Samacheer Kalvi Class 12 Maths Solutions Chapter 7 Applications of Differential Calculus Exercise 7.8

 

Question 1. Find two positive numbers whose sum is 12 and their product is maximum.
Answer: Let the two positive numbers be \( x \) and \( 12 - x \).
Their product, \( P = x(12 - x) = 12x - x^2 \).
To find the maximum product, we first find the derivative of \( P \) with respect to \( x \):
\( P'(x) = 12 - 2x \)
For maximum or minimum, set \( P'(x) = 0 \):
\( 12 - 2x = 0 \)
\( \implies 2x = 12 \)
\( \implies x = 6 \)
Next, find the second derivative:
\( P''(x) = -2 \)
At \( x = 6 \), \( P''(x) = -2 \), which is less than 0. This means the product is maximum when \( x = 6 \).
When \( x = 6 \), the first number is 6.
The second number is \( 12 - x = 12 - 6 = 6 \).
So, the two numbers are 6 and 6. These numbers are equal, which is often the case in optimization problems when symmetry is involved.
In simple words: We are looking for two numbers that add up to 12 and give the biggest possible answer when multiplied together. By using a bit of math to find the peak of their product, we discover that both numbers should be 6.

🎯 Exam Tip: For optimization problems, always check the second derivative to confirm whether it's a maximum (negative second derivative) or a minimum (positive second derivative).

 

Question 2. Find two positive numbers whose product is 20 and their sum is minimum.
Answer: Let the two positive numbers be \( x \) and \( y \).
Given that their product is 20, so \( xy = 20 \). This means \( y = \frac{20}{x} \).
Their sum, \( S = x + y = x + \frac{20}{x} \).
To find the minimum sum, we find the derivative of \( S \) with respect to \( x \):
\( \frac{dS}{dx} = 1 - \frac{20}{x^2} \)
For maximum or minimum, set \( \frac{dS}{dx} = 0 \):
\( 1 - \frac{20}{x^2} = 0 \)
\( \implies x^2 = 20 \)
\( \implies x = \pm\sqrt{20} = \pm2\sqrt{5} \)
Since we are looking for positive numbers, \( x = 2\sqrt{5} \).
Next, find the second derivative:
\( \frac{d^2S}{dx^2} = \frac{40}{x^3} \)
At \( x = 2\sqrt{5} \), \( \frac{d^2S}{dx^2} = \frac{40}{(2\sqrt{5})^3} = \frac{40}{8 \cdot 5\sqrt{5}} = \frac{40}{40\sqrt{5}} = \frac{1}{\sqrt{5}} \). This is greater than 0, meaning the sum is minimum.
When \( x = 2\sqrt{5} \), then \( y = \frac{20}{2\sqrt{5}} = \frac{10}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5} \).
The minimum sum is \( x + y = 2\sqrt{5} + 2\sqrt{5} = 4\sqrt{5} \). This type of problem often shows that for a fixed product, the sum is minimized when the numbers are equal.
In simple words: We need to find two positive numbers that multiply to 20, and their total sum should be the smallest possible. After doing the math, we find that both numbers are \( 2\sqrt{5} \), and their smallest sum is \( 4\sqrt{5} \).

🎯 Exam Tip: Remember to consider the domain of the variables (e.g., positive numbers) when solving and reject solutions that fall outside it.

 

Question 3. Find the smallest possible value of x² + y² given that x + y = 10.
Answer: Given \( x + y = 10 \), we can write \( y = 10 - x \).
We want to find the smallest value of \( A = x^2 + y^2 \).
Substitute \( y = 10 - x \) into the expression for \( A \):
\( A = x^2 + (10 - x)^2 \)
\( A = x^2 + (100 - 20x + x^2) \)
\( A = 2x^2 - 20x + 100 \)
To find the minimum value, we find the derivative of \( A \) with respect to \( x \):
\( \frac{dA}{dx} = 4x - 20 \)
For maximum or minimum, set \( \frac{dA}{dx} = 0 \):
\( 4x - 20 = 0 \)
\( \implies 4x = 20 \)
\( \implies x = 5 \)
Next, find the second derivative:
\( \frac{d^2A}{dx^2} = 4 \)
Since \( \frac{d^2A}{dx^2} = 4 \) (which is greater than 0), the value of \( A \) is minimum when \( x = 5 \).
When \( x = 5 \), then \( y = 10 - x = 10 - 5 = 5 \).
The smallest possible value of \( x^2 + y^2 \) is \( (5)^2 + (5)^2 = 25 + 25 = 50 \). This shows that for a fixed sum, the sum of squares is minimized when the numbers are equal.
In simple words: We have two numbers, \( x \) and \( y \), that add up to 10. We want to find the smallest value of \( x^2 \) plus \( y^2 \). By using calculus, we find that \( x \) should be 5 and \( y \) should also be 5, making the smallest sum of squares equal to 50.

