Getting stuck on a certain part of this question:
Q. A manufacturer wants to produce one litre (1000ml) cans from as small a quantity of metal sheeting as possible to reduce costs.
What I've done so far:
I've shown that the height of the cylinder is:
$h=\displaystyle\frac{1000}{\pi r^2}$
I've shown that the total surface area is:
$A = 2\pi r^2+\displaystyle\frac{2000}{r} $
Derivative:
$A' = 4\pi r - 2000r^{-2}$
Then found the zero or the x coordinate of the minimum value:
$r=\left(\displaystyle\frac{500}{\pi}\right)^{\frac{1}{3}}$
I am now trying to sub
$r=\left(\displaystyle\frac{500}{\pi}\right)^{\frac{1}{3}}$ into $A = 2\pi r^2+\displaystyle\frac{2000}{r} $ to find the minimum surface area of the 1000ml can. I'm just getting stuck with the algebra.
The answer I get is $A = 3000\left(\displaystyle\frac{500}{\pi}\right)^{\frac{-1}{3}}$ which is correct because I graphed it using a graphing calculator and both give $553.581$
The answer in the back of the book gives $A=6\pi\left(\displaystyle\frac{500}{\pi}\right)^{\frac{2}{3}} $
Can someone please show me the steps to get the answer in the back of the book when subbing $r$ into $A$
Your answers are both correct and identical. Since $-1/3=2/3-1$: $$A= 3000 \left(\frac{500}{\pi}\right)^{-1/3} = 3000 \left(\frac{500}{\pi}\right)^{-1}\left(\frac{500}{\pi}\right)^{2/3} = 3000 \left(\frac{\pi}{500}\right)\left(\frac{500}{\pi}\right)^{2/3} $$ $$=6\pi\left(\frac{500}{\pi}\right)^{2/3}$$