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?

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.

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 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:

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