# 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.

## Problem

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.

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.