Moe is deciding how much Regular Duff beer and how much Duff Strong beer to order each week. Regular Duff costs Moe \(1 per pint and he sells it at \)2 per pint; Duff Strong costs Moe $1.50 per pint and he sells it at per pint. However, as part of a complicated marketing scam, the Duff company will only sell a pint of Duff Strong for each two pints or more of Regular Duff that Moe buys. Furthermore, due to past events that are better left untold, Duff will not sell Moe more than 3,000 pints per week. Moe knows that he can sell however much beer he has. Formulate a linear program for deciding how much Regular Duff and how much Duff Strong to buy, so as to maximize Moe’s profit. Solve the program geometrically.

Let Moe order ‘R ’ regular duff beer per week.

It is given that Cost Price of Beer per Pint is=1$ and Selling Price of Beer per Pint is=2$

So profit obtain in regular duff beer per pints=2$-1$=1$

Let Moe order ‘’ Strong duff beer per week.

It is given that Cost Price of Beer per Pint is=1.5$ and Selling Price of Beer per Pint is=3$

So profit obtain in regular duff beer per pints=3$-1.5$=1.5$ .

This means that Moe earns profit by selling regular and strong duff beer=R+1.5S

It is said that, ‘Duff company will only sell a pint of Duff Strong for each two pints or more of Regular Duff that Moe buys’. This can be represented as: $R\ge 2S$.

Also, ‘Duff will not sell Moe more than 3,000 pints per week’. This can be express as:.

$R+S\le 3000$.

As we know that quantity can never be negative. So $R,S\ge 0$.

So, our objective is to maximize the profit i.e.,

Maximize$R+1.5S$.

Constraints:$R\ge 2S$

role="math" localid="1657275042140" $$\begin{gathered} R+S\le 3000 \\ R,S\ge 0 \end{gathered}$$

Constraints:

$$\begin{gathered} \left[1\right]R+S\le 3000 \\ \left[2\right]R\ge 2S \\ \left[3\right]R,S\ge 0 \end{gathered}$$

From graph we can clearly see by interaction of constraints [1] and [2] at point $\left(R,S\right)=\left(1000,2000\right)$we will get our MAX revenue.

On, putting value of $\left(R,S\right)$inrole="math" localid="1657275260361" $R+1.55$, we have:

$1000+1.5x2000=1000+3000=4000\$$

Thus our $maximunrevenue=4000\$$.