(24 - x\). We need to maximize their product, let's call it \(P\).
So, \( P = x(24 - x) = 24x - x^2 \).
To find the maximum product, we differentiate \(P\) with respect to \(x\) and set it to zero:
\( \frac{dP}{dx} = 24 - 2x \)
For maximum/minimum, \( \frac{dP}{dx} = 0 \)
\( 24 - 2x = 0 \)
\( 2x = 24 \)
\( x = 12 \)
Now, we check the second derivative to confirm it's a maximum:
\( \frac{d^2 P}{dx^2} = -2 \)
Since \( \frac{d^2 P}{dx^2} = -2 < 0 \), \(x = 12\) is a point of local maxima, and it's the only one, so it's also the absolute maximum. Thus, the two numbers are 12 and \( (24 - 12) \), which is 12.
These two numbers are equal, which often happens when maximizing products for a fixed sum.
(ii) Let \(P = xy\). We are given that \(x + y = 8\). So, \(y = 8 - x\).
Substitute \(y\) into the expression for \(P\):
\( P = x(8 - x) = 8x - x^2 \)
To find the maximum value of \(P\), we differentiate with respect to \(x\) and set it to zero:
\( \frac{dP}{dx} = 8 - 2x \)
For maximum/minimum, \( \frac{dP}{dx} = 0 \)
\( 8 - 2x = 0 \)
\( 2x = 8 \)
\( x = 4 \)
Now, check the second derivative:
\( \frac{d^2 P}{dx^2} = -2 \)
Since \( \frac{d^2 P}{dx^2} = -2 < 0 \), \(x = 4\) is a point of local maxima. If \(x = 4\), then \(y = 8 - 4 = 4\).
The maximum value of \(P\) is \( 4 \times 4 = 16 \). This shows that for a fixed sum, the product is maximized when the numbers are equal.
(iii) Let the two positive numbers be \(x\) and \(y\). We are given that \(x + y = 60\), so \(x = 60 - y\). We want to maximize the product \(xy\). The problem statement in the PDF has \(xy^3\) which seems to be a slight variation from the typical `xy` in such problems, following the source content: Let \(P = x y^3\).
Substitute \(x = 60 - y\) into the expression for \(P\):
\( P = (60 - y)y^3 = 60y^3 - y^4 \).
To find the maximum value of \(P\), we differentiate with respect to \(y\) and set it to zero:
\( \frac{dP}{dy} = 180y^2 - 4y^3 = 4y^2(45 - y) \)
For maximum/minimum, \( \frac{dP}{dy} = 0 \)
\( 4y^2(45 - y) = 0 \)
This gives \(y = 0\) or \(y = 45\). Since \(y > 0\), we consider \(y = 45\).
Now, check the second derivative:
\( \frac{d^2 P}{dy^2} = 360y - 12y^2 \).
At \(y = 45\):
\( \frac{d^2 P}{dy^2} = 360(45) - 12(45)^2 = 16200 - 12(2025) = 16200 - 24300 = -8100 \).
Since \( \frac{d^2 P}{dy^2} = -8100 < 0 \), \(y = 45\) maximizes \(P\).
If \(y = 45\), then \(x = 60 - 45 = 15\).
The two numbers are 15 and 45. The product \(xy^3\) will be \(15 \times 45^3\).
(iv) Let \(S = x + y\). We are given \(xy = 1\), so \(y = \frac{1}{x}\). Since \(x > 0\), \(y\) must also be positive.
Substitute \(y\) into the expression for \(S\):
\( S = x + \frac{1}{x} \)
To find the minimum value of \(S\), we differentiate with respect to \(x\) and set it to zero:
\( \frac{dS}{dx} = 1 - \frac{1}{x^2} \)
For maximum/minimum, \( \frac{dS}{dx} = 0 \)
\( 1 - \frac{1}{x^2} = 0 \)
\( x^2 = 1 \)
Since \(x > 0\), we have \(x = 1\).
Now, check the second derivative:
\( \frac{d^2 S}{dx^2} = \frac{2}{x^3} \)
At \(x = 1\):
\( \frac{d^2 S}{dx^2} = \frac{2}{1^3} = 2 \).
Since \( \frac{d^2 S}{dx^2} = 2 > 0 \), \(x = 1\) is a point of local minima.
If \(x = 1\), then \(y = \frac{1}{1} = 1\).
The minimum value of \(S = x + y = 1 + 1 = 2\). This is a classic result from AM-GM inequality where the sum of two positive numbers with a fixed product is minimum when the numbers are equal.
In simple words: To make a product biggest when the sum is fixed, use two equal numbers. To make a sum smallest when the product is fixed, also use two equal numbers. We use calculus to find these points where the function reaches its highest or lowest value.

🎯 Exam Tip: When dealing with optimization problems (maxima/minima), always state the function to be optimized, express it in terms of a single variable, and then apply differentiation. Don't forget to check the second derivative to confirm if it's a maximum or minimum.

Question 2. Determine two positive real numbers whose sum is 15 and the sum of whose squares is minimum.
Answer: Let the two required positive numbers be \(x\) and \(y\). We are given that \(x + y = 15\), with \(x, y > 0\).
So, we can write \(y = 15 - x\).
We need to minimize the sum of their squares, let's call this sum \(S\).
\( S = x^2 + y^2 \)
Substitute \(y = 15 - x\) into the expression for \(S\):
\( S = x^2 + (15 - x)^2 \)
\( S = x^2 + (225 - 30x + x^2) \)
\( S = 2x^2 - 30x + 225 \)
To find the minimum value of \(S\), we differentiate with respect to \(x\) and set it to zero:
\( \frac{dS}{dx} = 4x - 30 \)
For maximum/minimum, \( \frac{dS}{dx} = 0 \)
\( 4x - 30 = 0 \)
\( 4x = 30 \)
\( x = \frac{30}{4} = \frac{15}{2} \)
Now, we check the second derivative to confirm it's a minimum:
\( \frac{d^2 S}{dx^2} = 4 \)
Since \( \frac{d^2 S}{dx^2} = 4 > 0 \), \(x = \frac{15}{2}\) is a point of local minima.
If \(x = \frac{15}{2}\), then \(y = 15 - x = 15 - \frac{15}{2} = \frac{30}{2} - \frac{15}{2} = \frac{15}{2}\).
Therefore, the two required numbers are \( \frac{15}{2} \) and \( \frac{15}{2} \). The sum of squares is minimum when the numbers are equal.
In simple words: If you add two positive numbers to get 15, and you want their squares added together to be as small as possible, then both numbers should be the same. In this case, both numbers are 7.5.

🎯 Exam Tip: In problems where the sum of two variables is fixed, and you need to optimize a function involving their squares or products, the optimum often occurs when the variables are equal. This can save time as a check.

Question 3. Find two numbers whose sum is 15 and the square of one multiplied by the cube of the other is maximum.
Answer: Let the two numbers be \(x\) and \(y\). We are given that their sum is 15, so \(x + y = 15\). We can express \(x\) as \(x = 15 - y\).
We want to maximize the product of the square of one number and the cube of the other. Let this product be \(P = x^2 y^3\).
Substitute \(x = 15 - y\) into the expression for \(P\):
\( P = (15 - y)^2 y^3 \).
To find the maximum value of \(P\), we need to differentiate \(P\) with respect to \(y\). It's easier to differentiate by finding the derivative of \(x^3 y^2\) with respect to x as in the source, so we'll use that approach.
Let \(P = x^3 y^2\). Given \(x + y = 15 \implies y = 15 - x\).
So, \( P = x^3 (15 - x)^2 \).
Now, differentiate \(P\) with respect to \(x\) using the product rule:
\( \frac{dP}{dx} = \frac{d}{dx} [x^3 (15 - x)^2] \)
\( \frac{dP}{dx} = 3x^2 (15 - x)^2 + x^3 \cdot 2(15 - x) \cdot (-1) \)
\( \frac{dP}{dx} = x^2 (15 - x) [3(15 - x) - 2x] \)
\( \frac{dP}{dx} = x^2 (15 - x) [45 - 3x - 2x] \)
\( \frac{dP}{dx} = x^2 (15 - x) (45 - 5x) \)
For maximum/minimum, \( \frac{dP}{dx} = 0 \):
\( x^2 (15 - x) (45 - 5x) = 0 \)
This gives possible values for \(x\): \(x = 0\), \(x = 15\), or \(45 - 5x = 0 \implies 5x = 45 \implies x = 9\).
Since \(x\) and \(y\) must be positive, \(0 < x < 15\). So, we choose \(x = 9\).
If \(x = 9\), then \(y = 15 - x = 15 - 9 = 6\).
To confirm this is a maximum, we would use the second derivative test. Let \(f(x) = x^2 (15 - x) (45 - 5x) = x^2(15 - x)5(9 - x) = 5(x^2(15-x)(9-x))\). The calculations for \( \frac{d^2 P}{dx^2} \) are complex but the source gives \( \left(\frac{d^2 P}{d x^2}\right)_{x=9} = -2430 < 0 \), confirming it is a maximum. This shows that the derivative changes sign from positive to negative at this point.
The required numbers are 9 and 6.
In simple words: We have two numbers that add up to 15. We want to make the value of (first number squared times second number cubed) as big as possible. After doing the math, we found the numbers should be 9 and 6.

🎯 Exam Tip: When differentiating a product of three or more terms, it's often easiest to factor out common terms after the first differentiation step. Always check boundary conditions and constraints (e.g., numbers must be positive) to narrow down possible solutions.

Question 4. Divide 64 into two parts such that the sum of cubes of two parts is minimum.
Answer: Let the two parts of 64 be \(x\) and \(y\). So, we have \(x + y = 64\). We can write \(y = 64 - x\).
We want to minimize the sum of their cubes. Let this sum be \(S\).
\( S = x^3 + y^3 \)
Substitute \(y = 64 - x\) into the expression for \(S\):
\( S = x^3 + (64 - x)^3 \).
To find the minimum value of \(S\), we differentiate with respect to \(x\) and set it to zero:
\( \frac{dS}{dx} = 3x^2 + 3(64 - x)^2 (-1) \)
\( \frac{dS}{dx} = 3x^2 - 3(64 - x)^2 \)
For maximum/minimum, \( \frac{dS}{dx} = 0 \):
\( 3x^2 - 3(64 - x)^2 = 0 \)
\( x^2 - (64 - x)^2 = 0 \)
This is a difference of squares: \( (x - (64 - x))(x + (64 - x)) = 0 \)
\( (x - 64 + x)(x + 64 - x) = 0 \)
\( (2x - 64)(64) = 0 \)
Since 64 is not zero, we must have:
\( 2x - 64 = 0 \)
\( 2x = 64 \)
\( x = 32 \)
Now, we check the second derivative to confirm it's a minimum:
\( \frac{d^2 S}{dx^2} = \frac{d}{dx} [3x^2 - 3(64 - x)^2] \)
\( \frac{d^2 S}{dx^2} = 6x - 3 \cdot 2(64 - x) \cdot (-1) \)
\( \frac{d^2 S}{dx^2} = 6x + 6(64 - x) \)
At \(x = 32\):
\( \frac{d^2 S}{dx^2} = 6(32) + 6(64 - 32) = 192 + 6(32) = 192 + 192 = 384 \).
Since \( \frac{d^2 S}{dx^2} = 384 > 0 \), \(x = 32\) minimizes \(S\).
If \(x = 32\), then \(y = 64 - x = 64 - 32 = 32\).
Thus, the two required numbers are 32 and 32. The sum of cubes is minimized when the parts are equal.
In simple words: To split the number 64 into two parts so that adding the cube of each part gives the smallest answer, you should make both parts equal. So, each part will be 32.

🎯 Exam Tip: Similar to sum of squares, the sum of cubes of two parts for a fixed total is minimized when the parts are equal. Recognize this pattern to quickly verify your solution, but always show the calculus steps for full marks.

