Chapter 1.4, Problem 67PS

Solution

a.

To find: the girth of the package.

  $72\text{}\text{ in}$ .

Given information:

For the rectangular package, the sum of the length and the girth cannot exceed 108 inches.

The length of the package is given to be 36 inches.

Also, the girth is known to be as large as possible.

Calculation:

Let l be the length of the package and g be the girth of the package.

Now, as it is given that the sum of the length and the girth cannot exceed 108 inches, this implies that,

  $l+g\le 108$ ..... (1)

Also, it is given that the length of the package is 36 inches, that is $l=36\text{in}$ .

Substitute the value $l=36\text{in}$ in inequality (1),

  $36+g\le 108$

Subtract 36 from both sides of the above inequality,

  $\begin{array}{c}g\le 108-36\\ g\le 72\end{array}$

Also, it is given that the girth is as large as possible.

So, the maximal girth of the package is 72 in.

b.

To write: an expression for the package’s width w in terms of h and an equation giving the package’s volume V in terms of h .

  $\begin{array}{l}w=36-h\\ V=36h\left(36-h\right)\end{array}$

Given information:

For the rectangular package, the sum of the length and the girth cannot exceed 108 inches.

The length of the package is given to be 36 inches.

Also, the girth is known to be as large as possible.

Formula Used:

  $\begin{array}{c}\text{Girth of cuboid = 2}\left(\text{Width + Height}\right)\\ \text{Volume of cuboid = Length}\times \text{Width}\times \text{Height}\end{array}$

Calculation:

It is known that the girth of a cuboid is equal to 2 times the sum of the width and the height.

Let w denotes the width and h denotes the height of the package.

Then, $g=2\left(w+h\right)$ .

Also, from part (a),

  $g=72\text{ in}$

So,

  $\begin{array}{c}72=2\left(w+h\right)\\ 36=w+h\\ w=36-h\end{array}$

So, the expression for the package’s width w in terms of h is $w=36-h$ .

Now, find equation for volume of package,

  $\begin{array}{c}V=l\times w\times h\\ V=36\left(36-h\right)h\\ V=36h\left(36-h\right)\end{array}$

So, the equation giving the package’s volume V in terms of h is $V=36h\left(36-h\right)$ .

c.

To find: the height and width that maximizes the volume of the package and also the maximum volume.

The height and width that maximizes the volume of the package are both 18 inches and the maximum volume is $11,664{\text{ in}}^2$ .

Given information:

For the rectangular package, the sum of the length and the girth cannot exceed 108 inches.

The length of the package is given to be 36 inches.

Also, the girth is known to be as large as possible.

Property Used:

Factoring and Zeros:

To find the maximum or minimum value of a quadratic function, first use factoring to write the function in intercept form $y=a\left(x-p\right)\left(x-q\right)$ . Because the function’s vertex lies on the axis of symmetry $x=\frac{p+q}{2}$ , the maximum or minimum occurs at the average of the zeros p and q .

Calculation:

From part (b), the equation giving the package’s volume V in terms of the height h is $V\left(h\right)=36h\left(36-h\right)$ .

Now, it is clear that here the zeros of the above volume function are $h=0$ and $h=36$ .

The average of zeros is $\frac{0+36}{2}=18$ .

So, to maximize volume, the height should be $h=18\text{ in}$ .

Now, from part (b) $w=36-h$, so the corresponding width is:

  $\begin{array}{c}w=36-18\\ =18\text{ in}\end{array}$

The maximum volume is:

  $\begin{array}{c}V\left(18\right)=36\left(18\right)\left(36-18\right)\\ =36\times 18\times 18\\ =11,664{\text{ in}}^2\end{array}$