Question

The sum of the surface areas of a cuboid with sides x, 2x and \[\frac{x}{3}\] and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of sphere. Also find the minimum value of  the sum of their volumes.

Answer 1

Surface area of cuboid = 2 (lb + bh + hl)

\[= \left( 2 x^2 + \frac{2 x^2}{3} + \frac{x^2}{3} \right)\]

\[ = 6 x^2\]

Let radius of the sphere be r,
Surface area of sphere = \[4\pi r^2\]

Therefore,

\[6 x^2 + 4\pi r^2 = k(\text { constant })\] .....(i)
Now, volume of both figures will be \[V = \frac{2}{3} x^3 + \frac{4}{3}\pi r^3\]

Putting the value of r from the equation (i), 

\[V = \frac{2}{3} x^3 + \frac{4}{3}\pi \left( \frac{k - 6 x^2}{4\pi} \right)^\frac{3}{2}\]

For minimum volume \[\frac{dV}{dx} = 0\], so

\[\frac{dV}{dx} = 2 x^2 + \left( \frac{4}{3}\pi \right) \left( \frac{1}{4\pi} \right)^\frac{3}{2} . \frac{3}{2} \left( k - 6 x^2 \right)^\frac{1}{2} \left( - 12x \right) = 0\]

\[ \Rightarrow 2 x^2 = \left( \frac{1}{4\pi} \right)^\frac{1}{2} \left( k - 6 x^2 \right)^\frac{1}{2} \left( 6x \right)\]

\[ \Rightarrow 2 x^2 = \left( \frac{1}{4\pi} \right)^\frac{1}{2} \left( 4\pi r^2 \right)^\frac{1}{2} \left( 6x \right) \left[ \text { since }, k - 6 x^2 = 4\pi r^2 \right]\]

\[ \Rightarrow x = 3r\]

Hence proved.
Further, minimum value of sum of their volume is

\[V_\min = \frac{2}{3} x^3 + \frac{4}{3}\pi r^3 \]

\[ = \frac{2}{3} x^3 + \frac{4}{3}\pi \left( \frac{x}{3} \right)^3 \left[ r = \frac{x}{3} \right]\]

\[ = \frac{2}{3} x^3 + \frac{4}{3}\pi\frac{x^3}{27} \]

\[ = \frac{2}{3} x^3 \left( 1 + \frac{2}{27} \right)\]

\[ = \frac{58}{81} x^3 \]