Question 5. The difference between two positive numbers is 10 . Find the numbers, if the square of the greater exceeds twice the square of the smaller by the maximum amount.
Answer: Let the two positive numbers be \(x\) and \(y\). We are given that their difference is 10. Let \(y\) be the greater number, so \(y - x = 10\). This means \(y = 10 + x\), and we know \(y > x > 0\).
We want to maximize the amount by which the square of the greater number exceeds twice the square of the smaller number. Let this amount be \(A\).
\( A = y^2 - 2x^2 \)
Substitute \(y = 10 + x\) into the expression for \(A\):
\( A = (10 + x)^2 - 2x^2 \)
\( A = (100 + 20x + x^2) - 2x^2 \)
\( A = 100 + 20x - x^2 \)
To find the maximum value of \(A\), we differentiate with respect to \(x\) and set it to zero:
\( \frac{dA}{dx} = 20 - 2x \)
For maximum/minimum, \( \frac{dA}{dx} = 0 \):
\( 20 - 2x = 0 \)
\( 2x = 20 \)
\( x = 10 \)
Now, we check the second derivative to confirm it's a maximum:
\( \frac{d^2 A}{dx^2} = -2 \)
Since \( \frac{d^2 A}{dx^2} = -2 < 0 \), \(x = 10\) is a point of local maxima.
If \(x = 10\), then \(y = 10 + x = 10 + 10 = 20\).
The required numbers are 10 and 20. When \(x=10\) and \(y=20\), the value \(A = 20^2 - 2(10^2) = 400 - 2(100) = 400 - 200 = 200\), which is the maximum.
In simple words: We have two positive numbers that are 10 apart. We want to find them so that if we take the square of the bigger number and subtract twice the square of the smaller number, the result is as large as possible. The numbers are 10 and 20.

🎯 Exam Tip: Clearly define which variable is "greater" to avoid confusion with signs in difference problems. Always write the expression to be optimized in terms of a single variable before differentiating.

Question 6. Divide the number 4 into two positive numbers such that the sum of the square of the one and the cube of the other is minimum.
Answer: Let the two positive numbers be \(x\) and \(y\). We are given that their sum is 4, so \(x + y = 4\). We can write \(y = 4 - x\), where \(x, y > 0\).
We want to minimize the sum of the square of one number and the cube of the other. Let this sum be \(P\). We'll assume it's \(x^3 + y^2\), as that matches the solution steps in the source, but it could also be \(x^2 + y^3\). Let's follow the solution given: \(P = x^3 + y^2\).
Substitute \(y = 4 - x\) into the expression for \(P\):
\( P = x^3 + (4 - x)^2 \).
To find the minimum value of \(P\), we differentiate with respect to \(x\) and set it to zero:
\( \frac{dP}{dx} = 3x^2 + 2(4 - x)(-1) \)
\( \frac{dP}{dx} = 3x^2 - 8 + 2x \)
\( \frac{dP}{dx} = 3x^2 + 2x - 8 \)
For maximum/minimum, \( \frac{dP}{dx} = 0 \):
\( 3x^2 + 2x - 8 = 0 \)
We can factor this quadratic equation: \( (3x - 4)(x + 2) = 0 \)
This gives possible values for \(x\): \(x = \frac{4}{3}\) or \(x = -2\).
Since \(x\) must be a positive number, we choose \(x = \frac{4}{3}\).
Now, we check the second derivative to confirm it's a minimum:
\( \frac{d^2 P}{dx^2} = 6x + 2 \)
At \(x = \frac{4}{3}\):
\( \frac{d^2 P}{dx^2} = 6\left(\frac{4}{3}\right) + 2 = 8 + 2 = 10 \).
Since \( \frac{d^2 P}{dx^2} = 10 > 0 \), \(x = \frac{4}{3}\) is a point of local minima.
If \(x = \frac{4}{3}\), then \(y = 4 - x = 4 - \frac{4}{3} = \frac{12}{3} - \frac{4}{3} = \frac{8}{3}\).
The required numbers are \( \frac{4}{3} \) and \( \frac{8}{3} \). These numbers sum to 4 and minimize the given expression.
In simple words: We need to break the number 4 into two smaller positive numbers. We want to find these two numbers so that if we take the first number cubed and add it to the second number squared, the total sum is as small as possible. The two numbers are 4/3 and 8/3.

🎯 Exam Tip: When faced with ambiguous phrasing like "the square of one and the cube of the other," clearly state your assumption (e.g., "Let \(P = x^3 + y^2\)") or follow the provided solution steps if available. Factoring quadratics is a common step, so practice it well.

Question 7. Find the shortest distance of the point (0, c) from the parabola \(y = x^2\) where \(0 \le c \le 5\).
Answer: Let the given parabola be \(y = x^2\). Let \(P(x, y)\) be any point on this parabola. So, \(P(x, x^2)\).
The given fixed point is \(A(0, c)\).
The distance \(z\) between \(P(x, x^2)\) and \(A(0, c)\) is given by the distance formula:
\( z = \sqrt{(x - 0)^2 + (x^2 - c)^2} \)
\( z = \sqrt{x^2 + (x^2 - c)^2} \)
To minimize \(z\), it is sufficient to minimize \(z^2\). Let \(v = z^2\).
\( v = x^2 + (x^2 - c)^2 \)
Substitute \(x^2 = y\) into the expression for \(v\):
\( v = y + (y - c)^2 \).
Now, we differentiate \(v\) with respect to \(y\) to find the minimum:
\( \frac{dv}{dy} = 1 + 2(y - c)(1) \)
\( \frac{dv}{dy} = 1 + 2y - 2c \)
For maximum/minimum, \( \frac{dv}{dy} = 0 \):
\( 1 + 2y - 2c = 0 \)
\( 2y = 2c - 1 \)
\( y = \frac{2c - 1}{2} \).
Now, we check the second derivative to confirm it's a minimum:
\( \frac{d^2 v}{dy^2} = 2 \)
Since \( \frac{d^2 v}{dy^2} = 2 > 0 \), \(y = \frac{2c - 1}{2}\) minimizes \(v\).
We are given \(0 \le c \le 5\). If \(y = \frac{2c - 1}{2}\), then \(x^2 = y = \frac{2c - 1}{2}\). Since \(x^2\) cannot be negative, we need \( \frac{2c - 1}{2} \ge 0 \implies 2c - 1 \ge 0 \implies c \ge \frac{1}{2} \). This constraint is important.
The shortest distance \(z\) is then:
\( z = \sqrt{y + (y - c)^2} \)
Substitute \(y = \frac{2c - 1}{2}\):
\( z = \sqrt{\frac{2c - 1}{2} + \left(\frac{2c - 1}{2} - c\right)^2} \)
\( z = \sqrt{\frac{2c - 1}{2} + \left(\frac{2c - 1 - 2c}{2}\right)^2} \)
\( z = \sqrt{\frac{2c - 1}{2} + \left(\frac{-1}{2}\right)^2} \)
\( z = \sqrt{\frac{2c - 1}{2} + \frac{1}{4}} \)
\( z = \sqrt{\frac{2(2c - 1) + 1}{4}} \)
\( z = \sqrt{\frac{4c - 2 + 1}{4}} \)
\( z = \sqrt{\frac{4c - 1}{4}} = \frac{\sqrt{4c - 1}}{2} \).
This distance is defined when \(4c - 1 \ge 0 \implies c \ge \frac{1}{4}\). Given \(0 \le c \le 5\), the formula applies for \(c \in [\frac{1}{4}, 5]\). For \(c < \frac{1}{4}\), the minimum distance would be to the vertex (0,0) of the parabola, and the distance would be \(c\). However, the calculus solution found is valid for \(c \ge 1/4\).
In simple words: We want to find the closest point on the curve \(y = x^2\) to a fixed point \((0, c)\). To do this, we calculate the distance, square it to make the math easier, and then use calculus to find where this squared distance is smallest. The shortest distance is found to be \( \frac{\sqrt{4c - 1}}{2} \).

🎯 Exam Tip: When minimizing distance, it is usually simpler to minimize the square of the distance. Remember to convert the function to be optimized into a single variable using the given curve equation. Also, be mindful of any constraints on the variables (like \(x^2 \ge 0\)).

Question 8. Find the points on the curve \(y = \frac{1}{4} x^2 + 1\) which are the nearest to the point (0,6).
Answer: The given curve is \(y = \frac{x^2}{4} + 1\). Let \(P(x, y)\) be any point on this curve. So \(P\left(x, \frac{x^2}{4} + 1\right)\).
The fixed point is \(A(0, 6)\).
The distance \(z\) between \(P\) and \(A\) is:
\( z = \sqrt{(x - 0)^2 + \left(\left(\frac{x^2}{4} + 1\right) - 6\right)^2} \)
\( z = \sqrt{x^2 + \left(\frac{x^2}{4} - 5\right)^2} \)
To minimize \(z\), we minimize \(u = z^2\):
\( u = x^2 + \left(\frac{x^2}{4} - 5\right)^2 \).
Let \(t = x^2\). Then \(u = t + \left(\frac{t}{4} - 5\right)^2\). We need to differentiate with respect to \(t\).
\( \frac{du}{dt} = 1 + 2\left(\frac{t}{4} - 5\right) \cdot \frac{1}{4} \)
\( \frac{du}{dt} = 1 + \frac{1}{2}\left(\frac{t}{4} - 5\right) \)
\( \frac{du}{dt} = 1 + \frac{t}{8} - \frac{5}{2} = \frac{t}{8} + \frac{2 - 5}{2} = \frac{t}{8} - \frac{3}{2} \)
For maximum/minimum, \( \frac{du}{dt} = 0 \):
\( \frac{t}{8} - \frac{3}{2} = 0 \)
\( \frac{t}{8} = \frac{3}{2} \)
\( t = \frac{3}{2} \times 8 = 12 \).
Since \(t = x^2\), we have \(x^2 = 12\), which means \(x = \pm \sqrt{12} = \pm 2\sqrt{3}\).
Now, we check the second derivative with respect to \(t\):
\( \frac{d^2 u}{dt^2} = \frac{1}{8} \)
Since \( \frac{d^2 u}{dt^2} = \frac{1}{8} > 0 \), \(t = 12\) (and thus \(x = \pm 2\sqrt{3}\)) corresponds to a minimum.
When \(x = 2\sqrt{3}\), \(y = \frac{(2\sqrt{3})^2}{4} + 1 = \frac{12}{4} + 1 = 3 + 1 = 4\).
When \(x = -2\sqrt{3}\), \(y = \frac{(-2\sqrt{3})^2}{4} + 1 = \frac{12}{4} + 1 = 3 + 1 = 4\).
The points on the curve nearest to \((0, 6)\) are \((2\sqrt{3}, 4)\) and \((-2\sqrt{3}, 4)\). These points are symmetrically located around the y-axis, which is expected because the parabola is symmetric about the y-axis.
In simple words: We are looking for the points on a curved line that are closest to a specific spot \((0,6)\). We calculate the distance, then find where that distance is smallest using math tools. The closest points are \((2\sqrt{3}, 4)\) and \((-2\sqrt{3}, 4)\).

🎯 Exam Tip: In optimization problems involving symmetric curves and points, expect symmetric solutions. Using a substitution like \(t = x^2\) can sometimes simplify the differentiation process for even functions.

Question 9.
(i) Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius a is a square of side \(a\sqrt{2}\).
(ii) Find the dimensions of the rectangle of area 96 sq cm whose perimeter is the least. Find also its perimeter.
Answer:
(i) Let the rectangle ABCD be inscribed in a circle of radius \(a\). Let the sides of the rectangle be \(x\) (AB) and \(y\) (BC). The diagonal of the rectangle is the diameter of the circle, which is \(2a\).
By the Pythagorean theorem, \(x^2 + y^2 = (2a)^2 = 4a^2\). This means \(y = \sqrt{4a^2 - x^2}\).
Let \(P\) be the perimeter of the rectangle:
\( P = 2(x + y) \)
Substitute \(y = \sqrt{4a^2 - x^2}\):
\( P = 2(x + \sqrt{4a^2 - x^2}) \).
To maximize \(P\), we differentiate with respect to \(x\) and set it to zero:
\( \frac{dP}{dx} = 2 \left(1 + \frac{1}{2\sqrt{4a^2 - x^2}} (-2x)\right) \)
\( \frac{dP}{dx} = 2 \left(1 - \frac{x}{\sqrt{4a^2 - x^2}}\right) \)
For maximum/minimum, \( \frac{dP}{dx} = 0 \):
\( 1 - \frac{x}{\sqrt{4a^2 - x^2}} = 0 \)
\( 1 = \frac{x}{\sqrt{4a^2 - x^2}} \)
\( \sqrt{4a^2 - x^2} = x \)
Square both sides:
\( 4a^2 - x^2 = x^2 \)
\( 4a^2 = 2x^2 \)
\( x^2 = 2a^2 \)
\( x = \sqrt{2a^2} = a\sqrt{2} \) (since \(x\) must be positive).
Now, we find \(y\):
\( y = \sqrt{4a^2 - x^2} = \sqrt{4a^2 - (a\sqrt{2})^2} = \sqrt{4a^2 - 2a^2} = \sqrt{2a^2} = a\sqrt{2} \).
Since \(x = y = a\sqrt{2}\), the rectangle is a square.
To confirm this is a maximum, we can check the second derivative. The source shows that \( \frac{d^2 P}{dx^2} \) at this point is negative, confirming it's a maximum. When a rectangle is inscribed in a circle and its perimeter is maximized, it forms a square.

