## Solving Linear Programming Problems Graphically

To solve the linear programming problemMinimise

\[z=30a+40b\]

subject to the constraints\[2a+b \geq 12\]

\[a+b \geq 9\]

\[a+3b \geq 15\]

and of course

\[a, \: b \geq 0\]

:We plot these inequalities - as equalities - on a graph. We seek to maximise the objective function, so draw the line

\[3x+40y=C\]

for various values of \[C\]

, seeking to find the maximum value of \[C\]

for which the objective function is in or on a boundary of the feasible region (the part of the graph that satisfies all the constraints.This is at the intersection of the lines

\[2a+b=12\]

and \[a+3b=15\]

.Solving the equations

\[2a+b=12\]

\[a+3b=15\]

gives

\[a=4.2, \: b=3.6\]

The value of the objective is

\[30a+40b=30 \times 4.2+40 \times 3.6=270\]