🎯 Exam Tip: When minimizing the sum of squares with a fixed sum, the minimum often occurs when the variables are equal, demonstrating a principle of symmetry.

 

Question 4. A garden is to be laid out in a rectangular area and protected by a wire fence. What is the largest possible area of the fenced garden with 40 meters of wire?
Answer: Let the length of the rectangular garden be \( l \) and the width be \( b \). The total length of wire used for fencing is the perimeter.

Given perimeter \( P = 40 \) meters.
So, \( 2(l + b) = 40 \)
\( \implies l + b = 20 \)
Let \( l = x \) meters, then \( b = (20 - x) \) meters.
The area of the garden is \( A = l \times b = x(20 - x) = 20x - x^2 \).
To find the maximum area, we differentiate \( A \) with respect to \( x \):
\( A'(x) = 20 - 2x \)
Set \( A'(x) = 0 \) for maximum or minimum:
\( 20 - 2x = 0 \)
\( \implies 2x = 20 \)
\( \implies x = 10 \)
Now, find the second derivative:
\( A''(x) = -2 \)
Since \( A''(x) = -2 \) (which is less than 0), \( x = 10 \) corresponds to a maximum point.
Maximum Area \( = x(20 - x) = 10(20 - 10) = 10 \times 10 = 100 \) sq. m.
This demonstrates that a square shape maximizes the area for a given perimeter.
In simple words: We have 40 meters of wire to make a rectangular garden fence. We want to find the biggest area the garden can have. By using math to optimize the area, we find that a square garden with sides of 10 meters each will give the largest area of 100 square meters.

🎯 Exam Tip: Remember that for a fixed perimeter, a square will always yield the largest area among all rectangles.

 

Question 5. A rectangular page is to contain 24 cm² of print. The margins at the top and bottom of the page are 1.5 cm and the margins at the other sides of the page are 1 cm. What should be the dimensions of the page so that the area of the paper used is minimum?
Answer: Let the width of the printed part be \( x \) cm and the height of the printed part be \( y \) cm.
Given, the area of the printed part is 24 cm², so \( xy = 24 \). This means \( y = \frac{24}{x} \).

The margins at the top and bottom are 1.5 cm each, so total vertical margin is \( 1.5 + 1.5 = 3 \) cm.
The margins at the sides are 1 cm each, so total horizontal margin is \( 1 + 1 = 2 \) cm.
The total width of the page is \( x + 2(1) = x + 2 \) cm.
The total height of the page is \( y + 2(1.5) = y + 3 \) cm.
The total area of the paper used is \( A = (x + 2)(y + 3) \).
Substitute \( y = \frac{24}{x} \) into the area formula:
\( A = (x + 2)\left(\frac{24}{x} + 3\right) \)
\( A = 24 + 3x + \frac{48}{x} + 6 \)
\( A = 3x + \frac{48}{x} + 30 \)
To find the minimum area, differentiate \( A \) with respect to \( x \):
\( \frac{dA}{dx} = 3 - \frac{48}{x^2} \)
Set \( \frac{dA}{dx} = 0 \) for maximum or minimum:
\( 3 - \frac{48}{x^2} = 0 \)
\( \implies 3x^2 = 48 \)
\( \implies x^2 = 16 \)
\( \implies x = \pm4 \)
Since \( x \) represents a length, it cannot be negative, so \( x = 4 \) cm.
Now, find the second derivative:
\( \frac{d^2A}{dx^2} = \frac{96}{x^3} \)
At \( x = 4 \), \( \frac{d^2A}{dx^2} = \frac{96}{4^3} = \frac{96}{64} > 0 \). This means the area is minimum when \( x = 4 \) cm.
If \( x = 4 \) cm, then \( y = \frac{24}{4} = 6 \) cm.
The dimensions of the page for minimum area are:
Width \( = x + 2 = 4 + 2 = 6 \) cm.
Height \( = y + 3 = 6 + 3 = 9 \) cm.
These dimensions minimize the total paper used while maintaining the required print area.
In simple words: We have a paper where the printed part must be 24 square cm, with margins around it. We want to find the smallest possible size of the entire paper. The best dimensions for the full page are 6 cm wide and 9 cm high.

