Question

You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the​ side, but the top and bottom of radius r will be cut from squares that measure 4r units on a side. The total amount of aluminum used up by the can will therefore be Upper A equals 32 r squared plus 2 pi rh​, where h is the height of the right circular cylindrical can. What is the ratio of h to r for the most economical​ can?

Answer 1

Explanation:

It is given that,

The volume of a right circular cylindrical, $V=108\ cm^3$

We know that the volume of the cylinder is given by :

$V=\pi r^2 h$

$108=\pi r^2 h$    

$h=\dfrac{108}{\pi r^2}$............(1)

The upper area is given by :

$A=32r^2+2\pi rh$

$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

$r^3=\dfrac{216}{64}$

r = 1.83 m

Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

$\dfrac{h}{r}=\dfrac{108}{\pi 1.83}$

$\dfrac{h}{r}=59 \pi$

Hence, this is the required solution.