Question

Right circular cylindrical tin cans are to be manufactured to contain a given volume .The end pieces of the can are cut out from square pieces of tin with the corners wasted. If h is the height of the can and r its radius a)write down the least area of tin required (including the waste) in the making of each can. b)determine the ratio of the height to diameter for the most economical production.

Answer 1

\[
\text{a) Volume }V = \pi r^2 h \\
h = \frac{V}{\pi r^2} ~~~ \text{ ---- (1) } \\
\text{Total Surface Area A with two square end pieces }A = 2\pi rh + 2*2r*2r\\
A = 2\pi rh + 8r^2\\
\text{Substitute h from (1):} \\
A = 2\pi r \frac{V}{\pi r^2} + 8r^2\\
A = \frac{2V}{r} + 8r^2 ~~~ \text{ ---- (2) } \\\\
\text{To minimize A, set dA/dr = 0 and solve for r:}\\
\frac{dA}{dr} = -\frac{2V}{r^2} + 16r = 0 \\
16r = \frac{2V}{r^2} \\
r^3 = \frac{V}{8} = \frac{V}{2^3}\\
r = \frac{\sqrt[3]{V}}{2} \\
\text{From (2): }A = \frac{2V}{r} + 8r^2 = \frac{2V+8r^3}{r} =
\frac{2V+ V }{\frac{\sqrt[3]{V}}{2}} = \frac{6V}{V^{1/3}} = 6V^{2/3} = 6\sqrt[3]{V^2}\\
\text{ } \\
\text{b) height / diameter } = \frac{h}{2r} \\
\text{From (1): } \frac{h}{2r} = \frac{V}{\pi r^2 * 2r} =
\frac{V}{2\pi r^3} = \frac{V}{2\pi} * \frac 8V = \frac{4}{\pi}\\
\]