Chapter 2.5, Problem 104E

a.

To determine

Find the volume $V$ of the package as a function of $x$ .

Answer to Problem 104E

  $V\left(x\right)=60x^2-x^3$

Explanation of Solution

Calculation:

Consider the diagram,

  

Volume ,

  $\begin{array}{l}V\left(x\right)=y\times x\times x\\ =x^{2}y\end{array}$

Perimeter,

  $\begin{array}{l}2x+2y=120\\ x+y=60\\ y=60-x\\ put,y=60-x,in,V\left(x\right)=x^{2}y\\ V\left(x\right)=x^{2}y\\ V\left(x\right)=x^2\left(60-x\right)\\ V\left(x\right)=60x^2-x^3\end{array}$

Hence, the volume $V$ of the package as a function of $x$ is, $V\left(x\right)=60x^2-x^3$ .

b.

To determine

Use a graphing utility to graph the function and approximate the dimensions of the package that yield a maximum volume.

Answer to Problem 104E

  $x=40,y=20$

Explanation of Solution

Calculation:

Graph the function $V\left(x\right)$ ,

  

To find the maximum value $V$ , we have to find value of $x$ for which $\frac{dV}{dx}=0$ we get,

  $\begin{array}{l}120x-3x^2=0\\ x=0,40\end{array}$

By solving quadratic function ,consider positive value of $x=40$ because $x=0$ is not in domain,

Hence, the package that yield a maximum volume dimensions are $x=40,y=20$

c.

To determine

Find values of $x$ such that $V=13,500$ .

Answer to Problem 104E

  $18$

Explanation of Solution

Calculation:

Find values of $x$ such that $V=13,500$ .

  $\begin{array}{l}60x^2-x^3=13500\\ x^3-60x^2+13500=0\end{array}$

To find the value of $x$ for function $x^3-60x^2+13500=0$ by using zero rational test with leading coefficient is $1$ and constant term is $13500$

Possible rational zeroes,

  $\begin{array}{l}\frac{\text{Factor of 13500}}{\text{Factor of 1}}=\frac{\pm 1,\pm 2,\pm 4,\mathrm{...},\pm 13500}{\pm 1}\\ =\pm 1,\pm 2,\pm 4,\mathrm{...},\pm 13500\end{array}$

Therefor lot of zeroes possibility , now we will draw the graph of , $f\left(x\right)$

  

It is observed from the graph that, solution of $f\left(x\right)$ is $18$ .

Hence, values of $x$ such that $V=13,500$ is $18$ .