Chapter 5.4, Problem 7E

To determine

Find the dimensions of the box of largest volume.

Answer to Problem 7E

The maximum volume of the dimensions is $\frac{5}{3}in,\frac{14}{3}in,\frac{35}{3}in$ and the volume is $90.74i{n}^3$ .

Explanation of Solution

Given information: Given the dimensions of the rectangle carboard is $8by15in$

Calculation:

Let the upper vertices of the rectangle are $x,\text{ and }0<x<4$

Then the dimension of the base is $\left(8-2x\right)\text{ by }\left(15-2x\right)\text{in }$

Let the volume of the open box is V.

  $\begin{array}{l}V=x\left(8-2x\right)\left(15-2x\right)\\ =x\left(120-30x-16x+4x^2\right)\\ =4x^3-46x^2+120x\end{array}$

For any function $f\left(x\right)$ the condition of minimum and maximum is obtained by find $f^{\prime }\left(x\right)=0$ ,

Critical points and end points

  $\begin{array}{l}V=4x^3-46x^2+120x\\ \frac{dV}{dx}=12x^2-92x+120\end{array}$

Critical points occur at

  $\begin{array}{l}{V}^{\prime }=0\\ 12x^2-92x+120=0\\ 4\left(3x-5\right)\left(x-6\right)=0\\ x=\frac{5}{3},x=6\end{array}$

  $\text{since }0<x<4\text{ consider }x=\frac{5}{3}\approx 1.66$

solve it graphically, it has critical point occur at $x=\frac{5}{3}.$

Here $\frac{dV}{dx}>0\text{ for 0< }x<\frac{5}{3}\text{and}\frac{dV}{dx}<0\text{ for}\frac{5}{3}\text{< }x<4$ this corresponds to maximum area.

For

  $\begin{array}{l}x=\frac{5}{3}\\ 8-2x=8-2{\left(\frac{5}{3}\right)}^2\\ =\frac{14}{3}\\ \\ 15-2x=15-2\left(\frac{5}{3}\right)\\ =\frac{35}{3}\end{array}$

  $\begin{array}{l}V=x\left(8-2x\right)\left(15-2x\right)\\ =\left(\frac{5}{3}\right)\left(\frac{14}{3}\right)\left(\frac{35}{3}\right)\\ =\frac{2450}{27}\approx 90.74\end{array}$

Graphical support:

  

Thus, the maximum volume of the dimensionsare $\frac{5}{3}in,\frac{14}{3}in,\frac{35}{3}in$ and the volume is $90.74i{n}^3$ .