A cylindrical container is to be made from certain solid material with the following constraints:
It has fixed inner volume $V$ mm${}^3$ ,has a $2$ mm thick solid wall and is open at the top.
The bottom of the container is a solid circular disc of thickness $2$ mm and is of radius equal to the outer radius of the container.
If the volume of the material used to make the container is minimum when the inner radius of the container is $10$ mm, then find the value of $\frac{V}{250\pi}$.
My Attempt The total volume is volume of solid wall and the solid circular disk. So we have Volume of material $ =\pi.h((r+2)^2-r^2)+\pi.r^2.2$ so as volume is minimized at $10$, I differentiated the expression put value of $r$ to get value of $h=-12$ which is impossible. I think I have misinterpreted the question. Thanks!
$V = \pi r^2 h\\ M = \pi ( (r+2)^2 - r^2) h + \pi (r+2)^2$
$\frac{dM}{dr} = \pi [4h + (4r+4)\frac {dh}{dr} + 2r+4] = 0\\ \frac{dV}{dr} = \pi [2r h + r^2 \frac {dh}{dr}] = 0\\ \frac {dh}{dr} = \frac {-2h}{r}\\ 2h + (2r+2)\frac {-2h}{r} + r + 2= 0\\ (r+2) = (2h - \frac {4h}{r})\\ r(r+2) = 2h (r-4)\\ h = \frac {r(r+2)}{2(r-4)}$
$r = 10\\ h = \frac {120}{12} = 10\\ V = 1000\pi\\ \frac V{250\pi} = 4$