To solve the linear programming problem
Minimise  {jatex options:inline}z=30a+40b{/jatex}  subject to the constraints
{jatex options:inline}2a+b \geq 12{/jatex}
{jatex options:inline}a+b \geq 9{/jatex}
{jatex options:inline}a+3b \geq 15{/jatex}
and of course  {jatex options:inline}a, \: b \geq 0{/jatex}:
We plot these inequalities - as equalities - on a graph. We seek to maximise the objective function, so draw the line  {jatex options:inline}3x+40y=C{/jatex}  for various values of  {jatex options:inline}C{/jatex}, seeking to find the maximum value of  {jatex options:inline}C{/jatex}  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  {jatex options:inline}2a+b=12{/jatex}  and  {jatex options:inline}a+3b=15{/jatex}.
Solving the equations
{jatex options:inline}2a+b=12{/jatex}
{jatex options:inline}a+3b=15{/jatex}
gives {jatex options:inline}a=4.2, \: b=3.6{/jatex}
The value of the objective is {jatex options:inline}30a+40b=30 \times 4.2+40 \times 3.6=270{/jatex}