21) Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?

Answers (1)

Let r be the radius of base and h be the height of the cylinder
The volume of the cube (V) = \pi r^2 h
It is given that the volume of cylinder = 100 cm^3
\pi r^2 h = 100\Rightarrow h = \frac{100}{\pi r^2}
Surface area of cube(A) = 2\pi r(r+h)
A(r)= 2\pi r(r+\frac{100}{\pi r^2})
           = 2\pi r ( \frac{\pi r^3+100}{\pi r^2}) = \frac{2\pi r^3+200}{ r} = 2\pi r^2+\frac{200}{r}
A^{'}(r) = 4\pi r + \frac{(-200)}{r^2} \\ A^{'}(r)= 0\\ 4\pi r^3 = 200\\ r^3 = \frac{50}{\pi}\\ r = \left ( \frac{50}{\pi} \right )^{\frac{1}{3}} 
Hence, r = (\frac{50}{\pi})^\frac{1}{3} is the critical point
A^{''}(r) = 4\pi + \frac{400r}{r^3}\\ A^{''}\left ( (\frac{50}{\pi})^\frac{1}{3} \right )= 4\pi + \frac{400}{\left ( (\frac{50}{\pi})^\frac{1}{3} \right )^2} > 0
Hence, r = (\frac{50}{\pi})^\frac{1}{3} is the point of minima
h = \frac{100}{\pi r^2} = \frac{100}{\pi \left ( (\frac{50}{\pi})^\frac{1}{3} \right )^2} = 2.(\frac{50}{\pi})^\frac{1}{3}
Hence, r = (\frac{50}{\pi})^\frac{1}{3} and h = 2.(\frac{50}{\pi})^\frac{1}{3} are  the dimensions of the can which has the minimum surface area

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