RBSE Solutions Class 12 Maths Chapter 8 Application of Derivatives Exercise 8.6

 

Question 1. Show that the triangle of maximum area that can be inscribed in a circle is an equilateral triangle.
Answer: Let's consider a triangle \( \Delta ABC \) that is drawn inside a circle with a radius of \( r \). A line from point \( B \) is drawn perpendicular to side \( AC \). This line meets \( AC \) at point \( D \) and extends to meet the circle at point \( E \). We'll set the base of the triangle \( AC = 2x \) and its height \( BD = y \).
The area of the triangle \( A \) can be found using the formula: \( A = \frac {1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2x \times y = xy \). This is our first important equation (i).
Since \( AC \) and \( BE \) are lines (chords) inside the circle, we know that \( AD \times DC = BD \times DE \). Also, the radius of the circle is \( r \). We can write \( DE \) as \( BE - DB \). Since \( BE \) is the diameter, \( BE = 2r \). So, \( DE = 2r - y \).
We have \( x \times x = y \times (2r - y) \), which simplifies to \( x^2 = 2ry - y^2 \).
\( \implies x = \sqrt{2ry - y^2} \). This is our second important equation (ii).
Now, we substitute the value of \( x \) from equation (ii) into equation (i) to express the area \( A \) in terms of \( y \):
\( A = y \sqrt{2ry - y^2} \). To make differentiation easier, we can square both sides:
\( A^2 = y^2 (2ry - y^2) = 2ry^3 - y^4 \).
Let \( Z = A^2 = 2ry^3 - y^4 \). We want to find the maximum area, so we differentiate \( Z \) with respect to \( y \) and set it to zero:
\( \frac{dZ}{dy} = \frac{d}{dy} (2ry^3 - y^4) = 6ry^2 - 4y^3 \).
To find the maximum, set \( \frac{dZ}{dy} = 0 \):
\( 6ry^2 - 4y^3 = 0 \)
\( 2y^2 (3r - 2y) = 0 \)
Since \( y \neq 0 \) (height cannot be zero), we must have \( 3r - 2y = 0 \).
\( \implies 3r = 2y \)
\( \implies y = \frac{3}{2}r \).
Now, we need to check if this value gives a maximum by finding the second derivative of \( Z \) with respect to \( y \):
\( \frac{d^2Z}{dy^2} = \frac{d}{dy} (6ry^2 - 4y^3) = 12ry - 12y^2 \).
Substitute \( y = \frac{3}{2}r \) into the second derivative:
\( \frac{d^2Z}{dy^2} = 12r \left(\frac{3}{2}r\right) - 12 \left(\frac{3}{2}r\right)^2 \)
\( = 18r^2 - 12 \left(\frac{9}{4}r^2\right) \)
\( = 18r^2 - 27r^2 = -9r^2 \).
Since \( \frac{d^2Z}{dy^2} = -9r^2 < 0 \), the area is maximum when \( y = \frac{3}{2}r \).
Now we find \( x \) using \( x^2 = 2ry - y^2 \):
\( x^2 = 2r \left(\frac{3}{2}r\right) - \left(\frac{3}{2}r\right)^2 \)
\( x^2 = 3r^2 - \frac{9}{4}r^2 = \frac{12r^2 - 9r^2}{4} = \frac{3r^2}{4} \).
\( \implies x = \sqrt{\frac{3r^2}{4}} = \frac{\sqrt{3}r}{2} \).
In the right-angled triangle \( \Delta ABD \), we can find the angle \( A \). The side \( BD = y = \frac{3}{2}r \) and \( AD = x = \frac{\sqrt{3}r}{2} \).
\( \tan A = \frac{BD}{AD} = \frac{y}{x} = \frac{\frac{3}{2}r}{\frac{\sqrt{3}}{2}r} = \frac{3}{\sqrt{3}} = \sqrt{3} \).
\( \implies \tan A = \tan 60^\circ \). So, \( A = 60^\circ \).
Similarly, in \( \Delta CBD \), \( \tan C = \frac{BD}{DC} = \frac{y}{x} = \sqrt{3} \). So, \( C = 60^\circ \).
If two angles of the triangle are \( 60^\circ \), then the third angle \( B \) must also be \( 180^\circ - (60^\circ + 60^\circ) = 60^\circ \).
Therefore, \( \angle A = \angle B = \angle C = 60^\circ \), which means the triangle is equilateral. This shows that an equilateral triangle has the maximum area when inscribed in a circle. This geometric property is fundamental in optimizing inscribed shapes.
In simple words: To find the largest possible triangle that fits inside a circle, we used math to figure out its height and base. It turns out that when the triangle has all three angles equal to 60 degrees, making it an equilateral triangle, its area is the biggest.

