Dayaan M. answered • 08/26/26
Earned A’s in Calc 1/AB & Calc 2/BC | 5 Years of Tutoring Experience
In order to work an optimization problem, the routine is always the same. Name your variables, write the thing you want to minimize, use the other piece of information to get it down to ONE variable, then take a derivative and set it to zero.
Since the base is a square, let x be the side of the base and h be the height.
The volume constraint gives us:
x^2 h = 3456
which we can solve for h so we can get rid of it later:
h = 3456/x^2
Now build the cost. The top and bottom are each a square of area x^2, so together that is 2x^2 of material at $2 per square inch, giving 4x^2. The four sides are each a rectangle of area xh, so that is 4xh of material at $1 per square inch, giving 4xh:
C = 4x^2 + 4xh
Remember, we cannot take a derivative while there are two variables, so this is where we substitute the h we found:
C = 4x^2 + 4x(3456/x^2) = 4x^2 + 13824/x
Now differentiate. It helps to rewrite 13824/x as 13824x^(-1) first:
C' = 8x - 13824/x^2
Setting that equal to zero:
8x = 13824/x^2
8x^3 = 13824
x^3 = 1728
x = 12
Then the height comes from our constraint:
h = 3456/144 = 24
It is worth confirming this really is a minimum and not a maximum. The second derivative is C'' = 8 + 27648/x^3, which is positive for any positive x, so the graph is concave up and we have found a minimum.
So, our final answer is a base of 12 inches by 12 inches with a height of 24 inches, which comes out to a cost of $1,728.
If you notice, the optimal box is twice as tall as it is wide. That is not a coincidence. Since the top and bottom cost twice as much per square inch, it pays to shrink them and stretch the cheaper sides upward instead.
