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# Surface Area Problem | TOMATO BStat Objective 725

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Surface Area.

## Surface Area Problem (B.Stat Objective Question )

A right circular cylindrical container closed on both sides is to contain a fixed volume of motor oil. Suppose its base has diameter d and height is h. The overall surface area of the container is minimum when

• h=$\frac{4d\pi}{3}$
• h=d
• h=2d
• conditions other than the foregoing are satisfied

### Key Concepts

Equation

Area and Volume

Algebra

B.Stat Objective Problem 725

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

V=$\frac{d^{2}h\pi}{4}$

or,$\frac{4V}{h\pi}=d^{2}$

or, $d=\sqrt{\frac{4V}{h\pi}}$ positive value taken

S=$\frac{2d^{2}\pi}{4}+{\pi}dh=(\frac{2\pi}{4})(\frac{4V}{h\pi})+{h\pi\sqrt{\frac{4V}{h\pi}}}$

=$\frac{2V}{h}+(\sqrt{4V\pi})(\sqrt{h})$

Second Hint

for minimum surface area

$\frac{dS}{dh}$=$\frac{-2V}{h^{2}}$+$\frac{\sqrt{4V\pi}}{2\sqrt{h}}$=0

or, $\frac{2V}{h^{2}}$=$\frac{\sqrt{4V\pi}}{2\sqrt{h}}$

or, $h^{\frac{3}{2}}=2\sqrt{\frac{V}{\pi}}$

or, $h^{3}=\frac{4V}{\pi}$

Final Step

or,$h^{3}=\frac{4V}{\pi}$ where $V=\frac{hd^{2}\pi}{4}$

or,$h^{3}=\frac{4d^{2}h\pi}{4\pi}=d^{2}h$

or, $h^{2}=d^{2}$

or, h=d (since h,d both positive)