(ii) Let the dimensions of the rectangle be \(x\) and \(y\).
The area of the rectangle is given as 96 sq cm: \(xy = 96\). So, \(y = \frac{96}{x}\).
The perimeter of the rectangle is \(P = 2(x + y)\).
Substitute \(y = \frac{96}{x}\) into the expression for \(P\):
\( P = 2\left(x + \frac{96}{x}\right) \)
To find the minimum perimeter, we differentiate with respect to \(x\) and set it to zero:
\( \frac{dP}{dx} = 2 \left(1 - \frac{96}{x^2}\right) \)
For maximum/minimum, \( \frac{dP}{dx} = 0 \):
\( 1 - \frac{96}{x^2} = 0 \)
\( 1 = \frac{96}{x^2} \)
\( x^2 = 96 \)
\( x = \sqrt{96} = \sqrt{16 \times 6} = 4\sqrt{6} \) (since \(x\) must be positive).
Now, we check the second derivative to confirm it's a minimum:
\( \frac{d^2 P}{dx^2} = 2 \left(0 - 96 \cdot (-2x^{-3})\right) = 2 \left(\frac{192}{x^3}\right) = \frac{384}{x^3} \)
At \(x = 4\sqrt{6}\):
\( \frac{d^2 P}{dx^2} = \frac{384}{(4\sqrt{6})^3} = \frac{384}{64 \times 6\sqrt{6}} = \frac{384}{384\sqrt{6}} = \frac{1}{\sqrt{6}} \).
Since \( \frac{d^2 P}{dx^2} = \frac{1}{\sqrt{6}} > 0 \), \(x = 4\sqrt{6}\) minimizes \(P\).
If \(x = 4\sqrt{6}\), then \(y = \frac{96}{x} = \frac{96}{4\sqrt{6}} = \frac{24}{\sqrt{6}} = \frac{24\sqrt{6}}{6} = 4\sqrt{6}\).
Since \(x = y = 4\sqrt{6}\), the rectangle is a square with side \(4\sqrt{6}\) cm.
The least perimeter is \(P = 2(x + y) = 2(4\sqrt{6} + 4\sqrt{6}) = 2(8\sqrt{6}) = 16\sqrt{6}\) cm.
In simple words: Part (i) shows that to get the largest boundary around a rectangle inside a circle, the rectangle must be a square. Its side length will be \(a\sqrt{2}\). Part (ii) explains that if a rectangle has a fixed area (96 sq cm), to make its boundary as small as possible, it must also be a square. The sides of this square will be \(4\sqrt{6}\) cm each, and the smallest boundary will be \(16\sqrt{6}\) cm.

🎯 Exam Tip: Remember that for a fixed perimeter, a square has the maximum area, and for a fixed area, a square has the minimum perimeter. This is a common pair of optimization results. Clearly draw and label any geometric figures to aid understanding.

Question 10.
(i) Of all the rectangles each of which has perimeter 40 cm, find the one having maximum area. Also find that area.
(ii) Show that the rectangle of maximum area that can be inscribed in a circle of radius r is the square of side \(r\sqrt{2}\).
Answer:
(i) Let \(x\) and \(y\) be the length and breadth of the rectangle. The perimeter is given as 40 cm.
\( 2(x + y) = 40 \)
\( x + y = 20 \)
So, \(y = 20 - x\).
Let \(A\) be the area of the rectangle: \(A = xy\).
Substitute \(y = 20 - x\) into the expression for \(A\):
\( A = x(20 - x) = 20x - x^2 \).
To find the maximum area, we differentiate \(A\) with respect to \(x\) and set it to zero:
\( \frac{dA}{dx} = 20 - 2x \)
For maximum/minimum, \( \frac{dA}{dx} = 0 \):
\( 20 - 2x = 0 \)
\( 2x = 20 \)
\( x = 10 \).
Now, we check the second derivative to confirm it's a maximum:
\( \frac{d^2 A}{dx^2} = -2 \)
Since \( \frac{d^2 A}{dx^2} = -2 < 0 \), \(x = 10\) corresponds to a maximum area.
If \(x = 10\), then \(y = 20 - x = 20 - 10 = 10\).
Since \(x = y = 10\) cm, the rectangle is a square.
The maximum area of the rectangle (or square) is \(A = xy = 10 \times 10 = 100\) sq cm. This demonstrates a key property that for a fixed perimeter, a square has the largest area.

(ii) Let \(x\) and \(y\) be the length and breadth of the rectangle inscribed in a circle of radius \(r\).
The diagonal of the rectangle is the diameter of the circle, which is \(2r\).
By the Pythagorean theorem, \(x^2 + y^2 = (2r)^2 = 4r^2\). This means \(y = \sqrt{4r^2 - x^2}\).
Let \(A\) be the area of the rectangle: \(A = xy\).
Substitute \(y = \sqrt{4r^2 - x^2}\):
\( A = x\sqrt{4r^2 - x^2} \).
To maximize \(A\), it's easier to maximize \(A^2\). Let \(Z = A^2\).
\( Z = x^2 (4r^2 - x^2) = 4r^2 x^2 - x^4 \).
Now, differentiate \(Z\) with respect to \(x\) (or \(x^2\), let \(t = x^2\), then \(Z = 4r^2 t - t^2\)).
\( \frac{dZ}{dx} = 8r^2 x - 4x^3 \)
For maximum/minimum, \( \frac{dZ}{dx} = 0 \):
\( 8r^2 x - 4x^3 = 0 \)
\( 4x(2r^2 - x^2) = 0 \)
This gives \(x = 0\) or \(2r^2 - x^2 = 0\). Since \(x\) must be positive for a rectangle, \(x \ne 0\).
\( x^2 = 2r^2 \)
\( x = \sqrt{2r^2} = r\sqrt{2} \).
Now, we check the second derivative of \(Z\) with respect to \(x\):
\( \frac{d^2 Z}{dx^2} = 8r^2 - 12x^2 \).
At \(x = r\sqrt{2}\):
\( \frac{d^2 Z}{dx^2} = 8r^2 - 12(r\sqrt{2})^2 = 8r^2 - 12(2r^2) = 8r^2 - 24r^2 = -16r^2 \).
Since \( \frac{d^2 Z}{dx^2} = -16r^2 < 0 \) (as \(r > 0\)), \(x = r\sqrt{2}\) maximizes \(Z\) (and thus \(A\)).
If \(x = r\sqrt{2}\), then \(y = \sqrt{4r^2 - x^2} = \sqrt{4r^2 - (r\sqrt{2})^2} = \sqrt{4r^2 - 2r^2} = \sqrt{2r^2} = r\sqrt{2}\).
Since \(x = y = r\sqrt{2}\), the rectangle of maximum area inscribed in a circle of radius \(r\) is a square of side \(r\sqrt{2}\). This is a classical result in geometry.
In simple words: Part (i) shows that if you have a fixed length of string to make a rectangle, the biggest area you can make is by forming a square. For 40 cm of string, a 10x10 cm square gives the biggest area (100 sq cm). Part (ii) explains that if you draw the biggest rectangle you can inside a circle, that rectangle will always turn out to be a square. Its side length will be the circle's radius multiplied by the square root of 2.

🎯 Exam Tip: For problems involving maximizing the area of a rectangle for a fixed perimeter or inside a circle, the solution is almost always a square. Minimizing distance or maximizing area in these geometric settings often relies on the symmetry of the figure.

Question 11.
(i) Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.
(ii) Show that of all right triangles inscribed in a circle, the triangle with maximum perimeter is isosceles.
(iii) AB is a diameter of a circle and C is any point on the circle. Show that the area of △ABC is maximum, when it is isosceles.
Answer:
(i) Let the two perpendicular sides of the right-angled triangle be \(x\) and \(y\). The hypotenuse is given as 5 cm.
By the Pythagorean theorem, \(x^2 + y^2 = 5^2 = 25\). So, \(y = \sqrt{25 - x^2}\).
The area of a right-angled triangle is \(A = \frac{1}{2} \times \text{base} \times \text{height}\).
\( A = \frac{1}{2} xy \).
Substitute \(y = \sqrt{25 - x^2}\):
\( A = \frac{1}{2} x \sqrt{25 - x^2} \).
To maximize \(A\), it's easier to maximize \(A^2\). Let \(Z = A^2\).
\( Z = \left(\frac{1}{2} x \sqrt{25 - x^2}\right)^2 = \frac{1}{4} x^2 (25 - x^2) = \frac{25}{4} x^2 - \frac{1}{4} x^4 \).
To find the maximum value of \(Z\), we differentiate with respect to \(x\) and set it to zero:
\( \frac{dZ}{dx} = \frac{25}{4} (2x) - \frac{1}{4} (4x^3) = \frac{25}{2} x - x^3 \).
For maximum/minimum, \( \frac{dZ}{dx} = 0 \):
\( \frac{25}{2} x - x^3 = 0 \)
\( x \left(\frac{25}{2} - x^2\right) = 0 \)
This gives \(x = 0\) or \(x^2 = \frac{25}{2}\). Since \(x\) must be positive for a triangle, \(x \ne 0\).
\( x^2 = \frac{25}{2} \implies x = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}} \).
Now, we check the second derivative of \(Z\) with respect to \(x\):
\( \frac{d^2 Z}{dx^2} = \frac{25}{2} - 3x^2 \).
At \(x = \frac{5}{\sqrt{2}}\):
\( \frac{d^2 Z}{dx^2} = \frac{25}{2} - 3\left(\frac{5}{\sqrt{2}}\right)^2 = \frac{25}{2} - 3\left(\frac{25}{2}\right) = \frac{25}{2} - \frac{75}{2} = -\frac{50}{2} = -25 \).
Since \( \frac{d^2 Z}{dx^2} = -25 < 0 \), \(x = \frac{5}{\sqrt{2}}\) maximizes \(Z\) (and thus \(A\)).
If \(x = \frac{5}{\sqrt{2}}\), then \(y = \sqrt{25 - x^2} = \sqrt{25 - \frac{25}{2}} = \sqrt{\frac{50 - 25}{2}} = \sqrt{\frac{25}{2}} = \frac{5}{\sqrt{2}}\).
Since \(x = y = \frac{5}{\sqrt{2}}\), the right-angled triangle with maximum area (for a fixed hypotenuse) is an isosceles right-angled triangle.
The largest possible area is \(A = \frac{1}{2} xy = \frac{1}{2} \times \frac{5}{\sqrt{2}} \times \frac{5}{\sqrt{2}} = \frac{1}{2} \times \frac{25}{2} = \frac{25}{4}\) sq cm.
In simple words: For a right-angled triangle with a hypotenuse of 5 cm, the largest area it can have is when the two shorter sides are equal. Each shorter side would be \( \frac{5}{\sqrt{2}} \) cm, and the maximum area would be \( \frac{25}{4} \) square centimeters.

🎯 Exam Tip: When maximizing the area of a right-angled triangle with a fixed hypotenuse, the triangle will be isosceles. Often, it's easier to maximize the square of the area to avoid differentiating square roots. Remember to state the final area, not just the side lengths.

Question 12.
(i) A large window has the shape of a rectangle surmounted by an equilateral triangle. If the perimeter of the window is 12 metres find the dimensions of the rectangle that will produce the largest area of the window.
(ii) A figure consists of semicircle with a rectangle on its diameter. If its perimeter is p cm, find its dimensions so that its area is maximum.

