Q. 45 - Calculus

Question

The U.S. Postal Service ships a package under large-package rates if the sum of the length and the girth of the package is greater than $84$ inches and less than or equal to $108$ inches. The length of a package is considered to be the length of its longest side, and the girth of the package is the distance around the package perpendicular to its length. In the below question, Linda wants to ship packages under the USPS large-package rates.

      Linda needs to mail a rectangular package with square ends (in other words, with equal width and height). What is the largest volume that her package can have? What is the largest surface area that her package can have? 

Step-by-Step Solution

Answer

Ans:  The largest volume of the package is $11664$ cube inches and, the largest surface area of the package is $3332.6$ square inches.

Step 1. Given inforation.

given, 

        The figure above shows a box of length $l$ inches, width $w$ inches, and height $h$ inches.

Consider the ends of the box are square.

Therefore, $w=h$

The sum of the length and girth of the box is at least $84$ inches and no greater than $108$ inches.

Step 2. The objective is to find the largest possible volume and the largest possible surface area.

The girth of the box is, $w+w+w+w=4w$.

The sum of the girth and the length is, $4w+l$.

The range of values of this sum is,

     $84\le 4w+l\le 108$

The volume of the box is,

        $V=l{w}^2$

Substitute $l$ in terms of $w$ for maximum value,

       $l=108-4w$

Therefore, the volume is,

        $\begin{array}{r}V\left(w\right)=(108-4w)w^2\\ =108w^2-4w^3\end{array}$

Step 3. Differentiate the function with respect to w ,

       $V^{\mathrm{\prime }}\left(w\right)=216w-12w^2$

Equate this derivative to zero and solve for $w$,

       $\begin{array}{r}216w-12w^2=0\\ w=0,18\end{array}$

The length is,

        $\begin{array}{r}l=108-4\left(18\right)\\ =36\end{array}$

Therefore, the volume is,

         $\begin{array}{r}V=\left({18}^2\right)\left(36\right)\\ =11664\end{array}$

Therefore, the maximum volume of the box is $11664$ cubic inches.

Step 4. The surface area of the box is,

     $S=2w^2+4lw$

Substitute $l$ in terms of $w$ for maximum value,

     $\begin{array}{r}S=2w^2+4(108-4w)w\\ =2w^2+432w-16w^2\\ =432w-14w^2\end{array}$

Step 5. Differentiate the function with respect to w ,

      $S^{\mathrm{\prime }}\left(w\right)=432-28w$

Equate this derivative to zero and solve for $w$,

      $\begin{array}{r}432-28w=0\\ w\approx 15.43\end{array}$

The length is, 

       $\begin{array}{r}l=108-4\left(15.43\right)\\ \approx 46.29\end{array}$

The surface area of the box is,

      $\begin{array}{r}S=2(15.43{)}^2+4(46.29\left)\right(15.43)\\ \approx 3332.6\end{array}$

Therefore, the maximum surface area of the box is $3332.6$ square inches.