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