Question 564691
Assume the cylinder has a circular base. We want to minimize


*[tex \LARGE A = 2\pi r^2 + 2\pi rh], given *[tex \LARGE V = \pi r^2h = 1200]. Couple ways to do this:



Solution 1: Let *[tex \LARGE h = \frac{1200}{\pi r^2}]. Then *[tex \LARGE A = 2\pi r^2 + 2\pi r(\frac{1200}{\pi r^2}) = 2\pi r^2 + \frac{2400}{r}]. Taking derivative with respect to r,


*[tex \LARGE \frac{dA}{dr} = 4\pi r - \frac{2400}{r^2}]. This equals zero when *[tex \LARGE 4\pi r = \frac{2400}{r^2}], or when *[tex \LARGE r = \sqrt[3]{\frac{600}{\pi}}]. ∎



Solution 2: Using the method of Lagrange multipliers, there exists a constant *[tex \LARGE \lambda] with A(r,h) and V(r,h) such that


*[tex \LARGE \bigtriangledown A = \lambda \bigtriangledown V]


*[tex \LARGE \langle 4r\pi + 2\pi h, 2\pi r \rangle = \lambda \langle 2rh, r^2 \rangle], therefore


*[tex \LARGE 4r\pi + 2\pi h = 2rh\lambda \Rightarrow \lambda = \frac{2\pi}{h} + \frac{\pi}{r}]
*[tex \LARGE 2\pi r = r^2\lambda \Rightarrow \lambda = \frac{2\pi}{r}]


Equating these two,


*[tex \LARGE \frac{2\pi}{h} + \frac{\pi}{r} = \frac{2\pi}{r} \Rightarrow \frac{2}{h} = \frac{1}{r} \Rightarrow h = 2r], finding the actual dimensions follows. ∎


There are probably other solutions involving AM-GM inequality or other techniques. The preferred solution to use is probably dependent on the course you are taking.