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

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

Equation

Area and Volume

Algebra

But try the problem first...

Answer:h=d

Source

Suggested Reading

B.Stat Objective Problem 725

Challenges and Thrills of Pre-College Mathematics by University Press

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)

is required answer.

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