A cylindrical can is to hold $4 \pi$ volume orange juice. The cost of constructing per inch of the metal top and bottom is twice the cost per square inch of constructing the cardboard side. We need to find the dimensions of the least expensive can.
What I tried :
Let $x$ and $y$ denote the radius and the height of the cylindrical can. Given its volume : $\pi x^{2}y = 4 \pi$ => $x^{2}y=4$.
Now let $a$ be the per unit cost for constructing top and bottom , and $b$ for the remaining body..
So , I wrote the total cost function as follows :
$C(a,b) = (2\pi x^{2})a + (2\pi xy)b$ , also , according to the question , $a=2b$,
So , I applied Lagrange multiplier technique on the cost function with the constraint $g(a,b)=a-2b$ as follows :
$C_{a}= \lambda g_{a}$ and $C_{b}= \lambda g_{b}$ ,
=>$2 \pi x^{2} = \lambda$ and $2 \pi xy = -2\lambda$ ,
Clearly we can't conclude anything from this. Could anyone tell what am I doing wrong ?
Since the volume is fixed, you can express the height as a function of the radius, $y = 4/x^2$. The costs $a$ and $b$ are also fixed, so you can express the cost of can as a function of $x$. In particular, $$ C(x) = 4 \pi b x^2 + \frac{8 \pi b}{x}, $$ so $$C'(x) = 8 \pi b \left( x - \frac{1}{x^2} \right). $$ Solving $C'(x) = 0$ gives $ x = 1$. The derivative is negative for values less than one and positive for values larger than one, so the critical point corresponds to minimum. Hence the dimensions that minimize the cost are $x = 1$ and $y = 4$.