Chapter 2.3, Problem 68E

Solution

To determine the possible values of $x$ so that the function $V=2666x-210x^2+4x^3$ represents the volume of a box.

The possible values of $x$ are:

  $0\le x\le 21.5$

Given the function representing the volume:

  $V=2666x-210x^2+4x^3$

Concept Used:

The volume of a box crafted from a rectangular piece of material with length $l$ and width $w$ , by cutting squares of side length $x$ from each corner, is given by:

  $V=x(l-2x)(w-2x)$ .

Now, since $x$ is the side length of the square, it must be non-negative.

That is, $x\ge 0$ .

Also, the length of the material should not exceed 2 times $x$ .

That is, the half-length of the material should not exceed $x$ .

That is, $x\le \frac{l}{2}$ .

Finally, the width of the material also should not exceed 2 times $x$ .

That is, the half-width of the material should not exceed $x$ .

That is, $x\le \frac{w}{2}$ .

Concluding, $x$ must satisfy:

  $x\ge 0$ , $x\le \frac{l}{2}$ and $x\le \frac{w}{2}$ .

Calculations:

Equate $V=2666x-210x^2+4x^3$ with $V=x(l-2x)(w-2x)$ and find $l$ and $w$ :

  $\begin{array}{l}x(l-2x)(w-2x)=2666x-210x^2+4x^3\\ x(l-2x)(w-2x)=x\left(2666-210x+4x^2\right)\\ (l-2x)(w-2x)=2666-210x+4x^2\\ lw-2\left(l+w\right)x+4x^2=2666-210x+4x^2\\ lw-2\left(l+w\right)x=2666-210x\end{array}$ .

Compare the coefficients of $lw-2\left(l+w\right)x=2666-210x$ :

  $lw=2666$ and $-2(l+w)=-210$ .

Solve the equation $-2(l+w)=-210$ for $w$ :

  $\begin{array}{l}-2(l+w)=-210\\ (l+w)=105\\ w=105-l\end{array}$ .

Substitute $w=105-l$ in the equation $lw=2666$ :

  $\begin{array}{l}l(105-l)=2666\\ 105l-l^2=2666\\ 0=l^2-105l+2666\end{array}$ .

Solve the quadratic equation $l^2-105l+2666=0$ :

  $\begin{array}{l}l=\frac{-\left(-105\right)\pm \sqrt{{\left(-105\right)}^2-4\cdot 1\cdot 2666}}{2\cdot 1}\\ l=\frac{105\pm \sqrt{361}}{2}\\ l=\frac{124}{2},\frac{86}{2}\\ l=62,43\end{array}$ .

Substitute $l=62$ in $lw=2666$ :

  $w=43$ .

Substitute $l=43$ in $lw=2666$ :

  $w=62$ .

That is, the length and width of the material are interchangeable.

Now, without loss of generality, let $l=62$ and $w=43$ .

Then, it follows that:

  $x\ge 0$ , $x\le \frac{62}{2}$ and $x\le \frac{43}{2}$ .

That is:

  $x\ge 0$ , $x\le 31$ and $x\le 21.5$ .

That is:

  $x\ge 0$ and $x\le 21.5$ .

That is:

  $0\le x\le 21.5$ .

Thus, the possible values of x are:

  $0\le x\le 21.5$

Conclusion:

The possible values of $x$ are:

  $0\le x\le 21.5$