Question

A box with a square base and open top must have a volume of 2048 c m 3 . We wish to find the dimensions of the box that minimize the amount of material used. The length of the base is x and the height is h. Since the base is a square, the surface area of just the base would be: Area = The surface area of just one side would be: Area = The surface area of all 4 sides would be: Area = The surface area of the base and all four sides would be: Area = The volume of a box is V = length*width*height. We are calling the length, x, and the height, h. The box has a square base, so we have V = x 2 h . The volume is 2048 cm3. So we have the equation 2048 = x 2 h . Solving the volume equation for h , we get: h = Get the surface area in terms of just x, by substituting your previous answer in for h. Do not get a common denominator. Area = Now write your formula with a common denominator. Area = Use a graphing calculator or Excel to look at the graph of the function to find the value of x that minimizes the surface area. The length of the box that minimizes the surface area is x = The minimum surface area is What would the height of the box ned to be to minimize the surface area?

Answer 1

The Volume of a box with a square base x by x cm and height h cm is V=(x^2)h

The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.

The surface area of the box described is A=x^2+4xh

We need A as a function of x alone, so we'll use the fact that
V=(x^2)h= 2048 cm^3

which gives us h= 2048/x^2, so the area becomes:

A=x^2+4x(2048/x^2)=x^2 + 8192/x

We want to minimize A, so

A'=2x- 8192/x2=0 when (2x3− 8182)/x2=0

Which occurs when 2x^3−8182=0 or x^3 = 4091

The only critical number is x= 15.99 cm.

The second derivative test verifies that A has a minimum at this critical number:
A''=2+ 16364/x3 which is positive at x= 15.99

The box should have base 15.99 cm by 15.99 cm and height 8.01 cm .

(use h= 2048/x2 and x= 15.99)