We have 350 m2 of material to build a box whose base width is four times the base length. Determine the dimensions of the box that will maximize the enclosed volume.

Question

I set up the objective equation and the constraint equation.

I think the objective equation is: (4w * l) = 350

I think the constraint is lwh.

How would you go about solving this? You would solve for one variable in the objective and then plug it into the constraint for that variable right?

Answer 1

Assuming closed box, use the fact that you have 350 square meters of material to create the box, where x is the width of the base, 4x is the length of the base, and h is the height of the box.

The box will have two sides with dimensions x by h, so 2xh material will be consumed.

The box will have two sides with dimensions 4x by h, so 2(4xh) or 8xh material will be consumed.

The area of the bottom of the box will be 4x(x) or 4x². Since the assumption is both top and bottom

8x² represents the material consumed.

So we have 8x² + 10xh = 350. This equation is used to find an expression for h in terms of x. Solve for x.

The function to maximize is V = 4x (x) (h). Replace the h with the expression determined above.

Now you can find dV/dx, set equal to zero to find critical numbers for the function. Determine if the value(s) generate a max or min volume. The problem suggests that a max will be generated.