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Solving Linear Programming Problems Graphically

To solve the linear programming problem
Minimise  
\[z=30a+40b\]
  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
\[30a+40b=30 \times 4.2+40 \times 3.6=270\]