Question

The Canine Products company offers two dog foods, Frisky Pup and Husky Hound, that are made from a blend of cereal and meat. A package of Frisky Pup requires 1 pound of cereal and $\mathbf{1}\mathbf{.}\mathbf{5}\mathbf{}$pounds of meat, and sells for $\mathbf{\backslash (}\mathbf{7}$. A package of Husky Hound uses 2 pounds of cereal and 1 pound of meat, and sells for $\mathbf{\backslash )}\mathbf{6}$. Raw cereal costs$\mathbf{\backslash (}\mathbf{1}$per pound and raw meat costs$\mathbf{\backslash )}\mathbf{2}$per pound. It also costslocalid="1658981348093" $\mathbf{\backslash (}\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{}$to package the Frisky Pup and localid="1658981352345" $\mathbf{\backslash )}\mathbf{0}\mathbf{.}\mathbf{60}\mathbf{}$to package the Husky Hound. A total of localid="1658981356694" $\mathbf{240}\mathbf{,}\mathbf{000}\mathbf{}$pounds of cereal and pounds of meat are available each month. The only production bottleneck is that the factory can only package $\mathbf{110}\mathbf{,}\mathbf{000}$bags of Frisky Pup per month. Needless to say, management would like to maximize profit.

(a) Formulate the problem as a linear program in two variables.

(b) Graph the feasible region, give the coordinates of every vertex, and circle the vertex maximizing profit. What is the maximum profit possible?

Short answer

a)The problem as a linear program in two variables is given below.

b). The graph from above constraints is showing below of feasible region, give the coordinates of every vertex, and circle the vertex maximizing profit.

Step-by-step solution

Defining Maximum Profit

Let the $packageproducebyFriskyPuppermonth=F.$The package contains 1 pound of cereal and cost $1\$/pound$and 1.5 pound of meat and cost $2\$/pound.$

So,we will achieve maximum profit if we have:$1.6F+1.4H$

Here,

On solving above conditions in the question, maximum profit is:$22.2\times {10}^4$

So total production cost include cost of production of cereal and meat and package cost($1.40\$$). This givelocalid="1658924309919" $\mathrm{total}\mathrm{cost}\mathrm{of}\mathrm{production}\mathrm{of}\mathrm{a}\mathrm{package}=1+2\times 1.5+1.40=5.4\$.$.

Hence,localid="1658924239764" $\mathrm{profit}\mathrm{obtain}\mathrm{in}\mathrm{selling}\mathrm{a}\mathrm{package}=\mathrm{Selling}\mathrm{price}-\mathrm{Cost}\mathrm{of}\mathrm{production}.$

$Profit=7-5.4=1.6\$$

So,$maximumprofitofF=1.6F$

Let $\mathrm{the}\mathrm{package}\mathrm{produce}\mathrm{by}\mathrm{Husky}\mathrm{Hound}\mathrm{per}\mathrm{month}=\mathrm{H}$.The package contain 2 pound of cereal and cost $1\$/pound$and 1 pound of meat and costlocalid="1658981373747" $2\$/pound.$So total production cost include cost of production of cereal and meat and package cost $(0.6\$)$.This give localid="1658981380385" $\mathrm{total}\mathrm{cost}\mathrm{of}\mathrm{production}\mathrm{of}\mathrm{a}\mathrm{package}=2\times 1+2+0.6=4.6\$.$Hence, $profitobtaininsellingapackage=Sellingprice-Costofproduction.$$1.6F+1.4H$.

$Profit=6-4.6=1.4\$$

So, max profit$ofH=1.4H$

Therefore, max profitlocalid="1658923761946" $=1.6F+1.4H$

This is our optimal goal to achieve.

Define constraints

Total cereal available in a month $F+2H\le 240000$pounds. Also, Frisky Pup and Husky Hound contain 1 pound and 2 pound cereal respectively.

So, the constraint is: $F+2H\le 240000$

Total meat available in a month$H\le 180000$pounds. Also, Frisky Pup and Husky Hound contain $1.5$ pound and 1 pound meat respectively. So,

the constraint is: $H\le 180000$.

Frisky Pug can produce at the most localid="1658982614215" $F\le 110000$package.

This can be expressed as:$F\le 110000$ .

And the package quantity can never be negative.

Thus, our required linear program is:

$$\begin{gathered} \mathrm{Maximum}: \\ 1.6\mathrm{H}+1.4\mathrm{H} \end{gathered}$$

Constraints

$$\begin{gathered} \left[1\right]F+2H\le 240000 \\ \left[2\right]1.5F+H\le 180000 \\ \left[3\right]F\le 110000 \\ \left[4\right]F\ge 0 \\ \left[5\right]H\ge 0 \end{gathered}$$

The graph is given as:

The graph from above constraints is:

In this graph,$\left("\right)$denote power of 4 .

From the graph, it is clear that we will obtain maximum profit from

$(\mathrm{F},\mathrm{H})=(6\times {10}^4,9\times {10}^4)$

So, our maximum profit will be:

$$\begin{gathered} 1.6F+1.4H=1.6\times 6\times {10}^4+1.4\times 9\times {10}^4 \\ =22.2\times {10}^4 \end{gathered}$$