🎯 Exam Tip: When dealing with margins, carefully add the margin widths to both sides (e.g., top and bottom, or left and right) of the inner dimension to get the total outer dimension.

 

Question 6. A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 1,80,000 sq. mtrs in order to provide enough grass for herds. No fencing is needed along the river. What is the length of the minimum needed fencing material?
Answer: Let the length of the pasture parallel to the river be \( x \) meters and the breadth perpendicular to the river be \( y \) meters.

The given area is 1,80,000 sq. m, so \( xy = 1,80,000 \). This implies \( y = \frac{180000}{x} \).
Since one side is adjacent to the river, no fencing is needed along that side. The total length of fencing material needed is \( P = 2y + x \).
Substitute \( y = \frac{180000}{x} \) into the perimeter formula:
\( P = 2\left(\frac{180000}{x}\right) + x \)
\( P = \frac{360000}{x} + x \)
To find the minimum fencing material, differentiate \( P \) with respect to \( x \):
\( \frac{dP}{dx} = -\frac{360000}{x^2} + 1 \)
Set \( \frac{dP}{dx} = 0 \) for maximum or minimum:
\( -\frac{360000}{x^2} + 1 = 0 \)
\( \implies 1 = \frac{360000}{x^2} \)
\( \implies x^2 = 360000 \)
\( \implies x = \pm\sqrt{360000} \)
\( \implies x = \pm600 \)
Since length cannot be negative, \( x = 600 \) meters.
Next, find the second derivative:
\( \frac{d^2P}{dx^2} = \frac{720000}{x^3} \)
At \( x = 600 \), \( \frac{d^2P}{dx^2} = \frac{720000}{(600)^3} > 0 \). This means the perimeter \( P \) is minimum when \( x = 600 \) meters.
If \( x = 600 \) m, then \( y = \frac{180000}{600} = 300 \) m.
The length of the minimum needed fencing material is \( P = 2y + x = 2(300) + 600 = 600 + 600 = 1200 \) m. Minimizing the fence while maximizing space is an important skill in planning.
In simple words: A farmer wants to fence a rectangular field next to a river. No fence is needed on the river side. The field must be 1,80,000 square meters. We need to find the least amount of fence material. The minimum fence needed is 1200 meters, which happens when the side parallel to the river is 600 meters and the sides perpendicular to the river are 300 meters each.

🎯 Exam Tip: Always draw a simple diagram for geometry optimization problems to clearly visualize which sides require fencing or are constrained.

 

Question 7. Find the dimensions of the rectangle with the maximum area that can be inscribed in a circle of a radius of 10 cm.
Answer: Let the length of the rectangle be \( x \) cm and the breadth be \( y \) cm.

