Question

Solve the following linear programming problem graphically: Maximise $Z=7x+10y$ subject to the constraints
$4x+6y\le 2406x+3y\le 240x\ge 10x\ge 0,y\ge 0$

Answer 1

Here objective function of the given LPP is :
Maximise $Z=7x+10y$
Subject to the constraints:
$4x+6y\le 240or2x+3y\le 1206x+3y\le 240or2x+y\le 80x\ge 10x\ge 0,y\ge 0$
Consider $2x+3y=120$
Table of solutions is :

$\begin{array}{ccc}x& 60& 0\\ y& 0& 40\end{array}$
Consider $2x+y=80$

Table of solutions is :

$\begin{array}{ccc}x& 40& 0\\ y& 0& 80\end{array}$

And $x=10$

To solve the LPP, we draw the graph of the equations and get the feasible solution shown (shaded) in the graph. Corner points of the common shaded region are $A(10,0),B(40,0),C(30,20)$ and D $(10,\frac{100}{3})$.

Value of $Z$ at each corner point is given as:

$\begin{array}{cc}AtA(10,0)& Z=7\left(10\right)+0=70\\ AtB(40,0)& Z=7\left(40\right)+0=280\\ AtC(30,20)& Z=7\left(30\right)+10\left(20\right)=410\\ AtD(-10,\frac{100}{3})& Z=7\left(10\right)+10\left(\frac{100}{3}\right)\\ & =403.33\end{array}$



Hence, $Z=410$ is the maximum value obtained, when $x=30$ and $y=20$