🎯 Exam Tip: When maximizing areas or volumes in geometry, always express the quantity to be maximized as a function of one variable, then use calculus (first and second derivatives) to find the critical points and confirm if it's a maximum.

 

Question 2. Given the sum of the perimeters of a square and a circle, then prove that the sum of their area is least when one side of the square is equal to diameter of the circle.
Answer: Let's say the side of the square is \( x \) and the radius of the circle is \( r \).
The perimeter of the square is \( 4x \).
The perimeter of the circle (circumference) is \( 2\pi r \).
The problem states that the sum of their perimeters is a constant, let's call it \( k \):
\( 4x + 2\pi r = k \). This is our first equation (i).
Now, let's look at the areas:
Area of the square is \( A_s = x^2 \).
Area of the circle is \( A_c = \pi r^2 \).
The total sum of areas is \( A = \pi r^2 + x^2 \).
We need to minimize this total area \( A \). From equation (i), we can express \( x \) in terms of \( r \) and \( k \):
\( 4x = k - 2\pi r \)
\( x = \frac{k - 2\pi r}{4} \).
Now, substitute this value of \( x \) into the total area equation:
\( A = \pi r^2 + \left(\frac{k - 2\pi r}{4}\right)^2 \)
\( A = \pi r^2 + \frac{(k - 2\pi r)^2}{16} \).
To simplify differentiation, let's keep the denominator constant:
\( A = \pi r^2 + \frac{k^2 - 4k\pi r + 4\pi^2 r^2}{16} \).
Now, differentiate \( A \) with respect to \( r \) to find the minimum area:
\( \frac{dA}{dr} = 2\pi r + \frac{1}{16} (-4k\pi + 8\pi^2 r) \).
Set \( \frac{dA}{dr} = 0 \) for minimum area:
\( 2\pi r + \frac{1}{16} (-4k\pi + 8\pi^2 r) = 0 \)
Multiply by 16 to clear the fraction:
\( 32\pi r - 4k\pi + 8\pi^2 r = 0 \)
Divide by \( 4\pi \) (since \( 4\pi \neq 0 \)):
\( 8r - k + 2\pi r = 0 \)
\( 8r + 2\pi r = k \)
\( r(8 + 2\pi) = k \)
\( r = \frac{k}{8 + 2\pi} \).
Now, let's substitute this value of \( k \) back into the expression for \( x \) from equation (i):
\( 4x = k - 2\pi r \)
\( 4x = r(8 + 2\pi) - 2\pi r \)
\( 4x = 8r + 2\pi r - 2\pi r \)
\( 4x = 8r \)
\( \implies x = 2r \).
This means that the side of the square \( x \) is equal to \( 2r \), which is the diameter of the circle. This condition ensures that the sum of their areas is minimized. This principle highlights how different geometric shapes can interact in optimization problems.
To confirm it's a minimum, we find the second derivative of \( A \):
\( \frac{d^2A}{dr^2} = \frac{d}{dr} \left(2\pi r - \frac{4k\pi}{16} + \frac{8\pi^2 r}{16}\right) \)
\( \frac{d^2A}{dr^2} = 2\pi + \frac{8\pi^2}{16} = 2\pi + \frac{\pi^2}{2} \).
Since \( \pi \) is positive, \( 2\pi + \frac{\pi^2}{2} \) is always positive, so \( \frac{d^2A}{dr^2} > 0 \), confirming that this is indeed a minimum.
In simple words: Imagine you have a fixed length of wire to make a square and a circle. To make the combined area of the square and circle as small as possible, you should make the side of the square exactly equal to the distance across the circle (its diameter).

🎯 Exam Tip: For optimization problems, remember to express the quantity to be optimized as a function of a single variable. Use the first derivative test to find critical points and the second derivative test to confirm if it's a maximum or minimum.