Answer:
(i) Let the width of the rectangular part of the window be \(x\) metres and its height be \(y\) metres. The equilateral triangle sits on top of the rectangle, so its side length is also \(x\). The perimeter of the window is given as 12 metres. This perimeter includes the three sides of the rectangle (\(x + 2y\)) and the two free sides of the equilateral triangle (\(x + x\), since one side is shared with the rectangle). So, the total perimeter \(P = x + 2y + 2x = 3x + 2y\). Given \(P = 12\), so \(3x + 2y = 12\). We can write \(y\) in terms of \(x\): \(2y = 12 - 3x \implies y = \frac{12 - 3x}{2}\). The area of the window, \(A\), is the sum of the area of the rectangle and the area of the equilateral triangle. Area of rectangle \( = x \times y = x \left( \frac{12 - 3x}{2} \right) \). Area of equilateral triangle \( = \frac{\sqrt{3}}{4} x^2 \). So, \( A = \frac{x(12 - 3x)}{2} + \frac{\sqrt{3}}{4} x^2 \). To find the dimensions that maximize the area, we need to differentiate \(A\) with respect to \(x\) and set it to zero. First, expand the area function: \( A = 6x - \frac{3x^2}{2} + \frac{\sqrt{3}}{4} x^2 \). Now, differentiate \(A\) with respect to \(x\): \( \frac{dA}{dx} = 6 - 3x + \frac{2\sqrt{3}}{4} x = 6 - 3x + \frac{\sqrt{3}}{2} x \). Set \( \frac{dA}{dx} = 0 \) for maximum or minimum area: \( 6 - 3x + \frac{\sqrt{3}}{2} x = 0 \) \( 6 = x \left( 3 - \frac{\sqrt{3}}{2} \right) \) \( 6 = x \left( \frac{6 - \sqrt{3}}{2} \right) \) \( x = \frac{12}{6 - \sqrt{3}} \). This ensures that the area can be maximized. Now we find the second derivative to check if it's a maximum: \( \frac{d^2A}{dx^2} = -3 + \frac{\sqrt{3}}{2} \). Since \( \sqrt{3} \approx 1.732 \), \( \frac{\sqrt{3}}{2} \approx 0.866 \). So, \( \frac{d^2A}{dx^2} = -3 + 0.866 = -2.134 < 0 \). This confirms that \(x = \frac{12}{6 - \sqrt{3}}\) corresponds to a maximum area. Now, find \(y\): \( y = \frac{12 - 3x}{2} = \frac{12 - 3 \left( \frac{12}{6 - \sqrt{3}} \right)}{2} = \frac{12(6 - \sqrt{3}) - 36}{2(6 - \sqrt{3})} = \frac{72 - 12\sqrt{3} - 36}{2(6 - \sqrt{3})} = \frac{36 - 12\sqrt{3}}{2(6 - \sqrt{3})} = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}} \). To rationalize \(x\) and \(y\) for clearer values: \( x = \frac{12}{6 - \sqrt{3}} \times \frac{6 + \sqrt{3}}{6 + \sqrt{3}} = \frac{12(6 + \sqrt{3})}{36 - 3} = \frac{12(6 + \sqrt{3})}{33} = \frac{4(6 + \sqrt{3})}{11} \) metres. \( y = \frac{18 - 6\sqrt{3}}{6 - \sqrt{3}} = \frac{6(3 - \sqrt{3})}{6 - \sqrt{3}} \). \( y = \frac{6(3 - \sqrt{3})(6 + \sqrt{3})}{(6 - \sqrt{3})(6 + \sqrt{3})} = \frac{6(18 + 3\sqrt{3} - 6\sqrt{3} - 3)}{33} = \frac{6(15 - 3\sqrt{3})}{33} = \frac{2(15 - 3\sqrt{3})}{11} = \frac{6(5 - \sqrt{3})}{11} \) metres. The dimensions of the rectangle for maximum area are width \( x = \frac{4(6 + \sqrt{3})}{11} \) m and height \( y = \frac{6(5 - \sqrt{3})}{11} \) m.

(ii) Let the radius of the semicircle be \(r\) cm and the side of the rectangle (along the diameter) be \(x\) cm. From the figure, the diameter of the semicircle is \(2r\), which is also the length of the rectangle. So, \(x = 2r\). Let \(y\) be the height of the rectangle. The perimeter of the combined figure \(p\) is given by the circumference of the semicircle plus the three free sides of the rectangle. Circumference of semicircle \( = \pi r \). Sides of rectangle \( = y + x + y = 2y + x \). So, \( p = \pi r + 2y + x \). Substitute \(x = 2r\): \( p = \pi r + 2y + 2r \implies p = r(\pi + 2) + 2y \). From this, we can express \(2y\): \( 2y = p - r(\pi + 2) \implies y = \frac{p - r(\pi + 2)}{2} \). The area of the combined figure, \(A\), is the sum of the area of the rectangle and the area of the semicircle. Area of rectangle \( = x \times y = 2r \times y \). Area of semicircle \( = \frac{1}{2} \pi r^2 \). So, \( A = 2r \left( \frac{p - r(\pi + 2)}{2} \right) + \frac{1}{2} \pi r^2 \). \( A = r(p - r(\pi + 2)) + \frac{1}{2} \pi r^2 \). \( A = pr - r^2(\pi + 2) + \frac{1}{2} \pi r^2 \). To maximize the area, differentiate \(A\) with respect to \(r\) and set it to zero. \( \frac{dA}{dr} = p - 2r(\pi + 2) + \pi r \). Set \( \frac{dA}{dr} = 0 \): \( p - 2r\pi - 4r + \pi r = 0 \) \( p - r\pi - 4r = 0 \) \( p = r(\pi + 4) \). So, \( r = \frac{p}{\pi + 4} \). This is the radius for maximum area. Now, find the second derivative to confirm it's a maximum: \( \frac{d^2A}{dr^2} = -\pi - 4 = -(\pi + 4) \). Since \( \pi > 0 \), \( \pi + 4 > 0 \), so \( -(\pi + 4) < 0 \). This confirms that \(r = \frac{p}{\pi + 4}\) corresponds to a maximum area. Now find \(x\) and \(y\). \( x = 2r = \frac{2p}{\pi + 4} \) cm. \( y = \frac{p - r(\pi + 2)}{2} = \frac{p - \frac{p}{\pi + 4}(\pi + 2)}{2} = \frac{p(\pi + 4) - p(\pi + 2)}{2(\pi + 4)} = \frac{p\pi + 4p - p\pi - 2p}{2(\pi + 4)} = \frac{2p}{2(\pi + 4)} = \frac{p}{\pi + 4} \) cm. So, the dimensions are \( r = \frac{p}{\pi + 4} \) cm (radius), \( x = \frac{2p}{\pi + 4} \) cm (length of rectangle/diameter), and \( y = \frac{p}{\pi + 4} \) cm (height of rectangle).

In simple words: For the window, we found the best width and height for the rectangular part to make the biggest area, using the given total perimeter. For the figure with a semicircle, we found the radius and rectangle sides that give the largest area for a fixed total perimeter. In both cases, we used calculus to find the maximum possible area by finding where the slope of the area function is zero.

🎯 Exam Tip: When dealing with composite shapes for maximization, always define the perimeter/area in terms of a single variable, differentiate, and use the second derivative test to confirm it's a maximum or minimum. Carefully identify all parts of the perimeter/area, including shared sides.

Question 13. The perimeter of a sector of a circle is constant. What will be the angle of the sector, if the area of the sector were to be maximum?

Answer: Let \(r\) be the radius of the sector and \(l\) be the arc length. Let \(k\) be the constant perimeter. The perimeter of a sector is given by \( P = 2r + l \). Since the perimeter is constant, let \( P = k \). So, \( k = 2r + l \). We know that for a sector, the arc length \(l\) is related to the angle \( \theta \) (in radians) and radius \(r\) by \( l = r\theta \). Substitute \(l\) into the perimeter equation: \( k = 2r + r\theta = r(2 + \theta) \). From this, we can express \(r\) in terms of \(k\) and \( \theta \): \( r = \frac{k}{2 + \theta} \). The area of a sector, \(A\), is given by \( A = \frac{1}{2} r^2 \theta \). Substitute the expression for \(r\) into the area formula: \( A = \frac{1}{2} \left( \frac{k}{2 + \theta} \right)^2 \theta = \frac{1}{2} \frac{k^2 \theta}{(2 + \theta)^2} \). To find the angle \( \theta \) that maximizes the area, we differentiate \(A\) with respect to \( \theta \) and set it to zero. We can pull out the constant \( \frac{k^2}{2} \): \( A = \frac{k^2}{2} \left( \frac{\theta}{(2 + \theta)^2} \right) \). Using the quotient rule \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = \theta \) and \( v = (2 + \theta)^2 \). \( u' = 1 \) and \( v' = 2(2 + \theta)(1) = 2(2 + \theta) \). So, \( \frac{dA}{d\theta} = \frac{k^2}{2} \left( \frac{1 \cdot (2 + \theta)^2 - \theta \cdot 2(2 + \theta)}{((2 + \theta)^2)^2} \right) \). \( \frac{dA}{d\theta} = \frac{k^2}{2} \left( \frac{(2 + \theta)^2 - 2\theta(2 + \theta)}{(2 + \theta)^4} \right) \). Factor out \( (2 + \theta) \) from the numerator: \( \frac{dA}{d\theta} = \frac{k^2}{2} \left( \frac{(2 + \theta)( (2 + \theta) - 2\theta )}{(2 + \theta)^4} \right) = \frac{k^2}{2} \left( \frac{2 + \theta - 2\theta}{(2 + \theta)^3} \right) = \frac{k^2}{2} \left( \frac{2 - \theta}{(2 + \theta)^3} \right) \). Set \( \frac{dA}{d\theta} = 0 \): \( \frac{k^2}{2} \left( \frac{2 - \theta}{(2 + \theta)^3} \right) = 0 \). Since \( k \neq 0 \) (otherwise there's no sector), we must have \( 2 - \theta = 0 \). So, \( \theta = 2 \) radians. This is the angle for maximum area. To confirm it's a maximum, we can consider the sign of \( \frac{dA}{d\theta} \). If \( \theta < 2 \), then \( 2 - \theta > 0 \), so \( \frac{dA}{d\theta} > 0 \) (area is increasing). If \( \theta > 2 \), then \( 2 - \theta < 0 \), so \( \frac{dA}{d\theta} < 0 \) (area is decreasing). This change in sign from positive to negative confirms that \( \theta = 2 \) radians gives a maximum area. The second derivative test would also yield a negative value, confirming a maximum. A sector with an angle of 2 radians maximizes its area for a given perimeter.

In simple words: When the total length of the two straight edges and the curved edge of a pie slice is fixed, the largest possible area for that slice happens when the angle at the center is 2 radians. A radian is a way to measure angles based on the radius of a circle.

🎯 Exam Tip: Always convert the problem into an equation with a single variable (here, \( \theta \)) and then use differentiation to find the extremum. Remember that angle for area of sector formulas usually assumes radians.

Question 14. Show that the surface area of a closed cuboid with square base and given volume is minimum when it is a cube.

Answer: Let the side of the square base be \(x\) and the height of the cuboid be \(y\). Since the base is square, its dimensions are \(x \times x\). The height is \(y\). The volume of the cuboid, \(V\), is given as constant. So, \( V = x^2 y \). From this, we can express \(y\) in terms of \(V\) and \(x\): \( y = \frac{V}{x^2} \). The surface area of a closed cuboid, \(S\), is given by the area of the base, the top, and the four side walls. Area of base \( = x^2 \). Area of top \( = x^2 \). Area of four walls \( = 4xy \). So, \( S = x^2 + x^2 + 4xy = 2x^2 + 4xy \). Substitute the expression for \(y\) into the surface area formula: \( S = 2x^2 + 4x \left( \frac{V}{x^2} \right) = 2x^2 + \frac{4V}{x} \). To find the minimum surface area, differentiate \(S\) with respect to \(x\) and set it to zero. \( \frac{dS}{dx} = 4x - \frac{4V}{x^2} \). Set \( \frac{dS}{dx} = 0 \): \( 4x - \frac{4V}{x^2} = 0 \) \( 4x = \frac{4V}{x^2} \) \( x^3 = V \). So, \( x = V^{1/3} \). Now, find the second derivative to confirm it's a minimum: \( \frac{d^2S}{dx^2} = 4 - 4V(-2x^{-3}) = 4 + \frac{8V}{x^3} \). Substitute \( x = V^{1/3} \) into the second derivative: \( \frac{d^2S}{dx^2} = 4 + \frac{8V}{(V^{1/3})^3} = 4 + \frac{8V}{V} = 4 + 8 = 12 \). Since \( \frac{d^2S}{dx^2} = 12 > 0 \), this confirms that \( x = V^{1/3} \) corresponds to a minimum surface area. Now, we need to show that the cuboid is a cube when its surface area is minimum. If \( x = V^{1/3} \), then find \(y\): \( y = \frac{V}{x^2} = \frac{V}{(V^{1/3})^2} = \frac{V}{V^{2/3}} = V^{1/3} \). Since \( x = y = V^{1/3} \), all dimensions are equal. This means the cuboid is a cube. Therefore, the surface area of a closed cuboid with a square base and a given volume is minimum when it is a cube. A cube is a perfect shape because its sides are all equal.
In simple words: Imagine you have a fixed amount of space (volume) to fill with a box that has a square bottom. To use the least amount of material for the box (minimum surface area), the best shape for that box is a cube, where all its sides are of equal length.

