Use Lagrange to find the minimum surface area of a cylinder with volume 100pi cm^3

Let r denote the radius of the cylinder and h denote the height.

Since the volume of the cylinder is 100π cm^2:
V = πr^2*h = 100π ==> r^2*h = 100 and r^2*h - 100 = 0.

We want to minimize:
SA = 2πr^2 + 2πrh = 2π(r^2 + rh) with respect to r^2*h - 100 = 0.

To do this, consider the following function:
F(r, h, λ) = 2πr^2 + 2πrh - λ(r^2*h - 100)
= 2πr^2 + 2πrh - λr^2*h + 100λ.

Differentiating with respect to r, h, and λ:
(a) df/dr = 4πr + 2πh - 2λrh
(b) df/dh = 2πr - λr^2
(c) df/dλ = -r^2*h + 100.

Setting these all equal to zero gives:
4πr + 2πh - 2λrh = 0, 2πr - λr^2 = 0, and -r^2*h + 100 = 0
==> 2πr + πh - λrh = 0, 2πr - λr^2 = 0, and r^2*h = 100.

Note that r ≠ 0, so we can divide 2πr - λr^2 = 0 by r to get:
2π - λr = 0 ==> λr = 2π.

Substituting λr = 2π into the first equation:
2πr + πh - 2πh = 0
==> 2πr - πh = 0
==> h = 2r.

Note that h = 2r is not a coincidence; the surface area of any cylinder is minimized when the height is the diameter of the cylinder.

Since r^2*h = 100, 2r^3 = 100 ==> r = 50^(1/3) and h = 2*50^(1/3).

Therefore, the minimum surface area is:
SA(min) = 2πr^2 + 2πrh
= 2π*50^(2/3) + 4π*50^(2/3)
= 6π*50^(2/3)
≈ 225.83 cm^2.

I hope this helps!