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# Surface Area Problem | TOMATO BStat Objective 725

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Surface area. You may use sequential hints.

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

But try the problem first…

Source

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)