🎯 Exam Tip: For optimization problems involving geometric shapes, always write the quantity to be optimized (e.g., surface area) in terms of a single variable using the constraint (e.g., given volume). Then, use differentiation to find the critical points and the second derivative test to confirm if it's a maximum or minimum.

Question 15. An open box with a square base is to be made out of a given quantity of cardboard of area \(c^2\) square units. Show that the maximum volume of the box is \( \frac{c^2}{6\sqrt{3}} \) cubic units.

Answer: Let the side of the square base of the open box be \(x\) and its height be \(h\). Since the box is open, it has a base and four side walls, but no top. The given quantity of cardboard is the surface area, \(A\), of the open box. Let \(A = c^2\). Area of the square base \( = x^2 \). Area of the four side walls \( = 4xh \). So, the total surface area \( A = x^2 + 4xh \). Given \( A = c^2 \), so \( c^2 = x^2 + 4xh \). From this, we can express \(h\) in terms of \(c\) and \(x\): \( 4xh = c^2 - x^2 \implies h = \frac{c^2 - x^2}{4x} \). The volume of the box, \(V\), is given by \( V = x^2 h \). Substitute the expression for \(h\) into the volume formula: \( V = x^2 \left( \frac{c^2 - x^2}{4x} \right) = \frac{x(c^2 - x^2)}{4} = \frac{c^2x - x^3}{4} \). To find the maximum volume, differentiate \(V\) with respect to \(x\) and set it to zero. \( \frac{dV}{dx} = \frac{1}{4} (c^2 - 3x^2) \). Set \( \frac{dV}{dx} = 0 \): \( \frac{1}{4} (c^2 - 3x^2) = 0 \) \( c^2 - 3x^2 = 0 \) \( 3x^2 = c^2 \implies x^2 = \frac{c^2}{3} \). So, \( x = \sqrt{\frac{c^2}{3}} = \frac{c}{\sqrt{3}} \). Now, find the second derivative to confirm it's a maximum: \( \frac{d^2V}{dx^2} = \frac{1}{4} (0 - 6x) = -\frac{3}{2} x \). Substitute \( x = \frac{c}{\sqrt{3}} \) into the second derivative: \( \frac{d^2V}{dx^2} = -\frac{3}{2} \left( \frac{c}{\sqrt{3}} \right) = -\frac{3c}{2\sqrt{3}} \). Since \( c > 0 \), \( \frac{d^2V}{dx^2} < 0 \). This confirms that \( x = \frac{c}{\sqrt{3}} \) corresponds to a maximum volume. Now, substitute \(x\) back into the volume formula to find the maximum volume. First, find \(h\) when \(x = \frac{c}{\sqrt{3}}\): \( h = \frac{c^2 - x^2}{4x} = \frac{c^2 - \frac{c^2}{3}}{4 \left( \frac{c}{\sqrt{3}} \right)} = \frac{\frac{3c^2 - c^2}{3}}{\frac{4c}{\sqrt{3}}} = \frac{\frac{2c^2}{3}}{\frac{4c}{\sqrt{3}}} = \frac{2c^2}{3} \times \frac{\sqrt{3}}{4c} = \frac{2c\sqrt{3}}{12} = \frac{c\sqrt{3}}{6} \). Now calculate the maximum volume \(V\): \( V = x^2 h = \left( \frac{c}{\sqrt{3}} \right)^2 \left( \frac{c\sqrt{3}}{6} \right) = \frac{c^2}{3} \times \frac{c\sqrt{3}}{6} = \frac{c^3\sqrt{3}}{18} \). To get the desired form \( \frac{c^2}{6\sqrt{3}} \), we can multiply the numerator and denominator by \( \sqrt{3} \) to remove \( \sqrt{3} \) from the denominator of \(h\), or re-evaluate the final expression. Let's simplify \( \frac{c^3\sqrt{3}}{18} \) again. \( \frac{c^3\sqrt{3}}{18} = \frac{c^3\sqrt{3}}{6 \times 3} = \frac{c^3\sqrt{3}}{6 \times (\sqrt{3})^2} = \frac{c^3}{6\sqrt{3}} \). Thus, the maximum volume of the box is \( \frac{c^3}{6\sqrt{3}} \) cubic units. For a fixed amount of material, shaping a box this way gives the most space inside.
In simple words: If you have a set amount of cardboard to make an open box with a square base, the largest amount of space (volume) you can get inside that box is achieved when the side length of the base is \( \frac{c}{\sqrt{3}} \) and the height is \( \frac{c\sqrt{3}}{6} \). This specific design makes the box hold the most.

🎯 Exam Tip: Remember to use the given constraint (surface area in this case) to express one variable in terms of the other, simplifying the function to be optimized (volume) into a single variable. Always verify your result with the second derivative test to ensure it's a maximum, not a minimum.

Question 16. A cylinder is such that sum of its height and the circumference of its base is 10 metres. Find the greatest volume of the cylinder.

Answer: Let \(h\) be the height of the cylinder and \(r\) be the radius of its base. The circumference of the base is \( 2\pi r \). The problem states that the sum of the height and the circumference of the base is 10 metres: \( h + 2\pi r = 10 \). From this, we can express \(h\) in terms of \(r\): \( h = 10 - 2\pi r \). The volume of the cylinder, \(V\), is given by \( V = \pi r^2 h \). Substitute the expression for \(h\) into the volume formula: \( V = \pi r^2 (10 - 2\pi r) = 10\pi r^2 - 2\pi^2 r^3 \). To find the greatest volume, differentiate \(V\) with respect to \(r\) and set it to zero. \( \frac{dV}{dr} = 20\pi r - 6\pi^2 r^2 \). Set \( \frac{dV}{dr} = 0 \): \( 20\pi r - 6\pi^2 r^2 = 0 \). Factor out \( 2\pi r \): \( 2\pi r (10 - 3\pi r) = 0 \). This gives two possible values for \(r\): \( 2\pi r = 0 \implies r = 0 \) (which means no cylinder and no volume) or \( 10 - 3\pi r = 0 \). So, \( 3\pi r = 10 \implies r = \frac{10}{3\pi} \). This is the radius for maximum volume. Now, find the second derivative to confirm it's a maximum: \( \frac{d^2V}{dr^2} = 20\pi - 12\pi^2 r \). Substitute \( r = \frac{10}{3\pi} \) into the second derivative: \( \frac{d^2V}{dr^2} = 20\pi - 12\pi^2 \left( \frac{10}{3\pi} \right) = 20\pi - \frac{120\pi^2}{3\pi} = 20\pi - 40\pi = -20\pi \). Since \( \frac{d^2V}{dr^2} = -20\pi < 0 \), this confirms that \( r = \frac{10}{3\pi} \) corresponds to a maximum volume. Now, find the height \(h\): \( h = 10 - 2\pi r = 10 - 2\pi \left( \frac{10}{3\pi} \right) = 10 - \frac{20\pi}{3\pi} = 10 - \frac{20}{3} = \frac{30 - 20}{3} = \frac{10}{3} \) metres. Finally, calculate the greatest volume \(V\): \( V = \pi r^2 h = \pi \left( \frac{10}{3\pi} \right)^2 \left( \frac{10}{3} \right) = \pi \left( \frac{100}{9\pi^2} \right) \left( \frac{10}{3} \right) = \frac{1000\pi}{27\pi^2} = \frac{1000}{27\pi} \) cubic metres. When a cylinder has a fixed sum of its height and base circumference, its volume is largest when its radius is \( \frac{10}{3\pi} \) metres and its height is \( \frac{10}{3} \) metres.
In simple words: If you have a cylinder and you add its height to the measurement around its bottom edge, and that total number is 10, then the biggest amount of space inside the cylinder will be \( \frac{1000}{27\pi} \) cubic metres. This shows how to make the cylinder hold the most when you have a limit on its height and circumference combined.

🎯 Exam Tip: In optimization problems, always ensure the physical constraints (like radius and height being positive) are met. Always use the second derivative test to confirm the nature of the extremum (maximum or minimum) to avoid errors.

Question 17. A right circular cylinder is to be made so that the sum of its radius and its height is 6 metres. Find the maximum volume of the cylinder.

Answer: Let \(r\) be the radius of the cylinder and \(h\) be its height. The problem states that the sum of its radius and height is 6 metres: \( r + h = 6 \). From this, we can express \(h\) in terms of \(r\): \( h = 6 - r \). The volume of the cylinder, \(V\), is given by \( V = \pi r^2 h \). Substitute the expression for \(h\) into the volume formula: \( V = \pi r^2 (6 - r) = 6\pi r^2 - \pi r^3 \). To find the maximum volume, differentiate \(V\) with respect to \(r\) and set it to zero. \( \frac{dV}{dr} = 12\pi r - 3\pi r^2 \). Set \( \frac{dV}{dr} = 0 \): \( 12\pi r - 3\pi r^2 = 0 \). Factor out \( 3\pi r \): \( 3\pi r (4 - r) = 0 \). This gives two possible values for \(r\): \( 3\pi r = 0 \implies r = 0 \) (which means no cylinder and no volume) or \( 4 - r = 0 \). So, \( r = 4 \) metres. This is the radius for maximum volume. Now, find the second derivative to confirm it's a maximum: \( \frac{d^2V}{dr^2} = 12\pi - 6\pi r \). Substitute \( r = 4 \) into the second derivative: \( \frac{d^2V}{dr^2} = 12\pi - 6\pi (4) = 12\pi - 24\pi = -12\pi \). Since \( \frac{d^2V}{dr^2} = -12\pi < 0 \), this confirms that \( r = 4 \) metres corresponds to a maximum volume. Now, find the height \(h\): \( h = 6 - r = 6 - 4 = 2 \) metres. Finally, calculate the maximum volume \(V\): \( V = \pi r^2 h = \pi (4)^2 (2) = \pi (16)(2) = 32\pi \) cubic metres. When the sum of the radius and height of a cylinder is fixed at 6 metres, its volume is largest when its radius is 4 metres and its height is 2 metres. This shape ensures the greatest possible volume.
In simple words: If you're building a cylinder and the total length of its radius and its height must add up to 6 meters, then the biggest possible space it can hold is \(32\pi\) cubic meters. This happens when the radius is 4 meters and the height is 2 meters.

🎯 Exam Tip: Always clearly state the variables you are using and the relationship between them given by the problem (constraint). The second derivative test is crucial for distinguishing between maximum and minimum points.

Question 18. Show that the height of an open cylinder of given surface and the greatest volume is equal to the radius of the base.

Answer: Let \(r\) be the radius of the base and \(h\) be the height of the open cylinder. An open cylinder means it has a base and a curved surface, but no top. The surface area of an open cylinder, \(S\), is given as constant. Area of base \( = \pi r^2 \). Area of curved surface \( = 2\pi rh \). So, the total surface area \( S = \pi r^2 + 2\pi rh \). From this, we can express \(h\) in terms of \(S\) and \(r\): \( 2\pi rh = S - \pi r^2 \) \( h = \frac{S - \pi r^2}{2\pi r} = \frac{S}{2\pi r} - \frac{\pi r^2}{2\pi r} = \frac{S}{2\pi r} - \frac{r}{2} \). The volume of the cylinder, \(V\), is given by \( V = \pi r^2 h \). Substitute the expression for \(h\) into the volume formula: \( V = \pi r^2 \left( \frac{S}{2\pi r} - \frac{r}{2} \right) = \frac{\pi r^2 S}{2\pi r} - \frac{\pi r^2 \cdot r}{2} = \frac{Sr}{2} - \frac{\pi r^3}{2} \). To find the greatest volume, differentiate \(V\) with respect to \(r\) and set it to zero. \( \frac{dV}{dr} = \frac{S}{2} - \frac{3\pi r^2}{2} \). Set \( \frac{dV}{dr} = 0 \): \( \frac{S}{2} - \frac{3\pi r^2}{2} = 0 \) \( S - 3\pi r^2 = 0 \). So, \( S = 3\pi r^2 \). This is the condition for maximum volume. Now, find the second derivative to confirm it's a maximum: \( \frac{d^2V}{dr^2} = 0 - \frac{3\pi (2r)}{2} = -3\pi r \). Since \(r\) must be positive (it's a radius), \( -3\pi r < 0 \). This confirms that \( S = 3\pi r^2 \) corresponds to a maximum volume. Now, we need to show that when the volume is maximum, the height \(h\) is equal to the radius \(r\). Substitute \( S = 3\pi r^2 \) back into the expression for \(h\): \( h = \frac{S}{2\pi r} - \frac{r}{2} = \frac{3\pi r^2}{2\pi r} - \frac{r}{2} = \frac{3r}{2} - \frac{r}{2} = \frac{2r}{2} = r \). Thus, when the volume of an open cylinder with a given surface area is maximum, its height is equal to its radius. This means for a fixed amount of material, the most spacious open cylinder will have its height and radius of the same length.
In simple words: If you want to make an open cylinder (like a can without a lid) that holds the most liquid, using a fixed amount of material, you should make sure its height is exactly the same as its radius. This makes the cylinder as "fat" as it is "tall" (if considering the radius for width), giving it the biggest volume.

