
Optimization
Hello!
Here's a question that I couldn't understand;
A canvas tent is to be constructed in the shape of a rightcircular cone with the ground as base;
Using the volume V and curved surface area S of the cone,
$\displaystyle V = \frac{1}{3}\pi r^2 h$ and $\displaystyle S = \pi rl$Find the dimensions of the cone that maximises the volume for the $\displaystyle 4 \sqrt{3 \pi}$ mē of canvas material, and find this maximum volume.
My attempt:know the two formulas: $\displaystyle V = \frac{1}{3}\pi r^2 h$ and $\displaystyle S = \pi rl$, l, of course, is the "slant height". Using the Pythagorean theorem, l^2= r^2+ h^2.
Since we must have: $\displaystyle S= \pi rl= \pi r\sqrt{r^2+ h^2}= 4\sqrt{3\pi}$, Squaring both sides of that $\displaystyle \pi^2r^2(r^2+ h^2)= 48\pi$.Now we must solve for h and replace h in V= \pi r^2h/3 by that to get a problem in just the one variable r. (Then after that we find the derivate to find the largest possible value of that function i.e. maximized.)
My problem is here: How do we solve for h in $\displaystyle \pi^2r^2(r^2+ h^2)= 48\pi$??
Is it: $\displaystyle h = \sqrt\frac{{48 \pi  r}}{\pi^2 . r^2}$ ?
Thank you,
Regards