Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

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.

Accepted Answer

We have to prove Diameter of sphere $=$ Edge of cube

Let radius of sphere be $r$ and edge of cube be $x$

Surface area of cube $+$ Surface area of sphere $=$ Constant

$∴ 6 x^{2} + 4 π r^{2} = S$ $. . . (1)$

$x = \sqrt{\frac{S - 6 x^{2}}{4 π}}$

Now, total volume, $V = x^{3} + \frac{4}{3} π r^{3}$

$\Rightarrow V = x^{3} + \frac{4}{3} π {(\frac{S - 6 x^{2}}{4 π})}^{\frac{3}{2}}$

$V = x^{3} + \frac{1}{3 \sqrt{4 π}} {(S - 6 x^{2})}^{\frac{3}{2}}$ ... $(2)$

Differentiating it w.r.t. $x$ , we get

$\frac{d V}{d x} = 3 x^{2} + \frac{1}{3 \sqrt{4 π}} \frac{3}{2} \sqrt{S - 6 x^{2}} \times (- 12 x)$

$\frac{d V}{d x} = 3 x^{2} - (\frac{3 x}{\sqrt{π}}) \sqrt{S - 6 x^{2}}$ ... $(3)$

For $V$ to be minimum or maximum, $\frac{d V}{d x} = 0$

$\Rightarrow 3 x^{2} - (\frac{3 x}{\sqrt{π}}) \sqrt{S - 6 x^{2}} = 0$

$∴ x = \frac{1}{\sqrt{π}} \sqrt{S - 6 x^{2}}$

Squaring them we get,

$\Rightarrow x^{2} = \frac{1}{π} (S - 6 x^{2})$

$\Rightarrow x^{2} = \frac{S}{x + 6}$

$∴ x = \sqrt{\frac{S}{π + 6}}$

Again differentiating equation $(3)$ we get,

$\Rightarrow \frac{d^{2} V}{d x^{2}} = 6 x - \frac{3}{\sqrt{π}} \{\sqrt{S - 6 x^{2}} + x (\frac{1 \times (- 12 x)}{2 \sqrt{S - 6 x^{2}}})\}$

$\Rightarrow \frac{d^{2} V}{d x^{2}} = 6 x - \frac{3}{\sqrt{π}} \{\frac{S - 12 x^{2}}{\sqrt{S - 6 x^{2}}}\}$

At $x = \sqrt{\frac{S}{π + 6}}$

$\Rightarrow \frac{d^{2} V}{d x^{2}} = 6 \sqrt{\frac{S}{π + 6}} - \frac{3}{\sqrt{π}} \{\frac{S - \frac{12 S}{π + 6}}{\sqrt{S - \frac{6 S}{π + 6}}}\}$

$\Rightarrow \frac{d^{2} V}{d x^{2}} = 6 \sqrt{\frac{S}{π + 6}} - \frac{3}{\sqrt{π}} \{\frac{S π - 6 S}{\sqrt{S π} \times \sqrt{π + 6}}\}$

$\Rightarrow \frac{d^{2} V}{d x^{2}} = \frac{6 \sqrt{S}}{\sqrt{π + 6}} - \frac{3 \sqrt{S}}{\sqrt{π + 6}} + \frac{18 \sqrt{S}}{π \sqrt{π + 6}}$

$\Rightarrow \frac{d^{2} V}{d x^{2}} = \frac{3 \sqrt{S}}{\sqrt{π + 6}} + \frac{18 \sqrt{S}}{π \sqrt{π + 6}}$

$∵ \frac{d^{2} V}{d x^{2}}$ is $+ v e$

$∴ V$ is minimum at $x = \sqrt{\frac{S}{π + 6}}$

Using it in equation $(1)$

$\Rightarrow 6 x^{2} + 4 π r^{2} = x^{2} (π + 6)$

$\Rightarrow 4 π r^{2} = x^{2} π$

$∴ x = 2 r$

$∵$ Diameter of sphere $d = 2 r$

$∴ x = d$

Hence, it is proved that the diameter of sphere is equal to the edge of cube.

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.

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