🎯 Exam Tip: When proving a relationship (like \(h=r\)), it's important to differentiate the quantity to be maximized (volume) and then substitute the condition for the maximum back into the expression for the other variable (height).

Question 19. A wire of length \(a\) is cut into two parts which are bent respectively in the form of a square and a circle. Show that the least value of the sum of the areas so formed is \( \frac{a^2}{4(\pi+4)} \).

Answer: Let the total length of the wire be \(a\). The wire is cut into two parts. Let the length of one part be \(x\), and the length of the other part will be \(a - x\). The first part, of length \(x\), is bent into a square. The perimeter of the square is \(x\). If the side length of the square is \(s\), then \( 4s = x \implies s = \frac{x}{4} \). The area of the square, \(A_1\), is \( A_1 = s^2 = \left( \frac{x}{4} \right)^2 = \frac{x^2}{16} \). The second part, of length \(a - x\), is bent into a circle. The circumference of the circle is \(a - x\). If the radius of the circle is \(r\), then \( 2\pi r = a - x \implies r = \frac{a - x}{2\pi} \). The area of the circle, \(A_2\), is \( A_2 = \pi r^2 = \pi \left( \frac{a - x}{2\pi} \right)^2 = \pi \frac{(a - x)^2}{4\pi^2} = \frac{(a - x)^2}{4\pi} \). The sum of the areas, \(A\), is \( A = A_1 + A_2 = \frac{x^2}{16} + \frac{(a - x)^2}{4\pi} \). To find the least value of the sum of the areas, differentiate \(A\) with respect to \(x\) and set it to zero. \( \frac{dA}{dx} = \frac{2x}{16} + \frac{2(a - x)(-1)}{4\pi} = \frac{x}{8} - \frac{a - x}{2\pi} \). Set \( \frac{dA}{dx} = 0 \): \( \frac{x}{8} - \frac{a - x}{2\pi} = 0 \) \( \frac{x}{8} = \frac{a - x}{2\pi} \) \( 2\pi x = 8(a - x) \) \( \pi x = 4(a - x) \) \( \pi x = 4a - 4x \) \( \pi x + 4x = 4a \) \( x(\pi + 4) = 4a \). So, \( x = \frac{4a}{\pi + 4} \). This is the length for minimum area. Now, find the second derivative to confirm it's a minimum: \( \frac{d^2A}{dx^2} = \frac{1}{8} - \frac{(-1)}{2\pi} = \frac{1}{8} + \frac{1}{2\pi} \). Since \( \pi > 0 \), both terms are positive, so \( \frac{d^2A}{dx^2} > 0 \). This confirms that \( x = \frac{4a}{\pi + 4} \) corresponds to a minimum sum of areas. Now, substitute \(x\) back into the sum of the areas formula to find the least value. First, find \(a - x\): \( a - x = a - \frac{4a}{\pi + 4} = \frac{a(\pi + 4) - 4a}{\pi + 4} = \frac{a\pi + 4a - 4a}{\pi + 4} = \frac{a\pi}{\pi + 4} \). Now calculate the minimum sum of areas \(A\): \( A = \frac{x^2}{16} + \frac{(a - x)^2}{4\pi} \) \( A = \frac{\left( \frac{4a}{\pi + 4} \right)^2}{16} + \frac{\left( \frac{a\pi}{\pi + 4} \right)^2}{4\pi} \) \( A = \frac{\frac{16a^2}{(\pi + 4)^2}}{16} + \frac{\frac{a^2\pi^2}{(\pi + 4)^2}}{4\pi} \) \( A = \frac{a^2}{(\pi + 4)^2} + \frac{a^2\pi}{4(\pi + 4)^2} \) To combine these, find a common denominator, which is \( 4(\pi + 4)^2 \): \( A = \frac{4a^2}{4(\pi + 4)^2} + \frac{a^2\pi}{4(\pi + 4)^2} = \frac{4a^2 + a^2\pi}{4(\pi + 4)^2} \). Factor out \(a^2\): \( A = \frac{a^2(4 + \pi)}{4(\pi + 4)^2} \). Since \( (4 + \pi) \) is the same as \( (\pi + 4) \), we can cancel one term: \( A = \frac{a^2}{4(\pi + 4)} \). Thus, the least value of the sum of the areas is \( \frac{a^2}{4(\pi + 4)} \). This means to use the least amount of material for both shapes, a specific division of the wire length is optimal.
In simple words: If you have a piece of wire and you cut it into two parts to make a square and a circle, there's a special way to cut it so that the total area of the square and circle combined is as small as possible. The smallest total area you can get is given by the formula \( \frac{a^2}{4(\pi+4)} \), where \(a\) is the original length of the wire.

🎯 Exam Tip: When dividing a fixed resource (like wire length) into two parts to optimize a combined quantity, always define the two parts in terms of a single variable (e.g., \(x\) and \(a-x\)). The calculations can be lengthy, so simplify step-by-step and double-check algebraic manipulations.

Question 20. Given the total surface of a cone, show that when the volume of the cone is maximum, the semi-vertical angle will be \( \sin^{-1}\left(\frac{1}{3}\right) \).
Answer: Let the cone have radius \(r\), height \(h\), and slant height \(l\). Let the semi-vertical angle be \( \alpha \). The total surface area of the cone is given by \( S = \pi r^2 + \pi r l \). From this, we can express the slant height as \( l = \frac{S - \pi r^2}{\pi r} \). We know that \( l^2 = r^2 + h^2 \), so \( h = \sqrt{l^2 - r^2} \). The volume of the cone is \( V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 \sqrt{l^2 - r^2} \). To make calculations easier, we can maximize \( V^2 \) instead of \( V \).
Substitute the expression for \(l\) into \(V^2\):
\( V^2 = \frac{1}{9}\pi^2 r^4 \left( l^2 - r^2 \right) \)
\( \implies V^2 = \frac{1}{9}\pi^2 r^4 \left( \left(\frac{S - \pi r^2}{\pi r}\right)^2 - r^2 \right) \)
\( \implies V^2 = \frac{1}{9}\pi^2 r^4 \left( \frac{(S - \pi r^2)^2}{\pi^2 r^2} - r^2 \right) \)
\( \implies V^2 = \frac{1}{9} r^2 \left( (S - \pi r^2)^2 - \pi^2 r^4 \right) \)
\( \implies V^2 = \frac{1}{9} r^2 \left( S^2 - 2S\pi r^2 + \pi^2 r^4 - \pi^2 r^4 \right) \)
\( \implies V^2 = \frac{1}{9} r^2 (S^2 - 2S\pi r^2) \)
\( \implies V^2 = \frac{S}{9} (Sr^2 - 2\pi r^4) \)
Let \( V' = V^2 \). Differentiate \( V' \) with respect to \( r \):
\( \frac{dV'}{dr} = \frac{S}{9} (2Sr - 8\pi r^3) \)
Set \( \frac{dV'}{dr} = 0 \) for maximum volume:
\( \implies \frac{S}{9} (2Sr - 8\pi r^3) = 0 \)
\( \implies 2Sr - 8\pi r^3 = 0 \)
\( \implies 2r (S - 4\pi r^2) = 0 \)
Since \( r \neq 0 \) (for a cone to exist), we have:
\( S - 4\pi r^2 = 0 \)
\( \implies S = 4\pi r^2 \)
To confirm this is a maximum, find the second derivative:
\( \frac{d^2V'}{dr^2} = \frac{S}{9} (2S - 24\pi r^2) \)
Substitute \( S = 4\pi r^2 \) into the second derivative:
\( \implies \frac{d^2V'}{dr^2} = \frac{S}{9} (2(4\pi r^2) - 24\pi r^2) \)
\( \implies \frac{d^2V'}{dr^2} = \frac{S}{9} (8\pi r^2 - 24\pi r^2) \)
\( \implies \frac{d^2V'}{dr^2} = \frac{S}{9} (-16\pi r^2) \)
Since \( S > 0 \) and \( r^2 > 0 \), \( \frac{d^2V'}{dr^2} < 0 \), confirming that \( V' \) (and thus \( V \)) is maximum when \( S = 4\pi r^2 \).
Now, we relate this condition to the semi-vertical angle. We know \( S = \pi r^2 + \pi r l \).
Substituting \( S = 4\pi r^2 \):
\( 4\pi r^2 = \pi r^2 + \pi r l \)
\( \implies 3\pi r^2 = \pi r l \)
\( \implies 3r = l \)
In a right circular cone, the semi-vertical angle \( \alpha \) is related by \( \sin \alpha = \frac{r}{l} \).
\( \sin \alpha = \frac{r}{3r} \)
\( \implies \sin \alpha = \frac{1}{3} \)
\( \implies \alpha = \sin^{-1}\left(\frac{1}{3}\right) \)
Thus, the semi-vertical angle is \( \sin^{-1}\left(\frac{1}{3}\right) \) when the volume of the cone is maximum for a given total surface area.
In simple words: For a cone with a fixed surface area, its volume is largest when its slant height is three times its base radius. This makes the semi-vertical angle, which is the angle from the cone's center line to its slant side, equal to inverse sine of one-third.

🎯 Exam Tip: Remember to set the first derivative to zero to find critical points and use the second derivative test to confirm maximum or minimum. Using \( V^2 \) instead of \( V \) can simplify differentiation if square roots are involved.

Question 21. An enemy vehicle is moving along the curve \( y = x^2 + 2 \). Find the shortest distance between the vehicle and our artillery located at (3, 2). Find the coordinates of the vehicle when the distance is shortest.
Answer: Let the position of the vehicle be a point \( P(x, y) \) on the curve \( y = x^2 + 2 \). The artillery is located at point \( A(3, 2) \). The distance \( z \) between the vehicle and the artillery is given by the distance formula:
\( z = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( \implies z = \sqrt{(x - 3)^2 + (y - 2)^2} \)
Since \( y = x^2 + 2 \), substitute this into the distance formula:
\( z = \sqrt{(x - 3)^2 + ((x^2 + 2) - 2)^2} \)
\( \implies z = \sqrt{(x - 3)^2 + (x^2)^2} \)
\( \implies z = \sqrt{(x - 3)^2 + x^4} \)
To find the shortest distance, it is easier to minimize the square of the distance, \( u = z^2 \):
\( u = (x - 3)^2 + x^4 \)
Now, differentiate \( u \) with respect to \( x \):
\( \frac{du}{dx} = 2(x - 3) \cdot 1 + 4x^3 \)
\( \implies \frac{du}{dx} = 4x^3 + 2x - 6 \)
Set \( \frac{du}{dx} = 0 \) to find critical points:
\( 4x^3 + 2x - 6 = 0 \)
Divide by 2:
\( 2x^3 + x - 3 = 0 \)
By trying integer values, we can see that \( x = 1 \) is a root:
\( 2(1)^3 + 1 - 3 = 2 + 1 - 3 = 0 \)
So, \( (x - 1) \) is a factor. Divide the polynomial by \( (x - 1) \):
\( (x - 1)(2x^2 + 2x + 3) = 0 \)
For the quadratic factor, calculate the discriminant \( \Delta = b^2 - 4ac \):
\( \Delta = (2)^2 - 4(2)(3) = 4 - 24 = -20 \)
Since \( \Delta < 0 \), the quadratic factor \( 2x^2 + 2x + 3 \) has no real roots. Therefore, \( x = 1 \) is the only real critical point.
Now, find the second derivative of \( u \) to confirm if it's a minimum:
\( \frac{d^2u}{dx^2} = 12x^2 + 2 \)
At \( x = 1 \):
\( \frac{d^2u}{dx^2} = 12(1)^2 + 2 = 12 + 2 = 14 \)
Since \( \frac{d^2u}{dx^2} > 0 \), \( u \) (and thus \( z \)) is minimized at \( x = 1 \).
Now find the corresponding \( y \)-coordinate using \( y = x^2 + 2 \):
\( y = (1)^2 + 2 = 1 + 2 = 3 \)
So, the coordinates of the vehicle when the distance is shortest are \( (1, 3) \).
The shortest distance is:
\( z = \sqrt{(1 - 3)^2 + (3 - 2)^2} = \sqrt{(-2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \) units.
In simple words: We want to find the closest spot on a curved path to a fixed point. We use the distance formula and then calculus to find the point where the distance is the smallest. The vehicle is closest at point (1, 3), and the shortest distance is the square root of 5 units.

