PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# The Sum of the Surface Areas of a Sphere and a Cube is Given. Show that When the Sum of Their Volumes is Least, the Diameter of the Sphere is Equal to the Edge of the Cube. - PUC Karnataka Science Class 12 - Mathematics

#### Question

The sum of the surface areas of a sphere and a cube is given. Show that when the sum of their volumes is least, the diameter of the sphere is equal to the edge of the cube.

#### Solution

Let r be the radius of the sphere, x be the side of the cube and S be the sum of the surface area of both. Then, $S = 4\pi r^2 + 6 x^2$

$\Rightarrow$$x = \left( \frac{S - 4\pi r^2}{6} \right)^\frac{1}{2}$     ................(1)

Sum of volumes, V= $\frac{4}{3}\pi r^3 + x^3$

$\Rightarrow$ V = $\frac{4\pi r^3}{3} + \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{3}{2}$  [From eq. (1)]

$\Rightarrow \frac{dV}{dr} = 4\pi r^2 - 2\pi r \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{1}{2}$

For the minimum or maximum values of V, we must have $\frac{dV}{dr} = 0$       ..............(2)

$\Rightarrow 4\pi r^2 - 2\pi r \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{1}{2} = 0 \left[ \text { From eq } . \left( 2 \right) \right]$

$\Rightarrow 4\pi r^2 = 2\pi r \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{1}{2}$

$\Rightarrow 4\pi r^2 = 2\pi r x ..............\left[ \text { From eq }. \left( 1 \right) \right]$

$\Rightarrow x = 2r$

Now,

$\frac{d^2 V}{d r^2} = 8\pi r - 2\pi \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{1}{2} - \frac{2\pi r}{2} \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^{- \frac{1}{2}} \frac{\left( - 8\pi r \right)}{6}$

$\Rightarrow \frac{d^2 V}{d r^2} = 8\pi r - 2\pi \left[ \frac{\left( S - 4\pi r^2 \right)}{6} \right]^\frac{1}{2} + \frac{4}{3} \pi^2 r^2 \left[ \frac{6}{\left( S - 4\pi r^2 \right)} \right]^\frac{1}{2}$

$\Rightarrow \frac{d^2 V}{d r^2} = 8\pi r - 2\pi x + \frac{4}{3} \pi^2 r^2 \frac{1}{x} = 8\pi r - 4\pi r + \frac{2}{3} \pi^2 r$

$\Rightarrow \frac{d^2 V}{d r^2} = 4\pi r + \frac{2}{3} \pi^2 r > 0$

So, volume is minimum when x = 2r.

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Solution The Sum of the Surface Areas of a Sphere and a Cube is Given. Show that When the Sum of Their Volumes is Least, the Diameter of the Sphere is Equal to the Edge of the Cube. Concept: Graph of Maxima and Minima.
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