A container holding a fixed volume is made in the shape of a cylinder with a hemispherical top. The hemispherical top has the same radius as the cylinder. Find the ratio of height to radius of the cylinder which minimizes the cost of the container if
- the cost per unit area of the top is twice as great as the cost per unit area of the side, and the container is made with no bottom;
- the cost per unit area of the top is twice as great as the cost per unit area of the side, and the container is made with a circular bottom, for which the cost per unit area is 1.5 times the cost per unit area of the side.
Let radius of cylinder = radius of hemisphere = r
Let height of cylinder = h
Volume of container $= V = \pi r^2 h + \dfrac{2}{3}\pi r^3$
i.e. $V = \pi r^2 \left(h+\dfrac{2}{3}r\right)$
i.e. $h+\dfrac{2}{3}r = \dfrac{V}{\pi r^2}$
$\implies$ $h = \dfrac{V}{\pi r^2} - \dfrac{2}{3}r$ $\;\; \cdots$ (1)
-
Let cost per unit area of the side $= c$
Then, cost per unit area of the hemispherical top $=2c$
Area of the side of the container $=2\pi r h$
Area of the top of the container $= 2 \pi r^2$
$\therefore$ Cost of the container $= C = 2\pi r h c + 4 \pi r^2 c$ $\;\; \cdots$ (2)
Substituting the value of h from equation (1) in equation (2) gives
$C = 2 \pi r c \left(\dfrac{V}{\pi r^2} - \dfrac{2}{3}r\right) + 4 \pi r^2 c$
i.e. $C = \dfrac{2cV}{r} - \dfrac{4}{3}\pi r^2 c + 4\pi r^2 c$
i.e. $C = 2cV \times \dfrac{1}{r} + \dfrac{8\pi c}{3} \times r^2$ $\;\; \cdots$ (3)
For minimum cost, $\dfrac{dC}{dr}=0$
$\therefore$ From equation (3), $\dfrac{dC}{dr} = \dfrac{-2cV}{r^2}+\dfrac{16 \pi c r}{3} = 0$
i.e. $\dfrac{16 \pi c r}{3} = \dfrac{2cV}{r^2}$
i.e. $r^3 = \dfrac{3V}{8\pi}$
$\implies$ $\dfrac{V}{\pi r^3} = \dfrac{8}{3}$ $\;\; \cdots$ (4)
Now, from equation (1),
$\dfrac{h}{r} = \dfrac{V}{\pi r^3}- \dfrac{2}{3}$ $\;\; \cdots$ (5)
In view of equation (4), equation (5) can be written as
$\dfrac{h}{r}= \dfrac{8}{3}-\dfrac{2}{3} = 2$
$\therefore$ Ratio of height to radius of cylinder which minimizes the cost $= \dfrac{h}{r}=2:1$
-
Cost per unit area of the circular bottom of the container $= 1.5c$
Area of the bottom of the container $= \pi r^2$
$\therefore$ Cost of the container $=C = 4 \pi r^2 c + 2 \pi r h c + 1.5 \pi r^2 c$
i.e. $C = 2\pi rhc + 5.5 \pi r^2 c$ $\;\; \cdots$ (6)
Substituting the value of h from equation (1) in equation (6) gives
$C= 2 \pi r c \left(\dfrac{V}{\pi r^2}-\dfrac{2}{3}r\right) + 5.5 \pi r^2 c$
i.e. $C = \dfrac{2cV}{r} - \dfrac{4}{3}\pi r^2 c + \dfrac{11}{2}\pi r^2 c$
i.e. $C = 2cV \times\dfrac{1}{r} + \dfrac{25\pi c}{6} \times r^2$ $\;\; \cdots$ (7)
For minimum cost $\dfrac{dC}{dr} = 0$
$\therefore$ We have from equation (7) for minimum cost
$\dfrac{dC}{dr} = \dfrac{-2cV}{r^2} + \dfrac{50 \pi c r}{6} = 0$
i.e. $\dfrac{2cV}{r^2} = \dfrac{25 \pi c r}{3}$
$\implies$ $\dfrac{V}{\pi r^3} = \dfrac{25}{6}$ $\;\; \cdots$ (8)
Substituting equation (8) in equation (5) gives
$\dfrac{h}{r} = \dfrac{25}{6}- \dfrac{2}{3} = \dfrac{7}{2}$
$\therefore$ Ratio of height to radius of cylinder which minimizes the cost $= \dfrac{h}{r}=7:2$