So I'm doing these whole "applies maximum and minimum" problems, and it was going fine until I ran into ones where they abstract it... Like for example:
Find the height and radius of the cone of slant height L whose volume is as large as possible.
I know I should be able to find some values in terms of L, like if I make x equal to the radius, then solve for the height using x, I should then be able to find what x is in terms of L, but it just doesn't seem to work out how it should.
First, we're going to need an equation of one variable for the volume.
V = 1/3*pi*h*r^2
h = sqrt(L^2 - r^2)
Substitute the formula for h in:
V = 1/3*pi*r^2*sqrt(L^2-r^2)
L is a constant, so V has only one variable: r. So now, you need to find dV/dr:
dV/dr = 2/3*pi*r*sqrt(L^2-r^2) - 1/3*pi*r^3/sqrt(L^2-r^2) = (pi*r*(2*L^2-3*r^2))/(3*sqrt(L^2-r^2))
So now, find the critical values. A few sanity bounds need to be established:
r^2 < L^2
r =/= 0
The first bound allows us to ignore the denominator. The second bound is due to the fact that the radius of a cone must be positive. So now we have the following:
(pi*r*(2*L^2-3*r^2)) = 0
The pi disappears due to being a constant, and we can break up the remaining factors:
r = 0
2*L^2 - 3*r^2 = 0
The first is illegal by our bounds. The second can be rearranged into something more meaningful:
3*r^2 = 2*L^2
r^2 = 2/3*L^2
r = sqrt(2/3*L^2) = L*sqrt(2/3)
And there you have it, a value for r at an extreme value of the volume function. Because the minimum volume would be 0 (which cannot happen), you can safely conclude that this is a maximum.