From the figure, the diagonal of the inscribed rectangle is equal to the diameter of the circle. The radius of the circle is 10 cm, so the diameter is 20 cm.
Using the Pythagorean theorem for the rectangle's sides and diagonal:
\( x^2 + y^2 = (20)^2 \)
\( x^2 + y^2 = 400 \)
So, \( y^2 = 400 - x^2 \), which means \( y = \sqrt{400 - x^2} \) (since \( y \) must be positive).
The area of the rectangle is \( A = xy \).
Substitute \( y \):
\( A = x\sqrt{400 - x^2} \)
To maximize the area, we differentiate \( A \) with respect to \( x \):
\( \frac{dA}{dx} = 1 \cdot \sqrt{400 - x^2} + x \cdot \frac{1}{2\sqrt{400 - x^2}} (-2x) \)
\( \frac{dA}{dx} = \sqrt{400 - x^2} - \frac{x^2}{\sqrt{400 - x^2}} \)
\( \frac{dA}{dx} = \frac{(400 - x^2) - x^2}{\sqrt{400 - x^2}} \)
\( \frac{dA}{dx} = \frac{400 - 2x^2}{\sqrt{400 - x^2}} \)
Set \( \frac{dA}{dx} = 0 \) for maximum or minimum:
\( \frac{400 - 2x^2}{\sqrt{400 - x^2}} = 0 \)
\( \implies 400 - 2x^2 = 0 \)
\( \implies 2x^2 = 400 \)
\( \implies x^2 = 200 \)
\( \implies x = \pm\sqrt{200} = \pm10\sqrt{2} \)
Since \( x \) is a length, \( x = 10\sqrt{2} \) cm.
Now, find the second derivative \( \frac{d^2A}{dx^2} \). (The calculation is complex but leads to a negative value at \( x=10\sqrt{2} \), confirming a maximum).
Using the quotient rule \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = 400 - 2x^2 \) and \( v = (400 - x^2)^{1/2} \).
\( u' = -4x \)
\( v' = \frac{1}{2}(400 - x^2)^{-1/2}(-2x) = -x(400 - x^2)^{-1/2} \)
\( \frac{d^2A}{dx^2} = \frac{(-4x)\sqrt{400 - x^2} - (400 - 2x^2)\left(-x(400 - x^2)^{-1/2}\right)}{400 - x^2} \)
Multiplying numerator and denominator by \( \sqrt{400 - x^2} \):
\( \frac{d^2A}{dx^2} = \frac{-4x(400 - x^2) + x(400 - 2x^2)}{(400 - x^2)^{3/2}} \)
\( \frac{d^2A}{dx^2} = \frac{-1600x + 4x^3 + 400x - 2x^3}{(400 - x^2)^{3/2}} \)
\( \frac{d^2A}{dx^2} = \frac{2x^3 - 1200x}{(400 - x^2)^{3/2}} \)
At \( x = 10\sqrt{2} \):
\( x^2 = 200 \)
\( x^3 = 200 \cdot 10\sqrt{2} = 2000\sqrt{2} \)
Denominator \( (400 - 200)^{3/2} = (200)^{3/2} > 0 \).
Numerator \( = 2(2000\sqrt{2}) - 1200(10\sqrt{2}) = 4000\sqrt{2} - 12000\sqrt{2} = -8000\sqrt{2} \).
So, \( \frac{d^2A}{dx^2} < 0 \) at \( x = 10\sqrt{2} \), confirming a maximum.
When \( x = 10\sqrt{2} \) cm:
\( y = \sqrt{400 - (10\sqrt{2})^2} = \sqrt{400 - 200} = \sqrt{200} = 10\sqrt{2} \) cm.
Since \( x = y \), the rectangle with the maximum area inscribed in a circle is a square. Each side of the square will be \( 10\sqrt{2} \) cm. This is a common result in geometry optimization.
In simple words: We want to find the biggest rectangle that can fit inside a circle with a radius of 10 cm. Using mathematical steps, we find that the rectangle with the largest area is actually a square, and its sides are each \( 10\sqrt{2} \) cm long.

🎯 Exam Tip: Remember that the largest rectangle that can be inscribed in a circle is always a square. Knowing this can help you verify your answer quickly.

 

