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

Could you please explain this problem step by step, thank you!

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 11:14pm
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

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

## Answer this Question

## Related Questions

- Calculus-Applied Optimization Problem - If a total of 1900 square centimeters of...
- calculus optimization problem - by cutting away identical squares from each ...
- Calculus - Given: available material= 1200cm^2 Box w/ square base & open top ...
- Calulus/Optimization - I don't understand how to solve optimization problems, (...
- Calculus - What is the smallest possible slope for a tangent to y=x^3 - 3x^2 + ...
- Quick math help - In which step below does a mistake first appear in simplifying...
- math! - A rectangular box is twice as long as it is wide. The height of the box ...
- Calculus (Optimization) - A rectangular piece of cardboard, 8 inches by 14 ...
- algebra - would you help me write a logic application for andy has the cards1,5...
- Algebra II - Solving the formula for the indicated variable. I=prt,for r I need ...