91Ó°ÊÓ

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: $Râ‰\yen 2S$.

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

$R+S≤3000$.

As we know that quantity can never be negative. So $R,Sâ‰\yen 0$.

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

Maximize$R+1.5S$.

Constraints:$Râ‰\yen 2S$

role="math" localid="1657275042140" $$\begin{gathered} R+S≤3000 \\ R,Sâ‰\yen 0 \end{gathered}$$

Constraints:

$$\begin{gathered} \left[1\right]R+S≤3000 \\ \left[2\right]Râ‰\yen 2S \\ \left[3\right]R,Sâ‰\yen 0 \end{gathered}$$

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 $maximunrevenue=4000\$$.