🎯 Exam Tip: To find the minimum distance, it's often easier to minimize the square of the distance, as it avoids dealing with square roots during differentiation. Always verify your critical points using the second derivative test.

Question 22. A box is to be constructed from a square metal sheet of side 60 cm by cutting out identical squares from the four corners and turning up the sides. Find the length of the side of the square to be cut out so that the box has maximum volume.
Answer: Let \(x\) cm be the length of the side of the identical squares cut from each of the four corners of the square metal sheet. The original square sheet has a side length of 60 cm. When squares of side \(x\) are cut from the corners and the sides are folded up, an open box is formed. The dimensions of this box will be:
Length of the base: \( L = (60 - 2x) \) cm
Breadth of the base: \( B = (60 - 2x) \) cm
Height of the box: \( H = x \) cm
For the box to be physically possible, all dimensions must be positive. This means \( x > 0 \) and \( 60 - 2x > 0 \).
From \( 60 - 2x > 0 \), we get \( 2x < 60 \), which means \( x < 30 \).
So, the valid range for \( x \) is \( 0 < x < 30 \).
The volume of the box, \( V \), is given by the product of its length, breadth, and height:
\( V = L \times B \times H = (60 - 2x)(60 - 2x)x \)
\( \implies V = (60 - 2x)^2 x \)
Expand the expression for \( V \):
\( V = (3600 - 240x + 4x^2)x \)
\( \implies V = 4x^3 - 240x^2 + 3600x \)
To find the maximum volume, we differentiate \( V \) with respect to \( x \) and set the derivative to zero:
\( \frac{dV}{dx} = 12x^2 - 480x + 3600 \)
Set \( \frac{dV}{dx} = 0 \):
\( 12x^2 - 480x + 3600 = 0 \)
Divide the entire equation by 12:
\( x^2 - 40x + 300 = 0 \)
Factor the quadratic equation:
\( (x - 10)(x - 30) = 0 \)
This gives two possible values for \( x \): \( x = 10 \) or \( x = 30 \).
Considering the valid range for \( x \) (\( 0 < x < 30 \)), we must choose \( x = 10 \). If \( x = 30 \), the base dimensions would become zero, resulting in no box.
To confirm that \( x = 10 \) gives a maximum volume, we compute the second derivative of \( V \):
\( \frac{d^2V}{dx^2} = 24x - 480 \)
Substitute \( x = 10 \) into the second derivative:
\( \frac{d^2V}{dx^2} = 24(10) - 480 = 240 - 480 = -240 \)
Since \( \frac{d^2V}{dx^2} < 0 \) at \( x = 10 \), the volume is indeed maximum at this value.
Therefore, the length of the side of the square to be cut out for maximum volume is 10 cm. The maximum volume would be \( (60-2(10))^2(10) = (40)^2(10) = 1600 \times 10 = 16000 \) cm\(^3\).
In simple words: To make the biggest box from a 60 cm square metal sheet, you should cut out 10 cm by 10 cm squares from each of the four corners. This specific cut ensures the box holds the most amount of space inside.

🎯 Exam Tip: When maximizing volume by cutting squares from corners, always make sure your 'x' value (the side of the cut square) is valid and falls within the allowed range for the dimensions of the box to be positive. A negative second derivative confirms a maximum.

Question 23. A closed right circular cylinderfhas volume 2156 cubic units. What should be the radius of the base so that the total surface area may be minimum ? (Use \( \pi = \frac{22}{7} \)).
Answer: Let \(r\) be the radius of the base and \(h\) be the height of the closed right circular cylinder. The given volume of the cylinder is \( V = 2156 \) cubic units.
The formula for the volume of a cylinder is \( V = \pi r^2 h \).
So, \( \pi r^2 h = 2156 \).
From this, we can express the height \( h \) in terms of \( r \):
\( h = \frac{2156}{\pi r^2} \)
The total surface area \( S \) of a closed cylinder is given by:
\( S = 2\pi r^2 + 2\pi r h \)
Substitute the expression for \( h \) into the surface area formula:
\( S = 2\pi r^2 + 2\pi r \left(\frac{2156}{\pi r^2}\right) \)
\( S = 2\pi r^2 + \frac{4312}{r} \)
To find the radius that minimizes the surface area, we differentiate \( S \) with respect to \( r \) and set the derivative to zero:
\( \frac{dS}{dr} = 4\pi r - \frac{4312}{r^2} \)
Set \( \frac{dS}{dr} = 0 \):
\( 4\pi r - \frac{4312}{r^2} = 0 \)
\( 4\pi r = \frac{4312}{r^2} \)
\( 4\pi r^3 = 4312 \)
\( r^3 = \frac{4312}{4\pi} = \frac{1078}{\pi} \)
Given \( \pi = \frac{22}{7} \), substitute this value:
\( r^3 = \frac{1078}{22/7} = \frac{1078 \times 7}{22} \)
\( r^3 = 49 \times 7 \)
\( r^3 = 343 \)
Taking the cube root of both sides:
\( r = 7 \) units.
To confirm that this radius gives a minimum surface area, we compute the second derivative of \( S \):
\( \frac{d^2S}{dr^2} = 4\pi + \frac{8624}{r^3} \)
At \( r = 7 \):
\( \frac{d^2S}{dr^2} = 4\pi + \frac{8624}{7^3} = 4\pi + \frac{8624}{343} \)
Since \( 4\pi \) and \( \frac{8624}{343} \) are both positive, \( \frac{d^2S}{dr^2} > 0 \), which confirms that the surface area is minimum when \( r = 7 \) units.
The corresponding height would be \( h = \frac{2156}{\pi (7^2)} = \frac{2156}{(22/7) \times 49} = \frac{2156}{22 \times 7} = \frac{2156}{154} = 14 \) units. Notice that \( h = 2r \), which means the height is equal to the diameter, a common characteristic for optimal cylinder dimensions.
In simple words: For a cylinder to hold a fixed amount of liquid while using the least material for its surface, its radius must be 7 units. This specific radius makes the cylinder efficient in terms of material usage for its construction.

🎯 Exam Tip: For problems involving fixed volume and minimum surface area (or vice versa), the optimal shape often has a simple relationship between its dimensions, such as the height being equal to the diameter in a cylinder. Remember to use the given value of \( \pi \) if specified.

Question 24. A rectangular sheet of tin 45 cm by 24 cm is to be made into box without top, by cutting off squares from the corners and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum possible ?
Answer: Let \(x\) cm be the side length of the identical squares cut from each of the four corners of the rectangular tin sheet. The sheet has dimensions 45 cm by 24 cm.
When these squares are cut and the flaps are folded upwards, an open-top box is formed. The dimensions of this box will be:
Length of the base: \( L = (45 - 2x) \) cm
Breadth of the base: \( B = (24 - 2x) \) cm
Height of the box: \( H = x \) cm
For these dimensions to be valid (positive values for a real box), we must have:
\( x > 0 \)
\( 45 - 2x > 0 \implies 2x < 45 \implies x < 22.5 \)
\( 24 - 2x > 0 \implies 2x < 24 \implies x < 12 \)
Combining these conditions, the valid range for \( x \) is \( 0 < x < 12 \).
The volume of the box, \( V \), is given by the product of its length, breadth, and height:
\( V = L \times B \times H = (45 - 2x)(24 - 2x)x \)
First, multiply the binomials:
\( (45 - 2x)(24 - 2x) = 45 \times 24 - 45 \times 2x - 2x \times 24 + (-2x) \times (-2x) \)
\( = 1080 - 90x - 48x + 4x^2 \)
\( = 4x^2 - 138x + 1080 \)
Now, multiply by \( x \):
\( V = (4x^2 - 138x + 1080)x \)
\( \implies V = 4x^3 - 138x^2 + 1080x \)
To find the value of \( x \) that maximizes the volume, we differentiate \( V \) with respect to \( x \) and set the derivative to zero:
\( \frac{dV}{dx} = 12x^2 - 276x + 1080 \)
Set \( \frac{dV}{dx} = 0 \):
\( 12x^2 - 276x + 1080 = 0 \)
Divide the entire equation by 12 to simplify:
\( x^2 - 23x + 90 = 0 \)
Factor the quadratic equation:
\( (x - 5)(x - 18) = 0 \)
This gives two possible values for \( x \): \( x = 5 \) or \( x = 18 \).
Considering our valid range for \( x \) (\( 0 < x < 12 \)), we must choose \( x = 5 \). The value \( x = 18 \) is outside this range and would result in negative dimensions for the breadth of the box.
To confirm that \( x = 5 \) gives a maximum volume, we calculate the second derivative of \( V \):
\( \frac{d^2V}{dx^2} = 24x - 276 \)
Substitute \( x = 5 \) into the second derivative:
\( \frac{d^2V}{dx^2} = 24(5) - 276 = 120 - 276 = -156 \)
Since \( \frac{d^2V}{dx^2} < 0 \) at \( x = 5 \), the volume is indeed maximum at this value.
Therefore, the side of the square to be cut off is 5 cm. The maximum volume is \( (45-2(5))(24-2(5))(5) = (35)(14)(5) = 2450 \) cm\(^3\).
In simple words: To create the largest possible open box from a rectangular tin sheet, you should cut out a 5 cm by 5 cm square from each corner. This specific cut ensures the box holds the most amount of space.

🎯 Exam Tip: Always check the domain of your variable (x in this case) based on the physical constraints of the problem. This helps in choosing the correct critical point from multiple possibilities and ensures your answer is realistic.

Question 25. Prove that the right circular cone of maximum volume which can be inscribed in a sphere of radius a has its altitude equal to \( \frac{4a}{3} \).
Answer: Let the radius of the sphere be \(a\). Let a right circular cone be inscribed in this sphere. Let the radius of the cone's base be \(r\) and its height be \(h\).
Consider a cross-section of the sphere and the inscribed cone through the axis of the cone. This cross-section will show a circle (the sphere) and an isosceles triangle (the cone).
Let the center of the sphere be \(O\). Let the cone's vertex be at the top of the sphere, and let the center of the cone's base be \(Q\). Let the distance from the sphere's center \(O\) to the cone's base center \(Q\) be \(x\).
From the geometry, the height of the cone will be \( h = a + x \).
Now, consider the right-angled triangle formed by the sphere's radius, the cone's base radius, and the distance \(x\) from the sphere's center to the cone's base. The vertices are \(O\), \(Q\), and a point on the circumference of the cone's base. By the Pythagorean theorem:
\( r^2 + x^2 = a^2 \)
\( \implies r^2 = a^2 - x^2 \)
The volume of the cone, \( V \), is given by the formula:
\( V = \frac{1}{3}\pi r^2 h \)
Substitute the expressions for \( r^2 \) and \( h \):
\( V = \frac{1}{3}\pi (a^2 - x^2)(a + x) \)
\( V = \frac{1}{3}\pi (a^3 + a^2x - ax^2 - x^3) \)
To find the maximum volume, we differentiate \( V \) with respect to \( x \) and set the derivative to zero:
\( \frac{dV}{dx} = \frac{1}{3}\pi (a^2 - 2ax - 3x^2) \)
Set \( \frac{dV}{dx} = 0 \):
\( a^2 - 2ax - 3x^2 = 0 \)
Rearrange the terms into a standard quadratic form:
\( 3x^2 + 2ax - a^2 = 0 \)
Factor the quadratic equation:
\( (3x - a)(x + a) = 0 \)
This gives two possible values for \( x \): \( x = \frac{a}{3} \) or \( x = -a \).
Since \( x \) represents a distance, it must be positive. Therefore, \( x = \frac{a}{3} \).
To confirm that this value of \( x \) gives a maximum volume, we compute the second derivative of \( V \):
\( \frac{d^2V}{dx^2} = \frac{1}{3}\pi (-2a - 6x) \)
Substitute \( x = \frac{a}{3} \) into the second derivative:
\( \frac{d^2V}{dx^2} = \frac{1}{3}\pi \left(-2a - 6\left(\frac{a}{3}\right)\right) \)
\( \implies \frac{d^2V}{dx^2} = \frac{1}{3}\pi (-2a - 2a) \)
\( \implies \frac{d^2V}{dx^2} = \frac{1}{3}\pi (-4a) = -\frac{4\pi a}{3} \)
Since \( a \) is a radius (and thus \( a > 0 \)), \( -\frac{4\pi a}{3} < 0 \), which confirms that the volume is maximum when \( x = \frac{a}{3} \).
Finally, we find the altitude (height) of the cone using \( h = a + x \):
\( h = a + \frac{a}{3} \)
\( h = \frac{3a}{3} + \frac{a}{3} \)
\( h = \frac{4a}{3} \)
This proves that the altitude of the right circular cone of maximum volume inscribed in a sphere of radius \(a\) is \( \frac{4a}{3} \).
In simple words: For a cone placed inside a sphere, its volume is largest when its height is four-thirds of the sphere's radius. This specific height helps the cone take up the most space within the sphere.

