College Algebra & Trigonometry, 2018a

4.9. OPTIMIZATION 243
Geometry/Cost Optimization
A closed box with a square base is built from material that costs $1 per ft 2 for the
four sides and the bottom and $5 per ft 2 for the top. What are the dimensions of
the box of largest volume that can be constructed for $72?
There are two issues we are working with in this problem - the volume of the
box and the surface area of the box. The volume of the box is important because
that’s what we’re looking to maximize. The surface area of the box is important
because that’s what will control the cost of the box - notice that the costs are given
in terms of ft 2 or square feet which is related to the surface area.
The box has a square base, but the height is some other dimension - h. So the
volume of the box will be length ∗ width ∗ height, but since the box has a square
base the length and width will be the same - we’ll call them x. From this we see
that the volume can be expressed as:
V = x ∗ x ∗ h = x 2 h
For the cost of the box, we need to know the surface area. Each side of the box
should be included - the base and the top have the same area (x 2 ) because they
are both squares that are x units on each side. The four sides of the box are all
rectangles that are x by h units. That means each one has an area of x ∗ h -soall
four would be 4 ∗ x ∗ h or 4xh.
This makes the surface area of the box:
The cost of the box then will be:
S =2x 2 +4xh
C = x 2 ($1) + x 2 ($5) + 4xh($1)
Here we see the one x 2 is multiplied by $1, because the bottom will cost $1 per
ft 2 , but the other x 2 is multiplied by $5, because the top costs $5 per ft 2 . The four
sides : 4xh is also multiplied by $1.