Posted by Lisa on Tuesday, January 3, 2012 at 10:17pm.
For symmetry reasons, the optimum must have a square base. Let the height of the walls be x and the side length be y
Volume = x^2*y = 50
Area = x^2 +4xy = x^2 + 4x*50/x^2
= x^2 + 200/x
dA/dx = 0 = 2x -200/x^2
x^3 = 100
x = 4.64 ft
y = 2.32 ft
x = length
y = width
h = height
x y h = 50
area bottom = x y
area sides = 2 x h + 2 y h
total area a = x y + 2 x h + 2 y h
so
h = 50/xy
a = x y + 100/y + 100/x
da/dx=x dy/dx+y-100 dy/dx /y^2-100/x^2
for min = 0
dy/dx(x-100/y^2) + y-100/x^2 = 0
x y^2 = 100 and y x^2 = 100
y (100^2/y^4) = 100
y^3 = 100
y = 4.64
x = 100/y^2 = 4.64
h = 2.32
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