Question

Problem 2 (1.1.5, 4 points): Find the rectangular parallelepiped of unit volume that has the minimum surface area. Hint: By eliminating one of the dimensions, show that the problem is equivalent to the minimization over > 0 and y > 0 of f(x,y) =xy +-+- Show that the sets {(x,y) I f(x,y) x > 0, y > 0} are compact for all scalars γ.

Answer 1

Assume a parallelopipe with dimensions: x,y,z

$V=1= xyz$

$S= 2(xy+yz+zx)$

$z= 1/xy$ substitute z in S we get,

$S= 2(xy+ 1/x+1/y)$

$argmin. S= argmin.{2(xy+ 1/x+1/y)}= argmin.(xy+1/x+1/z)$

$argmin. S= argmin.(\gamma (xy+ 1/x+1/y))= argmin.(xy+1/x+1/z)$

for all $\gamma$ as scaling doesnot change the value of the minimizing function