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January 31, 2015

January 31, 2015

Posted by **Lisa** on Tuesday, January 3, 2012 at 10:17pm.

You are planning to make an open rectangular box that will hold a volume of 50 cubed feet. What are the dimensions of the box with minimum surface area?

- Calculus I Quick Optimization Problem -
**drwls**, Tuesday, January 3, 2012 at 11:14pmFor 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

- Calculus I Quick Optimization Problem -
**Damon**, Tuesday, January 3, 2012 at 11:22pmx = 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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