Calculus: im having so much trouble with optimization of a cylinder's volume and surface area

So optimization is about maximizing and minimizing right? How do you maximize/optimize the volume of a cylinder based on a surface area?? so let's say the surface area is 40 in^2. what would the radius and height have to be to have the maximum volume?? i know it has something to do with the derivative and using v=πr^2h and sa=2πr^2+2πrh

First , let's look at the formulas:

v = pi * r^2 * h
a = 2pi * r^2 + 2pi * r * h = 2 * pi * r * (r + h)

What we want to do is put r into terms of h. We want to do this before we do any differentiation (it just makes life easier when we do this first)

a = 40
a = 2 * pi * r * (r + h)
40 = 2 * pi * r * (r + h)
20 = pi * r * (r + h)
20 / (pi * r) = r + h
h = 20/(pi * r) - r
h = (20 - pi * r^2) / (pi * r)

Pretty simple, right? Now we just plug that into our volume formula:

V = pi * r^2 * h
V = pi * r^2 * (20 - pi * r^2) / (pi * r)
V = r * (20 - pi * r^2)
V = 20r - pi * r^3

Now we derive V with respect to r

dV/dr = 20 - 3 * pi * r^2

Find when dV/dr = 0

0 = 20 - 3 * pi * r^2
3 * pi * r^2 = 20
r^2 = 20 / (3pi)
r = sqrt(20 / (3pi))

There's the radius of our optimized cylinder. Plug that value for r back into our formula for h to find h:

h = (20 - pi * r^2) / (pi * r)
h = (20 - pi * (20/(3pi))) / (pi * sqrt(20 / (3pi)))
h = (20 - 20/3) / (pi * sqrt(20/(3pi)))
h = (40/3) / (pi * sqrt(20/(3pi)))
h = 40 * sqrt(20 / (3pi)) / (3 * pi * (20 / (3pi))
h = 40 * sqrt(20 / (3pi)) / (20)
h = 2 * sqrt(20 / (3pi))

In this case, the optimized cylinder will have a radius of sqrt(20 / (3pi)) and a height of 2 * sqrt(20 / (3pi)). If you'll notice, the height is twice the radius. Keep that in mind. All optimized cylinders share that feature: The height is twice the radius