Maximize Volume of a Box
Optimization Problem

How to maximize the volume of a box using the first derivative of the volume. A volume optimization problem with solution.


A sheet of metal 12 inches by 10 inches is to be used to make an open box. Squares of equal sides \(x\) are cut out of each corner then the sides are folded to make the box. Find the value of \(x\) that makes the volume maximum.
maximize volume problem 1

Solution to Problem 1:

We first use the formula of the volume of a rectangular box.
\(V = L \times W \times H\)
The box to be made has the following dimensions:
\(L = 12 - 2x\)
\(W = 10 - 2x\)
\(H = x\)
We now write the volume of the box to be made as follows:
\(V(x) = x(12 - 2x)(10 - 2x) = 4x(x^2 - 11x + 30) = 4x(x^2 - 11x + 30)\)
We now determine the domain of function \(V(x)\). All dimensions of the box must be positive or zero, hence the conditions
\(x \geq 0\) and \(6 - x \geq 0\) and \(5 - x \geq 0\)
Solve the above system of inequalities to find the domain of function \(V(x)\)
\(0 \leq x \leq 5\)
Let us now find the first derivative of \(V(x)\) using its last expression.
\(\dfrac{dV}{dx} = 4[3x^2 - 22x + 30]\)
Let us now find all values of \(x\) that make \(\frac{dV}{dx} = 0\) by solving the quadratic equation
\(3x^2 - 22x + 30 = 0\)
Two values make \(\frac{dV}{dx} = 0\):
\(x = 5.52\) and \(x = 1.81\), rounded to one decimal place.
\(x = 5.52\) is outside the domain and is therefore rejected.
Let us now examine the values of \(V(x)\) at \(x = 1.81\) and the endpoints of the domain.
\(V(0) = 0\), \(v(5) = 0\) and \(V(1.81) = 96.77\) (rounded to two decimal places)
So \(V(x)\) is maximum for \(x \approx 1.81\) inches. The graph of function \(V(x)\) is shown below and we can clearly see that there is a maximum very close to 1.8.

graph of V(x), problem 1

References and Links

calculus problems