Question 8. Prove that among all the rectangles of the given perimeter, the square has the maximum area.
Answer: Let \( x \) be the length and \( y \) be the breadth of a rectangle. Let \( P \) be the given perimeter.
So, \( 2(x + y) = P \). This means \( x + y = \frac{P}{2} \), and we can express \( y \) as \( y = \frac{P}{2} - x \).
The area of the rectangle is \( A = xy \).
Substitute \( y = \frac{P}{2} - x \) into the area formula:
\( A = x\left(\frac{P}{2} - x\right) \)
\( A = \frac{P}{2}x - x^2 \)
To find the maximum area, differentiate \( A \) with respect to \( x \):
\( \frac{dA}{dx} = \frac{P}{2} - 2x \)
Set \( \frac{dA}{dx} = 0 \) for maximum or minimum:
\( \frac{P}{2} - 2x = 0 \)
\( \implies 2x = \frac{P}{2} \)
\( \implies x = \frac{P}{4} \)
Next, find the second derivative:
\( \frac{d^2A}{dx^2} = -2 \)
Since \( \frac{d^2A}{dx^2} = -2 \) (which is less than 0), the area is maximum when \( x = \frac{P}{4} \).
Now, find the value of \( y \) when \( x = \frac{P}{4} \):
\( y = \frac{P}{2} - x = \frac{P}{2} - \frac{P}{4} = \frac{2P - P}{4} = \frac{P}{4} \).
Since \( x = \frac{P}{4} \) and \( y = \frac{P}{4} \), the length equals the breadth. Therefore, the rectangle is a square.
Hence, it is proved that among all rectangles with a given perimeter, the square has the maximum area. This is a fundamental concept in optimization and geometry.
In simple words: We want to show that if you have a certain amount of fence to make a rectangle, you will get the biggest possible area if you make a square. By using calculus to find the maximum area, we proved that the length and width must be equal, which means the shape is a square.

🎯 Exam Tip: This is a classic proof in calculus. Make sure to clearly define variables, set up the optimization function, find critical points, and use the second derivative test to confirm maximums.

 

Question 9. Find the dimensions of the largest rectangle that can be inscribed in a semi-circle of radius r cm.
Answer: Let the semi-circle have radius \( r \). Place the semi-circle on the x-axis with its center at the origin.

Let the dimensions of the rectangle be \( 2x \) for its length (along the diameter) and \( y \) for its breadth.
The vertices of the rectangle on the semi-circle lie on the circle \( x^2 + y^2 = r^2 \).
Consider the top-right corner of the rectangle, which has coordinates \( (x, y) \). This point lies on the circle.
So, \( x^2 + y^2 = r^2 \). This implies \( y^2 = r^2 - x^2 \), so \( y = \sqrt{r^2 - x^2} \).
The area of the rectangle is \( A = (\text{length}) \times (\text{breadth}) = (2x) \times y = 2xy \).
Substitute \( y \):
\( A = 2x\sqrt{r^2 - x^2} \)
To maximize \( A \), it's often easier to maximize \( A^2 \) when dealing with square roots. Let \( S = A^2 \).
\( S = (2x\sqrt{r^2 - x^2})^2 = 4x^2(r^2 - x^2) = 4r^2x^2 - 4x^4 \).
Differentiate \( S \) with respect to \( x \):
\( \frac{dS}{dx} = 8r^2x - 16x^3 \)
Set \( \frac{dS}{dx} = 0 \) for maximum or minimum:
\( 8r^2x - 16x^3 = 0 \)
\( 8x(r^2 - 2x^2) = 0 \)
This gives \( x = 0 \) or \( r^2 - 2x^2 = 0 \).
If \( x = 0 \), the area is 0, which is a minimum. So we consider \( r^2 - 2x^2 = 0 \).
\( \implies 2x^2 = r^2 \)
\( \implies x^2 = \frac{r^2}{2} \)
\( \implies x = \sqrt{\frac{r^2}{2}} = \frac{r}{\sqrt{2}} = \frac{r\sqrt{2}}{2} \)
Next, find the second derivative of \( S \):
\( \frac{d^2S}{dx^2} = 8r^2 - 48x^2 \)
At \( x = \frac{r\sqrt{2}}{2} \):
\( \frac{d^2S}{dx^2} = 8r^2 - 48\left(\frac{r^2}{2}\right) = 8r^2 - 24r^2 = -16r^2 \)
Since \( \frac{d^2S}{dx^2} < 0 \), \( x = \frac{r\sqrt{2}}{2} \) corresponds to a maximum area.
Now find the dimensions of the rectangle:
Length \( = 2x = 2\left(\frac{r\sqrt{2}}{2}\right) = r\sqrt{2} \) cm.
Breadth \( = y = \sqrt{r^2 - x^2} = \sqrt{r^2 - \left(\frac{r\sqrt{2}}{2}\right)^2} = \sqrt{r^2 - \frac{r^2}{2}} = \sqrt{\frac{r^2}{2}} = \frac{r\sqrt{2}}{2} \) cm.
The dimensions of the largest rectangle are \( r\sqrt{2} \) cm by \( \frac{r\sqrt{2}}{2} \) cm. This result shows the optimal proportions for maximum area.
In simple words: We want to fit the largest possible rectangle inside a semi-circle of a given radius \( r \). By using math to find the biggest area, we discover that the rectangle should have a length of \( r\sqrt{2} \) cm and a breadth of \( \frac{r\sqrt{2}}{2} \) cm.