 

Question 3. Show that the cone of the greatest volume which can be inscribed in a given sphere has an altitude equal to \( \frac {2}{3} \) of the diameter of the sphere.
Answer: Let's consider a sphere with radius \( R \). Inside this sphere, we inscribe a cone. Let the radius of the cone be \( r \) and its height be \( h \). If we place the sphere's center at the origin, the diameter of the sphere and the axis of the cone will line up for maximum height. Let \( x \) be the distance from the center of the sphere to the base of the cone. Then, the height of the cone will be \( h = R + x \). The radius of the cone \( r \) can be found using the Pythagorean theorem: \( r^2 = R^2 - x^2 \).
The volume of a cone is given by the formula \( V = \frac{1}{3} \pi r^2 h \).
Substitute \( r^2 = R^2 - x^2 \) and \( h = R + x \) into the volume formula:
\( V = \frac{1}{3} \pi (R^2 - x^2)(R + x) \).
\( V = \frac{1}{3} \pi (R^3 + R^2x - Rx^2 - x^3) \).
To find the maximum volume, we differentiate \( V \) with respect to \( x \) and set the derivative to zero:
\( \frac{dV}{dx} = \frac{1}{3} \pi (R^2 - 2Rx - 3x^2) \).
Set \( \frac{dV}{dx} = 0 \):
\( \frac{1}{3} \pi (R^2 - 2Rx - 3x^2) = 0 \)
\( \implies R^2 - 2Rx - 3x^2 = 0 \).
We can factor this quadratic equation: \( (R - 3x)(R + x) = 0 \).
This gives two possible values for \( x \): \( R - 3x = 0 \implies x = \frac{R}{3} \) or \( R + x = 0 \implies x = -R \).
Since \( x \) represents a distance, it cannot be negative. So, we take \( x = \frac{R}{3} \).
Now, we calculate the height of the cone \( h = R + x \):
\( h = R + \frac{R}{3} = \frac{3R + R}{3} = \frac{4R}{3} \).
The diameter of the sphere is \( D = 2R \). We need to show that \( h = \frac{2}{3}D \).
\( h = \frac{4R}{3} = \frac{2}{3} (2R) = \frac{2}{3}D \).
Thus, the altitude (height) of the cone with the greatest volume is \( \frac{2}{3} \) of the sphere's diameter. This shows a specific ratio between the dimensions of the sphere and the inscribed cone for maximum volume.
To ensure this is a maximum, we find the second derivative:
\( \frac{d^2V}{dx^2} = \frac{d}{dx} \left( \frac{1}{3} \pi (R^2 - 2Rx - 3x^2) \right) \)
\( = \frac{1}{3} \pi (-2R - 6x) \).
Substitute \( x = \frac{R}{3} \):
\( \frac{d^2V}{dx^2} = \frac{1}{3} \pi \left(-2R - 6\left(\frac{R}{3}\right)\right) \)
\( = \frac{1}{3} \pi (-2R - 2R) = \frac{1}{3} \pi (-4R) = -\frac{4\pi R}{3} \).
Since \( R \) is a positive radius, \( -\frac{4\pi R}{3} \) is negative, confirming that the volume is maximum at \( x = \frac{R}{3} \).
In simple words: To get the biggest possible cone that fits inside a ball, its height should be exactly two-thirds of the ball's diameter. This is the special height that gives the cone the most space inside the ball.

🎯 Exam Tip: Remember that for an inscribed figure, its dimensions are related to the dimensions of the larger figure. Use geometry (like the Pythagorean theorem) to establish these relationships, then use calculus for optimization.

 

