What dimensions should the box have?

Question

The packaging department of a corporation is designing a box with a square base and no top.The volume is to be 32m^3. to reduce cost, the box is to have minimum surface area.

Answer 1

Let sides = x

Let height = h

 

x²h = 32               eq1 ----> volume

 

SA =  x² + 4xh            eq2  ------> surface area

 

 

Substitute eq1 into eq2 so that eq2 is only in terms of x.

 

SA = x² + 4x(32/x²)

 

SA = x² + 128x^-1

 

 

Now just take the derivative of SA and set it equal to zero.  Solve for x.  Then evaluate the derivative of SA before and after the x value.  If the derivative changes from negative to positive, then you have a minimum surface area.

Answer 2

A'(x) = 2x - 128/x^2

 

0 = 2x - 128/x^2

 

128/x^2 = 2x

 

128 = 2x^3

 

64 = x^3

 

x =4 (critical point)

 

The 2nd derivative test might work just as easy

 

The 2nd derivative is 2+ 128/x^3 which is positive for any positive x including our critical point x = 4

 

When the 2nd derivative is positive at a crtitical point, the critical point is a MINIMUM.

 

Therefore x = 4 is a minimum and the height would be h = 32/x^2 = 32/16 = 2

 

base dimension  is 4 by 4

height = 2