🎯 Exam Tip: When a problem involves a square root in the function to be optimized, it's often easier to optimize the square of that function to simplify differentiation.

 

Question 10. A manufacturer wants to design an open box having a square base and a surface area of 108 sq. cm. Determine the dimensions of the box for the maximum volume.
Answer: Let the side length of the square base be \( x \) cm and the height of the box be \( y \) cm.
Since the box is open (no top), its surface area consists of the base and four side faces.
Surface Area \( SA = x^2 + 4xy \).
Given \( SA = 108 \) sq. cm.
So, \( x^2 + 4xy = 108 \).
From this, we can express \( y \) in terms of \( x \):
\( 4xy = 108 - x^2 \)
\( y = \frac{108 - x^2}{4x} \)
The volume of the box is \( V = x^2y \).
Substitute the expression for \( y \) into the volume formula:
\( V = x^2 \left(\frac{108 - x^2}{4x}\right) \)
\( V = \frac{x(108 - x^2)}{4} \)
\( V = \frac{108x - x^3}{4} \)
To find the maximum volume, differentiate \( V \) with respect to \( x \):
\( \frac{dV}{dx} = \frac{1}{4}(108 - 3x^2) \)
Set \( \frac{dV}{dx} = 0 \) for maximum or minimum:
\( \frac{1}{4}(108 - 3x^2) = 0 \)
\( \implies 108 - 3x^2 = 0 \)
\( \implies 3x^2 = 108 \)
\( \implies x^2 = 36 \)
\( \implies x = \pm6 \)
Since length cannot be negative, \( x = 6 \) cm.
Next, find the second derivative of \( V \):
\( \frac{d^2V}{dx^2} = \frac{1}{4}(-6x) = -\frac{3}{2}x \)
At \( x = 6 \):
\( \frac{d^2V}{dx^2} = -\frac{3}{2}(6) = -9 \)
Since \( \frac{d^2V}{dx^2} < 0 \), \( x = 6 \) corresponds to a maximum volume.
Now find the height \( y \):
\( y = \frac{108 - x^2}{4x} = \frac{108 - 6^2}{4(6)} = \frac{108 - 36}{24} = \frac{72}{24} = 3 \) cm.
The dimensions of the box for maximum volume are: length = 6 cm, breadth = 6 cm (square base), and height = 3 cm. This design uses material efficiently to maximize storage space.
In simple words: A company wants to make an open box with a square bottom. The total surface area of the box is 108 square cm. We need to find the size of the box that gives the largest possible inside space (volume). The best dimensions are a base of 6 cm by 6 cm and a height of 3 cm.

🎯 Exam Tip: Always remember to consider if the problem specifies an "open" or "closed" box, as this affects the surface area formula. An open box means no top surface area.

 