Question 5. If sum of hypotenuse and one side of right angled triangle is given, then area of triangle is maximum if angle between these sides is \( \frac {\pi }{3} \).
Answer: Let's consider a right-angled triangle. Let the hypotenuse be \( h \) and one side (base) be \( x \). The problem states that the sum of the hypotenuse and this side is a constant, let's call it \( k \):
\( h + x = k \). This means \( h = k - x \). (i)
For a right-angled triangle, if one side is \( x \) and the hypotenuse is \( h \), the other side (height) would be \( \sqrt{h^2 - x^2} \).
The area \( S \) of the triangle is \( S = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times x \times \sqrt{h^2 - x^2} \).
Substitute \( h = k - x \) into the area formula:
\( S = \frac{1}{2} x \sqrt{(k - x)^2 - x^2} \).
\( S = \frac{1}{2} x \sqrt{k^2 - 2kx + x^2 - x^2} \).
\( S = \frac{1}{2} x \sqrt{k^2 - 2kx} \).
To make differentiation easier, let's square the area:
\( S^2 = \frac{1}{4} x^2 (k^2 - 2kx) = \frac{1}{4} (k^2 x^2 - 2kx^3) \).
Let \( Z = S^2 = \frac{1}{4} (k^2 x^2 - 2kx^3) \). To find the maximum area, we differentiate \( Z \) with respect to \( x \) and set it to zero:
\( \frac{dZ}{dx} = \frac{1}{4} (2k^2 x - 6kx^2) \).
Set \( \frac{dZ}{dx} = 0 \):
\( \frac{1}{4} (2k^2 x - 6kx^2) = 0 \)
\( 2k^2 x - 6kx^2 = 0 \).
Factor out \( 2kx \):
\( 2kx (k - 3x) = 0 \).
Since \( x \neq 0 \) (a side cannot be zero) and \( k \neq 0 \) (sum of sides is not zero), we must have \( k - 3x = 0 \).
\( \implies k = 3x \).
Now, we relate this back to the initial condition \( h + x = k \):
\( h + x = 3x \)
\( h = 2x \).
So, the hypotenuse is twice the length of the base side. Let the angle between these sides (the base \( x \) and the adjacent height) be \( \theta \). In a right-angled triangle, the cosine of the angle \( \theta \) between the base \( x \) and the hypotenuse \( h \) is \( \frac{\text{base}}{\text{hypotenuse}} \).
\( \cos \theta = \frac{x}{h} = \frac{x}{2x} = \frac{1}{2} \).
The angle whose cosine is \( \frac{1}{2} \) is \( 60^\circ \) or \( \frac{\pi}{3} \) radians. This indicates that for maximum area, the angle between the given side and the hypotenuse should be \( \frac{\pi}{3} \). This relationship is a specific condition for maximizing area under a given perimeter constraint.
To confirm this is a maximum, we find the second derivative of \( Z \):
\( \frac{d^2Z}{dx^2} = \frac{d}{dx} \left( \frac{1}{4} (2k^2 x - 6kx^2) \right) \)
\( = \frac{1}{4} (2k^2 - 12kx) \).
Substitute \( k = 3x \):
\( \frac{d^2Z}{dx^2} = \frac{1}{4} (2(3x)^2 - 12(3x)x) \)
\( = \frac{1}{4} (18x^2 - 36x^2) = \frac{1}{4} (-18x^2) = -\frac{9}{2}x^2 \).
Since \( x \) is a side length, \( x^2 \) is positive, so \( -\frac{9}{2}x^2 < 0 \), confirming that it is a maximum.
In simple words: If you have a right-angled triangle where you know the total length of its longest side (hypotenuse) and one other side, the triangle will have the biggest possible area when the angle between those two sides is 60 degrees.

🎯 Exam Tip: When dealing with geometric optimization involving angles, trigonometric ratios can simplify the final step. Always remember to check the second derivative to confirm if the critical point yields a maximum or minimum.

 

Question 6. If a circle of radius 'a' is inside an equilateral triangle, then prove that minimum perimeter of triangle will be \( 6\sqrt {3}a \).
Answer: Let's assume \( \Delta ABC \) is an equilateral triangle. If a circle with radius \( a \) is inscribed inside this triangle, it is an incircle. For an equilateral triangle, the incenter (center of the incircle) is also the centroid and circumcenter.
Let \( O \) be the center of the inscribed circle, and \( D \) be the point where the circle touches side \( BC \). So, \( OD = a \). In \( \Delta OBD \), \( OD \) is perpendicular to \( BC \).
In an equilateral triangle, each angle is \( 60^\circ \). The line \( BO \) bisects \( \angle B \), so \( \angle OBD = 30^\circ \).
In right-angled \( \Delta OBD \):
\( \tan(\angle OBD) = \frac{OD}{BD} \)
\( \tan(30^\circ) = \frac{a}{BD} \)
\( \frac{1}{\sqrt{3}} = \frac{a}{BD} \)
\( \implies BD = a\sqrt{3} \).
Since \( D \) is the midpoint of \( BC \) in an equilateral triangle, \( BC = 2 \times BD = 2a\sqrt{3} \).
The perimeter of an equilateral triangle is \( 3 \times \text{side length} \).
Perimeter \( P = 3 \times BC = 3 \times (2a\sqrt{3}) = 6\sqrt{3}a \).
This calculation shows that for a circle of radius \( a \) inscribed in an equilateral triangle, the perimeter of the triangle is always \( 6\sqrt{3}a \). Since there's only one possible equilateral triangle for a given inradius, this value is automatically the minimum. The relationship between the inradius and the side length of an equilateral triangle is direct, leading to a fixed perimeter.
In simple words: If you have an equilateral triangle with a circle of radius 'a' perfectly fitting inside it, then the shortest possible total length of the triangle's sides (its perimeter) will always be \( 6\sqrt{3}a \). It's a fixed size for that specific circle.