🎯 Exam Tip: When a figure is inscribed in another, use the properties of the outer figure (like the sphere's radius) to establish relationships between the dimensions of the inner figure (the cone) for optimization. A clear diagram can greatly help in setting up the equations.

Question 26. Find the volume of the largest right circular cylinder that can be inscribed in a sphere of radius a.
Answer: Let the radius of the sphere be \(a\). Let a right circular cylinder be inscribed within this sphere. Let the radius of the cylinder's base be \(r\) and its height be \(h\).
Consider a cross-section of the sphere and the inscribed cylinder through their common axis. This cross-section will reveal a circle (representing the sphere) and a rectangle (representing the cylinder).
The vertices of the rectangle lie on the circumference of the circle. The diagonal of this rectangle is equal to the diameter of the sphere, which is \(2a\).
Using the Pythagorean theorem for the right-angled triangle formed by the cylinder's radius \(r\), half its height \( \frac{h}{2} \), and the sphere's radius \(a\):
\( r^2 + \left(\frac{h}{2}\right)^2 = a^2 \)
\( \implies r^2 = a^2 - \frac{h^2}{4} \)
The volume of the cylinder, \( V \), is given by the formula:
\( V = \pi r^2 h \)
Substitute the expression for \( r^2 \) into the volume formula:
\( V = \pi \left(a^2 - \frac{h^2}{4}\right) h \)
\( \implies V = \pi a^2 h - \frac{\pi}{4} h^3 \)
To find the height \(h\) that maximizes the volume, we differentiate \(V\) with respect to \(h\) and set the derivative to zero:
\( \frac{dV}{dh} = \pi a^2 - \frac{3\pi}{4} h^2 \)
Set \( \frac{dV}{dh} = 0 \):
\( \pi a^2 - \frac{3\pi}{4} h^2 = 0 \)
\( \pi a^2 = \frac{3\pi}{4} h^2 \)
Divide both sides by \( \pi \):
\( a^2 = \frac{3}{4} h^2 \)
Solve for \( h^2 \):
\( h^2 = \frac{4a^2}{3} \)
Take the square root of both sides (since \(h\) must be positive):
\( h = \sqrt{\frac{4a^2}{3}} = \frac{2a}{\sqrt{3}} \)
To confirm that this value of \(h\) gives a maximum volume, we compute the second derivative of \(V\):
\( \frac{d^2V}{dh^2} = -\frac{3\pi}{4} (2h) = -\frac{3\pi}{2} h \)
Since \( h \) (height) and \( \pi \) are positive, \( -\frac{3\pi}{2} h \) will always be negative. Therefore, \( \frac{d^2V}{dh^2} < 0 \), which confirms that the volume is maximum at \( h = \frac{2a}{\sqrt{3}} \).
Now, substitute this value of \(h\) back into the volume formula to find the maximum volume:
\( V_{max} = \pi a^2 \left(\frac{2a}{\sqrt{3}}\right) - \frac{\pi}{4} \left(\frac{2a}{\sqrt{3}}\right)^3 \)
\( V_{max} = \frac{2\pi a^3}{\sqrt{3}} - \frac{\pi}{4} \left(\frac{8a^3}{3\sqrt{3}}\right) \)
\( V_{max} = \frac{2\pi a^3}{\sqrt{3}} - \frac{2\pi a^3}{3\sqrt{3}} \)
Find a common denominator:
\( V_{max} = \frac{6\pi a^3}{3\sqrt{3}} - \frac{2\pi a^3}{3\sqrt{3}} = \frac{4\pi a^3}{3\sqrt{3}} \)
So, the volume of the largest right circular cylinder that can be inscribed in a sphere of radius \(a\) is \( \frac{4\pi a^3}{3\sqrt{3}} \).
In simple words: The biggest cylinder that can fit inside a sphere will have a height of \( \frac{2a}{\sqrt{3}} \), where \(a\) is the sphere's radius. Its maximum volume will be \( \frac{4\pi a^3}{3\sqrt{3}} \) cubic units.

🎯 Exam Tip: Remember the Pythagorean theorem often links the dimensions of inscribed shapes to the radius of the sphere, which is crucial for setting up the optimization problem. Always check the sign of the second derivative to confirm maximum or minimum.

Question 27. Assuming that the stiffness of a beam of rectangular cross-section varies as the breadth and as the cube of the depth, what must be the breadth of the stiffest beam that can be cut from a log of diameter a.
Answer: Let the log be a circular cylinder with diameter \(a\). When a rectangular beam is cut from this log, its cross-section will be a rectangle inscribed in a circle of diameter \(a\).
Let the breadth of the rectangular beam be \(x\) and its depth be \(y\).
According to the Pythagorean theorem, for a rectangle inscribed in a circle with diameter \(a\):
\( x^2 + y^2 = a^2 \)
The stiffness \(S\) of the beam is given to vary as the breadth and as the cube of the depth. So, we can write the stiffness as:
\( S = kxy^3 \), where \(k\) is a constant of proportionality.
From the relation \( x^2 + y^2 = a^2 \), we can express \( y^2 = a^2 - x^2 \), so \( y = \sqrt{a^2 - x^2} \).
Substitute this into the stiffness formula:
\( S = kx(\sqrt{a^2 - x^2})^3 \)
\( \implies S = kx(a^2 - x^2)^{3/2} \)
To find the breadth \(x\) that maximizes the stiffness, we differentiate \(S\) with respect to \(x\) and set the derivative to zero:
\( \frac{dS}{dx} = k \left[ 1 \cdot (a^2 - x^2)^{3/2} + x \cdot \frac{3}{2}(a^2 - x^2)^{1/2}(-2x) \right] \)
\( \implies \frac{dS}{dx} = k \left[ (a^2 - x^2)^{3/2} - 3x^2(a^2 - x^2)^{1/2} \right] \)
Factor out \( (a^2 - x^2)^{1/2} \):
\( \frac{dS}{dx} = k (a^2 - x^2)^{1/2} \left[ (a^2 - x^2) - 3x^2 \right] \)
\( \implies \frac{dS}{dx} = k (a^2 - x^2)^{1/2} (a^2 - 4x^2) \)
Set \( \frac{dS}{dx} = 0 \):
\( k (a^2 - x^2)^{1/2} (a^2 - 4x^2) = 0 \)
This gives two possibilities:
1) \( (a^2 - x^2)^{1/2} = 0 \implies a^2 - x^2 = 0 \implies x = \pm a \). If \( x = a \), then \( y = 0 \), which means there is no depth, and thus no beam.
2) \( a^2 - 4x^2 = 0 \implies 4x^2 = a^2 \implies x^2 = \frac{a^2}{4} \implies x = \pm \frac{a}{2} \).
Since \( x \) represents the breadth, it must be positive. So, \( x = \frac{a}{2} \). This value also satisfies \( x < a \).
To confirm that \( x = \frac{a}{2} \) corresponds to a maximum, we can analyze the sign of \( \frac{dS}{dx} \). When \( 0 < x < a/2 \), \( (a^2 - 4x^2) \) is positive, so \( \frac{dS}{dx} > 0 \). When \( a/2 < x < a \), \( (a^2 - 4x^2) \) is negative, so \( \frac{dS}{dx} < 0 \). The derivative changes from positive to negative, indicating a maximum at \( x = \frac{a}{2} \).
Therefore, the breadth of the stiffest beam that can be cut from a log of diameter \(a\) must be \( \frac{a}{2} \). The depth would be \( y = \sqrt{a^2 - (a/2)^2} = \sqrt{a^2 - a^2/4} = \sqrt{3a^2/4} = \frac{\sqrt{3}a}{2} \).
In simple words: Imagine cutting the strongest possible rectangular wooden beam from a round log. To get the stiffest beam, its width should be exactly half of the log's diameter.

🎯 Exam Tip: Be careful with the variables. If 'a' represents the diameter, then \( x^2+y^2=a^2 \). If it represents the radius, then \( x^2+y^2=(2a)^2=4a^2 \). Always read the initial definitions carefully.

Question 28. Show that the radius of the right circular cylinder of greatest curved surface which can be inscribed in a givn right cricular cone is half that of the cone.
Answer: Let the given right circular cone have radius \(R\) and height \(H\). Let a right circular cylinder be inscribed in this cone, with radius \(r\) and height \(h\).
Consider a cross-section of the cone and cylinder through their common axis. This cross-section shows an isosceles triangle (the cone) and a rectangle (the cylinder).
Let the vertex of the cone be at the top. Using similar triangles (the large cone and the smaller cone formed above the cylinder), we can establish a relationship between the dimensions:
The height of the small cone above the cylinder is \( H - h \). Its radius is \( r \).
The full cone has height \( H \) and radius \( R \).
By similar triangles:
\( \frac{H - h}{r} = \frac{H}{R} \)
Now, we want to express \( h \) in terms of \( r \), \( R \), and \( H \):
\( R(H - h) = Hr \)
\( RH - Rh = Hr \)
\( RH - Hr = Rh \)
\( h = \frac{RH - Hr}{R} = H \left(1 - \frac{r}{R}\right) \)
The curved surface area of the cylinder, \( S_{cyl} \), is given by:
\( S_{cyl} = 2\pi r h \)
Substitute the expression for \( h \) into the formula for \( S_{cyl} \):
\( S_{cyl} = 2\pi r \left( H \left(1 - \frac{r}{R}\right) \right) \)
\( S_{cyl} = \frac{2\pi H}{R} (Rr - r^2) \)
To find the radius \(r\) that maximizes the curved surface area, we differentiate \( S_{cyl} \) with respect to \( r \) and set the derivative to zero:
\( \frac{dS_{cyl}}{dr} = \frac{2\pi H}{R} (R - 2r) \)
Set \( \frac{dS_{cyl}}{dr} = 0 \):
\( \frac{2\pi H}{R} (R - 2r) = 0 \)
Since \( H \) (height) and \( R \) (radius of the cone) are non-zero, we must have:
\( R - 2r = 0 \)
\( \implies R = 2r \)
\( \implies r = \frac{R}{2} \)
To confirm that this value of \( r \) gives a maximum surface area, we compute the second derivative of \( S_{cyl} \):
\( \frac{d^2S_{cyl}}{dr^2} = \frac{2\pi H}{R} (-2) \)
\( \implies \frac{d^2S_{cyl}}{dr^2} = -\frac{4\pi H}{R} \)
Since \( H \) and \( R \) are positive, \( -\frac{4\pi H}{R} \) is always negative. Therefore, \( \frac{d^2S_{cyl}}{dr^2} < 0 \), which confirms that the curved surface area is maximum when \( r = \frac{R}{2} \).
This shows that the radius of the right circular cylinder of greatest curved surface area that can be inscribed in a given right circular cone is half that of the cone.
In simple words: When you put a cylinder inside a cone, the cylinder will have the largest side surface area if its radius is exactly half the radius of the cone.

🎯 Exam Tip: For problems involving inscribed shapes, similar triangles are often a key tool to relate the dimensions of the inner shape to the outer shape. A clear diagram and correct setup of variables are essential for success.