Question 11. The volume of a cylinder is given by the formula V = πr²h. Find the greatest and least values of V if r + h = 6.
Answer: Given the formula for the volume of a cylinder \( V = \pi r^2 h \).
Also given the constraint \( r + h = 6 \). We can express \( h \) in terms of \( r \): \( h = 6 - r \). (Alternatively, \( r = 6 - h \)). Let's use \( r = 6 - h \).
Substitute \( r = 6 - h \) into the volume formula:
\( V = \pi (6 - h)^2 h \)
\( V = \pi (36 - 12h + h^2) h \)
\( V = \pi (36h - 12h^2 + h^3) \)
To find the greatest and least values, we differentiate \( V \) with respect to \( h \):
\( \frac{dV}{dh} = \pi (36 - 24h + 3h^2) \)
Set \( \frac{dV}{dh} = 0 \) for critical points:
\( \pi (36 - 24h + 3h^2) = 0 \)
\( \implies 3h^2 - 24h + 36 = 0 \)
Divide by 3:
\( \implies h^2 - 8h + 12 = 0 \)
Factor the quadratic equation:
\( \implies (h - 2)(h - 6) = 0 \)
This gives critical values \( h = 2 \) or \( h = 6 \).
Now, find the second derivative of \( V \):
\( \frac{d^2V}{dh^2} = \pi (-24 + 6h) \)
Evaluate \( \frac{d^2V}{dh^2} \) at each critical point:
For \( h = 2 \):
\( \frac{d^2V}{dh^2} = \pi (-24 + 6(2)) = \pi (-24 + 12) = -12\pi \)
Since \( -12\pi < 0 \), \( h = 2 \) corresponds to a maximum volume.
If \( h = 2 \), then \( r = 6 - h = 6 - 2 = 4 \).
Maximum volume \( V = \pi (4)^2 (2) = 32\pi \).
For \( h = 6 \):
\( \frac{d^2V}{dh^2} = \pi (-24 + 6(6)) = \pi (-24 + 36) = 12\pi \)
Since \( 12\pi > 0 \), \( h = 6 \) corresponds to a minimum volume.
If \( h = 6 \), then \( r = 6 - h = 6 - 6 = 0 \).
Minimum volume \( V = \pi (0)^2 (6) = 0 \). (A cylinder with radius 0 has no volume, which is logically the least possible positive volume).
So, the greatest value of \( V \) is \( 32\pi \) and the least value is \( 0 \). These results highlight the range of possible volumes for a cylinder under this constraint.
In simple words: We have a cylinder where its radius \( r \) and height \( h \) add up to 6. We want to find the biggest and smallest possible amounts of space (volume) it can hold. The greatest volume is \( 32\pi \) when the radius is 4 and height is 2, and the least volume is 0 when the radius is 0 and height is 6 (which means no cylinder at all).

🎯 Exam Tip: When finding greatest and least values, always test the function at critical points and at the boundaries of the domain (if applicable), as the minimum/maximum can occur at either.

 

Question 12. A hollow cone with a base radius of a cm and height of b cm is placed on a table. Show that the volume of the largest cylinder that can be hidden underneath is \( \frac{4}{9} \) times the volume of the cone.
Answer: Let the cone have base radius \( a \) and height \( b \). The volume of the cone is \( V_{cone} = \frac{1}{3}\pi a^2 b \).
Let a cylinder be inscribed in this cone with radius \( r \) and height \( h \).

Using similar triangles (formed by the cone's axis, its radius, and the top of the cylinder):
\( \frac{b - h}{r} = \frac{b}{a} \)
\( \implies a(b - h) = br \)
\( \implies ab - ah = br \)
\( \implies h = b - \frac{br}{a} = b\left(1 - \frac{r}{a}\right) \)
The volume of the cylinder is \( V_{cylinder} = \pi r^2 h \).
Substitute the expression for \( h \):
\( V = \pi r^2 b\left(1 - \frac{r}{a}\right) \)
\( V = \pi b \left(r^2 - \frac{r^3}{a}\right) \)
To find the maximum volume, differentiate \( V \) with respect to \( r \):
\( \frac{dV}{dr} = \pi b \left(2r - \frac{3r^2}{a}\right) \)
Set \( \frac{dV}{dr} = 0 \) for critical points:
\( \pi b r \left(2 - \frac{3r}{a}\right) = 0 \)
This gives \( r = 0 \) or \( 2 - \frac{3r}{a} = 0 \).
If \( r = 0 \), the volume is 0, which is a minimum. So we consider \( 2 - \frac{3r}{a} = 0 \).
\( \implies 2 = \frac{3r}{a} \)
\( \implies 3r = 2a \)
\( \implies r = \frac{2a}{3} \)
Next, find the second derivative of \( V \):
\( \frac{d^2V}{dr^2} = \pi b \left(2 - \frac{6r}{a}\right) \)
At \( r = \frac{2a}{3} \):
\( \frac{d^2V}{dr^2} = \pi b \left(2 - \frac{6}{a}\left(\frac{2a}{3}\right)\right) \)
\( \frac{d^2V}{dr^2} = \pi b \left(2 - 4\right) = -2\pi b \)
Since \( b \) (height) is positive, \( -2\pi b < 0 \). This means \( r = \frac{2a}{3} \) corresponds to a maximum volume.
Now find the height \( h \) of the cylinder at this radius:
\( h = b\left(1 - \frac{r}{a}\right) = b\left(1 - \frac{2a/3}{a}\right) = b\left(1 - \frac{2}{3}\right) = b\left(\frac{1}{3}\right) = \frac{b}{3} \)
The maximum volume of the inscribed cylinder is:
\( V_{cylinder\_max} = \pi r^2 h = \pi \left(\frac{2a}{3}\right)^2 \left(\frac{b}{3}\right) \)
\( V_{cylinder\_max} = \pi \left(\frac{4a^2}{9}\right) \left(\frac{b}{3}\right) = \frac{4\pi a^2 b}{27} \)
We need to show this is \( \frac{4}{9} \) times the volume of the cone, \( V_{cone} = \frac{1}{3}\pi a^2 b \).
\( V_{cylinder\_max} = \frac{4}{9} \left(\frac{\pi a^2 b}{3}\right) = \frac{4}{9} V_{cone} \).
Hence, it is proved that the volume of the largest cylinder that can be inscribed in a cone is \( \frac{4}{9} \) times the volume of the cone. This ratio is a fixed constant, regardless of the cone's dimensions.
In simple words: We have a cone, and we want to fit the biggest possible cylinder inside it. Using math, we found that the maximum space the cylinder can take up is exactly \( \frac{4}{9} \) of the cone's total space. This means the cylinder fills a little less than half of the cone.

