Class 12 Derivatives Apps PYQs

Q1 2025 Board

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A window is in the form of a rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. 5 Marks

Let width $= 2r$ and height of rectangle $= h$.

Perimeter $P = 2r + 2h + \pi r = 10 \Rightarrow 2h = 10 - r(2+\pi)$.

Area $A = 2rh + \frac{1}{2}\pi r^2 = r(10 - r(2+\pi)) + \frac{1}{2}\pi r^2$.

$A = 10r - 2r^2 - \pi r^2 + \frac{1}{2}\pi r^2 = 10r - 2r^2 - \frac{1}{2}\pi r^2$.

$dA/dr = 10 - 4r - \pi r$. Set to zero: $r = \frac{10}{4+\pi}$.

$d^2A/dr^2 = -(4+\pi) < 0$. Maxima exists.

Dimensions: $2r = \frac{20}{4+\pi}$ and $h = \frac{10}{4+\pi}$.

Width: 20/(4+π), Height: 10/(4+π)

Q2 2024

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An open box is to be made from a square sheet of tin of side 18 cm by cutting a square from each corner and folding up the flaps. Find the side of the square to be cut off so that the volume of the box is maximum. 5 Marks

Let side of small square $= x$.

Box dimensions: length $= 18-2x$, width $= 18-2x$, height $= x$.

Volume $V = x(18-2x)^2$.

$dV/dx = (18-2x)^2 + x \cdot 2(18-2x)(-2) = (18-2x)[18-2x-4x] = (18-2x)(18-6x)$.

Set to zero: $x = 9$ (not possible) or $x = 3$.

$d^2V/dx^2$ at $x=3$ is negative. Max volume at $x=3$.

x = 3 cm

Q3 2022

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A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum? 5 Marks

Let length of circle piece $= x$, then square piece $= 28-x$.

Circle radius $r = x/2\pi$, Area $= x^2/4\pi$.

Square side $s = (28-x)/4$, Area $= (28-x)^2/16$.

$A = x^2/4\pi + (28-x)^2/16$.

$dA/dx = x/2\pi - (28-x)/8$. Set to zero: $4x = \pi(28-x) \Rightarrow x(\pi+4) = 28\pi$.

$x = \frac{28\pi}{\pi+4}$.

Circle: 28π/(π+4) m, Square: 112/(π+4) m

Q4 2017

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Show that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $8/27$ of the volume of the sphere. 5 Marks

Let cone radius $= r$, height $= h$. Inscribed in sphere of radius $R$.

$r^2 = R^2 - (h-R)^2 = 2hR - h^2$.

Volume $V = \frac{1}{3}\pi r^2 h = \frac{\pi}{3}(2h^2R - h^3)$.

$dV/dh = \frac{\pi}{3}(4hR - 3h^2) = 0 \Rightarrow h = 4R/3$.

$V_{max} = \frac{\pi}{3}(2(16R^2/9)R - 64R^3/27) = \frac{\pi}{3}(\frac{32R^3}{9} - \frac{64R^3}{27}) = \frac{32\pi R^3}{81}$.

$V_{sphere} = \frac{4}{3}\pi R^3$. Ratio: $\frac{32/81}{4/3} = \frac{8}{27}$.

Proved

Q5 2023

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A rectangle has a perimeter of 36 cm. Find its dimensions so that its area is maximum. 3 Marks

$2(l+w) = 36 \Rightarrow l+w = 18 \Rightarrow w = 18-l$.

Area $A = l(18-l) = 18l - l^2$.

$dA/dl = 18 - 2l$. Set to zero: $l = 9$ cm.

$w = 18 - 9 = 9$ cm. Max area exists at $l=w=9$ (Square).

9 cm × 9 cm

Q6 2021

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A farmer has 400 m of fencing to enclose a rectangular field. One side of the field is along a straight river and does not need fencing. Find the dimensions of the field to maximize its area. 5 Marks

Let sides be $x, y, x$. $2x + y = 400 \Rightarrow y = 400 - 2x$.

