Q.Graph the feasible region of the system of linear inequalities $x+y\le5$, $x\ge0$, $y\ge0$, and identify its corner points. If $Z=3x+5y$ is to be maximised over this region, find the maximum value of $Z$.

Combine the half-plane of $x+y\le5$ with the non-negative quadrant to get a triangular feasible region, then evaluate $Z$ at each of its three corner points.

Step 1 — Find the boundary line and its intercepts. $x+y=5$ meets the x-axis at $(5,0)$ (putting $y=0$) and the y-axis at $(0,5)$ (putting $x=0$).

Step 2 — Test the origin. $0+0=0\le5$, true, so the origin's side of the line is shaded.

Step 3 — Combine with $x\ge0, y\ge0$. Restricting to the first quadrant, the feasible region is the closed triangle bounded by the y-axis, the x-axis, and the line $x+y=5$, with corner points $O(0,0)$, $(5,0)$, $(0,5)$.

Step 4 — Evaluate $Z=3x+5y$ at each corner point.

The feasible region is bounded (a triangle), so by the Corner-Point Theorem, the maximum of $Z$ occurs at one of these three points. The largest value is $Z=25$, at $(0,5)$. …

The coefficient-comparison reasoning in the Independent check — when two corner points lie on the same constraint line, the one favouring the higher-coefficient variable gives the larg …

Assuming the maximum of $Z$ always occurs at the corner point with the largest $x$-value (here $(5,0)$) is a common but wrong shortcut — which corner gives the maximum depends entirely on the objective …