🎯 Exam Tip: Remember that similar triangles are crucial for relating the dimensions of the inscribed figure (cylinder) to the outer figure (cone) in these types of problems.

 

Question 12. A hollow cone with a base radius of a cm and height of b cm is placed on a table. Show that the volume of the largest cylinder that can be hidden underneath is \( \frac { 4 }{ 9 } \) times the volume of the cone.
Answer:
(Solution continues from previous page where \( r = \frac{2a}{3} \) was found for maximum/minimum volume.)
To check if it is a maximum, we calculate the second derivative:
\( \frac{d^2V}{dr^2} = \pi \left( 2b - \frac{6br}{a} \right) \)
Now, we put \( r = \frac{2a}{3} \) into the second derivative:
\( \frac{d^2V}{dr^2} = \pi \left( 2b - \frac{6b}{a} \left( \frac{2a}{3} \right) \right) \)
\( \frac{d^2V}{dr^2} = \pi (2b - 4b) \)
\( \frac{d^2V}{dr^2} = -2b\pi \)
Since \( -2b\pi \) is less than 0 (as b and \( \pi \) are positive), the volume is maximum when \( r = \frac{2a}{3} \).
Next, we find the height 'h' of the cylinder using the relation \( h = b - \frac{b}{a}r \):
\( h = b - \frac{b}{a} \left( \frac{2a}{3} \right) \)
\( h = b - \frac{2b}{3} \)
\( h = \frac{3b - 2b}{3} = \frac{b}{3} \)
Now we find the maximum volume of the cylinder using \( V = \pi r^2 h \):
\( V = \pi \left( \frac{2a}{3} \right)^2 \left( \frac{b}{3} \right) \)
\( V = \pi \left( \frac{4a^2}{9} \right) \left( \frac{b}{3} \right) \)
\( V = \frac{4 \pi a^2 b}{27} \)
We know the volume of the cone is \( V_{cone} = \frac{1}{3} \pi a^2 b \).
So, we can write the cylinder's maximum volume as:
\( V = \frac{4}{9} \left( \frac{1}{3} \pi a^2 b \right) \)
\( V = \frac{4}{9} V_{cone} \)
Therefore, the volume of the largest cylinder is \( \frac{4}{9} \) times the volume of the cone. This type of problem often involves optimization using calculus to find maximum or minimum values under certain conditions.
Hence proved.
In simple words: We used calculus to find the best radius and height for a cylinder inside a cone so it has the biggest possible volume. The math showed that this maximum volume is exactly 4/9 of the cone's total volume.

🎯 Exam Tip: When proving relationships like this, clearly state the formulas, show each step of differentiation and substitution, and always verify if it's a maximum or minimum using the second derivative test.