Erik M. answered • 05/02/23
Experienced Mathematics Tutor for All Levels
First thing we notice is that this is an optimization question, and we are trying to optimize (find the maximum) the volume. That means that ultimately, we want to find an equation for the volume in terms of the side lengths. First, the problem states that the box has a square base. That means that if we draw our box, we would have two sides labeled as L, and a height of H. Knowing this, our volume equation must be length times width times height, which would be
L*L*H, or L2H=V
We want to take the derivative of this function, however, there are two variables, which makes the derivative more complex. In order to create a volume equation in terms of either H or L, we will use a constraining equation. In this case, the price of the box must not exceed 300$. We know that the sides cost 10$, and the top cost 14$, and the bottom cost 11$. This means that
300=10$*(Area of sides)*(Number of sides) + 14$*(Area of top)+11$*(Area of bottom).
Filling in what we know about the box, we get
300=10$(LH)(4)+14$(L2)+11$(L2) = 40LH+25L2.
Using this new question, we can solve for the variable H:
H = (300-25L2)/40L.
We can then plug this into the volume equation, giving us
V = L2*(300-25L2)/40L = 7.5L + .625L3.
Now we have a volume equation with respect to one variable. We can optimize functions by setting their derivative equal to 0. In this case, the derivative would be
V’ = 7.5 + 1.875L2.
Setting it equal to 0 we get
L = +2, -2.
In this case, we get two values when we take the square root. -2 doesn’t make sense for our problem, so we can ignore it. We now know the length at which the volume of this shape is optimized. We can continue forward to solve for H as well.
H = (300-25(2)2)/40(2) = 2.5
Therefore, our volume is optimized when our box with a square base with side lengths L has a dimension of 2, and a height H of 2.5
