CUET UG 2025 — Mathematics Calculus

For a closed cylinder with radius $r$r and height $h$h:

Surface Area: $S = 2\pi r^2 + 2\pi rh$S=2πr2+2πrh

Volume: $V = \pi r^2 h$V=πr2h

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Since the surface area $S$S is fixed:

$$S = 2\pi r^2 + 2\pi rh$$S=2πr2+2πrh

$$2\pi rh = S - 2\pi r^2$$2πrh=S−2πr2

$$h = \frac{S - 2\pi r^2}{2\pi r}$$h=2πrS−2πr2​

$$h = \frac{S}{2\pi r} - r$$h=2πrS​−r

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Substituting into the volume formula:

$$V = \pi r^2 h$$V=πr2h

$$V = \pi r^2 \left(\frac{S}{2\pi r} - r\right)$$V=πr2(2πrS​−r)

$$V = \frac{Sr}{2} - \pi r^3$$V=2Sr​−πr3

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For maximum volume, $\frac{dV}{dr} = 0$drdV​=0:

$$\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2$$drdV​=2S​−3πr2

$$\frac{S}{2} - 3\pi r^2 = 0$$2S​−3πr2=0

$$S = 6\pi r^2$$S=6πr2

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Substituting $S = 6\pi r^2$S=6πr2 into the surface area equation:

$$6\pi r^2 = 2\pi r^2 + 2\pi rh$$6πr2=2πr2+2πrh

$$4\pi r^2 = 2\pi rh$$4πr2=2πrh

$$h = \frac{4\pi r^2}{2\pi r}$$h=2πr4πr2​

$$h = 2r$$h=2r

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Therefore, the volume of the cylinder is maximum when its height is $2r$2r.