Converting inequations into equations and drawing the corresponding lines.
3x + 5y = 15, 5x + 2y = 20 i.e. \(\frac{x}{5}+\frac {y}{3}=1,\,\frac {x}{4}+\frac {y}{10} =1\)
As x ≥ 0, y ≥ 0 solution lies in first quadrant.
Let us draw the graph of the above equations.
B is the point of intersection of the lines 3x + 5y = 15 and
5x + 2y = 20, i.e. \(B= \left(\frac {70}{19},\frac {15}{19}\right )\)
∴ z has maximum value 12.63 at only one point i.e. \(B\left(\frac {70}{19},\frac {15}{19}\right)\).
