Q.Solve the linear programming problem formulated in Worked Example 4 by the graphical (corner-point) method, and state the maximum profit.

Graph the feasible region of the LPP from Worked Example 4, find all its corner points, and evaluate $Z=40x+30y$ at each — the largest value is the maximum profit.

Finding the corner points. The feasible region is bounded by $2x+y=100$, $x+3y=120$, the x-axis, and the y-axis.

On the x-axis ($y=0$): $2x+y=100$ gives $x=50$; $x+3y=120$ gives $x=120$. Since both constraints must hold, the smaller value governs, so the region reaches only as far as $(50,0)$ along the x-axis (checking: at $(50,0)$, $x+3y=50\le120$, true).

On the y-axis ($x=0$): $2x+y=100$ gives $y=100$; $x+3y=120$ gives $y=40$. The smaller value governs, so the region reaches only as far as $(0,40)$ along the y-axis (checking: at $(0,40)$, $2x+y=40\le100$, true).

Intersection of the two constraint lines. Solving $2x+y=100$ and $x+3y=120$ simultaneously: from the first, $y=100-2x$; substituting,

So the two lines meet at $(36,28)$, which lies between the two axis points found above and is therefore also a corner point of the feasible region.

Evaluating $Z=40x+30y$ at each corner point: …

The single most common error at this stage is assuming the maximum must occur at one of the two axis intercepts (the all-tables or all-chairs extremes) and never actually computing the interior corner point $(36,28) …