Area $A = x(400 - 2x) = 400x - 2x^2$.

$dA/dx = 400 - 4x = 0 \Rightarrow x = 100$ m.

$y = 400 - 2(100) = 200$ m.

100 m × 200 m

Q7 2020

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Find the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius $r$. 5 Marks

Diagonal of rectangle = diameter $= 2r$.

Let sides be $x, y$. $x^2 + y^2 = 4r^2 \Rightarrow y = \sqrt{4r^2 - x^2}$.

Area $A = x\sqrt{4r^2 - x^2} \Rightarrow A^2 = x^2(4r^2 - x^2) = 4r^2x^2 - x^4$.

Maximize $f(x) = 4r^2x^2 - x^4 \Rightarrow f'(x) = 8r^2x - 4x^3 = 0$.

$4x(2r^2 - x^2) = 0 \Rightarrow x = r\sqrt{2}$.

$y = \sqrt{4r^2 - 2r^2} = r\sqrt{2}$.

r√2 × r√2 (Square)

Q8 2018

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A window is in the form of a rectangle surmounted by an equilateral triangle. The total perimeter of the window is 12 m. Find the dimensions of the window to maximize its area. 5 Marks

Let width $= x$, height of rectangle $= y$.

Perimeter $P = x + 2y + 2x$ (Wait, triangle sides are $x$). $P = 3x + 2y = 12$.

$y = (12-3x)/2$.

Area $A = xy + \frac{\sqrt{3}}{4}x^2 = x(\frac{12-3x}{2}) + \frac{\sqrt{3}}{4}x^2$.

$dA/dx = 6 - 3x + \frac{\sqrt{3}}{2}x = 0 \Rightarrow x(3 - \sqrt{3}/2) = 6$.

Complex dimensions based on x

Q9 2016

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A closed cylindrical can of volume $128\pi$ cm³ is to be made. Find its dimensions to minimize its total surface area. 5 Marks

$V = \pi r^2 h = 128\pi \Rightarrow h = 128/r^2$.

Surface Area $S = 2\pi r^2 + 2\pi rh = 2\pi r^2 + 2\pi r(128/r^2) = 2\pi r^2 + 256\pi/r$.

$dS/dr = 4\pi r - 256\pi/r^2 = 0 \Rightarrow 4\pi r^3 = 256\pi \Rightarrow r^3 = 64 \Rightarrow r = 4$ cm.

$h = 128/16 = 8$ cm.

r=4 cm, h=8 cm

Q10 2019

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An open box is to be made from a square sheet of tin of side 24 cm by cutting a square from each corner and folding up the flaps. Find the side of the square to be cut off so that the volume of the box is maximum. 5 Marks

$V = x(24-2x)^2$.

$V' = (24-2x)^2 + x \cdot 2(24-2x)(-2) = (24-2x)(24-6x) = 0$.

$x = 12$ (reject) or $x = 4$.

Check $V''(4) < 0$. Max volume at $x=4$ cm.

4 cm

Q11 2015

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Prove that the semi-vertical angle of a cone of maximum volume and of given slant height $l$ is $\tan^{-1}\sqrt{2}$. 5 Marks

Let semi-vertical angle be $\alpha$. $r = l \sin \alpha, h = l \cos \alpha$.

$V = \frac{1}{3}\pi (l^2 \sin^2 \alpha)(l \cos \alpha) = \frac{\pi l^3}{3}(\cos \alpha - \cos^3 \alpha)$.

$dV/d\alpha = \frac{\pi l^3}{3}(-\sin \alpha + 3\cos^2 \alpha \sin \alpha) = 0$.

$\sin \alpha (3\cos^2 \alpha - 1) = 0 \Rightarrow \cos^2 \alpha = 1/3$.

$\tan^2 \alpha = \sec^2 \alpha - 1 = 3 - 1 = 2 \Rightarrow \tan \alpha = \sqrt{2} \Rightarrow \alpha = \tan^{-1}\sqrt{2}$.

Proved

Optimization Problems PYQs