Question

a. write a function v that represents the volume of the box.
b. determine the domain of the function v.
c. sketch a graph of the function and estimate the value of x for which v(x) is a maximum.

Explanation:

Step1: Find the volume function

The side length of the base of the box after cutting out the squares of side - length \(x\) is \(l = 24 - 2x\) (since we cut out \(x\) from both ends of each side). The height of the box is \(h=x\).
The volume formula for a rectangular box is \(V=l\times w\times h\). Here, \(l = w=24 - 2x\) and \(h=x\). So, \(V(x)=(24 - 2x)(24 - 2x)x\).
Expand \((24 - 2x)(24 - 2x)=4x^{2}-96x + 576\). Then \(V(x)=4x^{3}-96x^{2}+576x\).

Step2: Determine the domain

For the physical dimensions:

1. \(x>0\) (because the height cannot be non - positive).
2. \(24-2x>0\) (because the side length of the base cannot be non - positive). Solve \(24-2x>0\) for \(x\):

\(24>2x\), then \(x < 12\).
So the domain of \(V(x)\) is \(0

Step3: Sketch the graph (conceptually)

We can use the first - derivative test. First, find the derivative \(V^\prime(x)=12x^{2}-192x + 576\).
Set \(V^\prime(x)=0\), then \(12x^{2}-192x + 576 = 0\). Divide through by \(12\): \(x^{2}-16x + 48=0\).
Factor: \((x - 4)(x - 12)=0\). The critical points are \(x = 4\) and \(x = 12\).
We can also use a graphing utility. When \(x = 4\), \(V(4)=4\times4^{3}-96\times4^{2}+576\times4=4\times64-96\times16 + 576\times4=256-1536 + 2304=1024\).
As \(x\to0^{+}\), \(V(x)\to0\); as \(x\to12^{-}\), \(V(x)\to0\).

Answer:

a. \(V(x)=4x^{3}-96x^{2}+576x\)
b. Domain: \((0,12)\)
c. The value of \(x\) for which \(V(x)\) is a maximum is \(x = 4\) (by first - derivative test or graphing utility analysis).