🎯 Exam Tip: When dealing with inscribed circles in equilateral triangles, remember that the incenter is also the centroid. This means properties like angle bisection and median relationships can simplify calculations involving radius and side lengths.

 

Question 7. If a normal is drawn a point 'P' of ellipse \( \frac {x^2}{a^2} + \frac {y^2}{b^2} = 1 \), then Prove that the maximum distance from centre of ellipse will be \( a-b \).
Answer: Let's consider an ellipse with the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). A point \( P \) on this ellipse can be represented by coordinates \( (a\cos\theta, b\sin\theta) \). The equation of the normal to the ellipse at point \( P \) is \( ax\sec\theta - by\operatorname{cosec}\theta = a^2 - b^2 \).
We want to find the maximum distance of this normal from the center of the ellipse, which is the origin \( (0,0) \).
The formula for the perpendicular distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \) is \( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \).
In our case, the point is \( (0,0) \) and the line is \( (a\sec\theta)x - (b\operatorname{cosec}\theta)y - (a^2 - b^2) = 0 \).
So, the distance \( p \) from the center \( (0,0) \) to the normal is:
\( p = \frac{|(a\sec\theta)(0) - (b\operatorname{cosec}\theta)(0) - (a^2 - b^2)|}{\sqrt{(a\sec\theta)^2 + (-b\operatorname{cosec}\theta)^2}} \)
\( p = \frac{|-(a^2 - b^2)|}{\sqrt{a^2\sec^2\theta + b^2\operatorname{cosec}^2\theta}} \)
\( p = \frac{a^2 - b^2}{\sqrt{a^2\sec^2\theta + b^2\operatornamecosec^2\theta}} \).
To maximize \( p \), we need to minimize the denominator \( \sqrt{a^2\sec^2\theta + b^2\operatornamecosec^2\theta} \). This is equivalent to minimizing the expression inside the square root: \( y = a^2\sec^2\theta + b^2\operatornamecosec^2\theta \).
We can rewrite \( \sec^2\theta \) as \( 1 + \tan^2\theta \) and \( \operatorname{cosec}^2\theta \) as \( 1 + \cot^2\theta \).
So, \( y = a^2(1 + \tan^2\theta) + b^2(1 + \cot^2\theta) \).
\( y = a^2 + a^2\tan^2\theta + b^2 + b^2\cot^2\theta \).
\( y = (a^2 + b^2) + (a^2\tan^2\theta + b^2\cot^2\theta) \).
To minimize \( (a^2\tan^2\theta + b^2\cot^2\theta) \), we can use the AM-GM inequality, which states that \( \frac{A+B}{2} \ge \sqrt{AB} \). So, \( A+B \ge 2\sqrt{AB} \).
Let \( A = a^2\tan^2\theta \) and \( B = b^2\cot^2\theta \).
\( a^2\tan^2\theta + b^2\cot^2\theta \ge 2\sqrt{(a^2\tan^2\theta)(b^2\cot^2\theta)} \)
\( \ge 2\sqrt{a^2b^2\tan^2\theta\cot^2\theta} \)
\( \ge 2\sqrt{a^2b^2 \cdot 1} \)
\( \ge 2ab \).
The minimum value of \( a^2\tan^2\theta + b^2\cot^2\theta \) is \( 2ab \). This occurs when \( a^2\tan^2\theta = b^2\cot^2\theta \).
So, the minimum value of \( y \) is \( (a^2 + b^2) + 2ab = (a+b)^2 \).
Now, substitute this minimum value back into the expression for \( p \):
\( p_{\text{max}} = \frac{a^2 - b^2}{\sqrt{(a+b)^2}} \)
\( p_{\text{max}} = \frac{a^2 - b^2}{|a+b|} \).
Since \( a \) and \( b \) are lengths (radii), they are positive, so \( a+b > 0 \).
\( p_{\text{max}} = \frac{(a-b)(a+b)}{a+b} = a-b \).
Thus, the maximum distance of the normal from the center of the ellipse is \( a-b \). This implies that the furthest a normal can be from the center is directly related to the difference between the semi-major and semi-minor axes, illustrating a key property of the ellipse.

In simple words: For an ellipse, if you draw a line that is perpendicular to the curve at any point (this is called a "normal"), the furthest this line can be from the very center of the ellipse is found by subtracting the shorter radius (b) from the longer radius (a).

🎯 Exam Tip: Remember the equation for the normal to an ellipse and the formula for the perpendicular distance from a point to a line. Optimization problems often require minimizing the denominator of a fraction to maximize the entire expression.

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