Maximize: \(z=60x+50y\)Subject to: \(x+2y\le 40\)\(3x+2y\le 60\)\(x\ge 0,\ y\ge 0\)

Answer: Maximum \(z=1350\) at \(x=10,\ y=15\)

Line \(x+2y=40\) meets the axes at \((40,0)\) and \((0,20)\). Line \(3x+2y=60\) meets the axes at \((20,0)\) and \((0,30)\). Both inequalities are satisfied by the origin, so the feasible region lies on the origin side of both lines, in the first quadrant.

[Figure: rough sketch: the feasible region is the quadrilateral OABC with O(0, 0), A(20, 0), B(10, 15), C(0, 20).]

Point B: subtracting \(x+2y=40\) from \(3x+2y=60\) gives \(2x=20\Rightarrow x=10\), \(y=15\).

\(z\) is maximum at B: \(z_{\max}=1350\) when \(x=10,\ y=15\).