Q.Two students solve the linear programming problem: Maximize $Z = 5x + 3y$ subject to $x + y \le 6,\ 2x + y \le 9,\ x, y \ge 0$. Student A evaluates $Z$ only at $(0,0)$ and $(6,0)$ and concludes the maximum value is $30$. Student B finds all the corner points correctly. Find the true maximum value of $Z$, and explain the error in Student A's working.

Constraints: $x+y\le6,\ 2x+y\le9,\ x,y\ge0$.

Checking Student A's point $(6,0)$: $x+y=6\le6$ ✓, but $2x+y=12+0=12$, and $12>9$ — this violates $2x+y\le9$. So $(6,0)$ is not even a feasible point, let alone the correct corner at the x-axis; Student A used only the first constraint's x-intercept and never checked it against the second constraint.

Finding the true corner points. At $y=0$: first constraint gives $x\le6$, second gives $x\le4.5$; the binding (smaller) bound is $x\le4.5$, giving the true vertex $(4.5,0)$ — check first: $4.5\le6$, slack.

At $x=0$: first gives $y\le6$, second gives $y\le9$; binding is $y\le6$, giving vertex $(0,6)$ — check second: $6\le9$, slack.

Intersection: $x+y=6,\ 2x+y=9$; subtracting, $x=3,\ y=3$, giving $(3,3)$ — this is exactly the corner point Student A never computed.

Corners: $(0,0),\ (4.5,0),\ (3,3),\ (0,6)$. …

Never accept a candidate corner point without checking it against EVERY constraint of the problem, not just the one it happens to lie on; always include the inter …

Using only one constraint's axis intercept as 'the' corner point without checking it against the other constraint (Student A's exact mistake); stop …