Q18

1 markMCQSection A

The maximum value of the function $z = 7x + 5y$ , subject to the constraints $x \le 3, y \le 2, x \ge 0, y \ge 0$ is :

Q: The maximum value of the function $z = 7x + 5y$, subject to the constraints $x \le 3, y \le 2, x \ge 0, y \ge 0$ is :

The correct option is **(C) 31**. **Feasible region:** - The constraints $0 \le x \le 3$ and $0 \le y \le 2$ form a rectangle with corners $(0,0)$, $(3,0)$, $(3,2)$, $(0,2)$. **Evaluate $z = 7x + 5y$ at each corner:** - $(0,0)$: 0 - $(3,0)$: 21 - $(0,2)$: 10 - $(3,2)$: $21 + 10$ = **31 (maximum)**

Q: How is the feasible region formed from these constraints?

The constraints $x \ge 0$, $y \ge 0$ restrict us to the first quadrant, while $x \le 3$ and $y \le 2$ cap the variables. Together they enclose a rectangle with vertices (0,0), (3,0), (3,2) and (0,2), which is the feasible region.

Linear Programming

Linear Programming — Maximum at a Corner Point

Options

(A)10

(B)21

(D)29

Official Answer

The correct option is (C) 31.

Feasible region:

The constraints $0 \le x \le 3$ and $0 \le y \le 2$ form a rectangle with corners $(0,0)$ , $(3,0)$ , $(3,2)$ , $(0,2)$ .

Evaluate $z = 7x + 5y$ at each corner:

$(0,0)$ : 0
$(3,0)$ : 21
$(0,2)$ : 10
$(3,2)$ : $21 + 10$ = 31 (maximum)

linear programmingobjective functionfeasible regioncorner point theoremmaximum valueconstraintsvertex evaluation

Marking Scheme

11 mark: correct option (C) 31, ideally with z evaluated at the corner $(3,2)$ .
2Accept the working $7(3)+5(2)=31$ as sufficient justification.

Hint

The region is a rectangle $0 \le x \le 3$ , $0 \le y \le 2$ ; test z at each corner and pick the largest.

Quick Oral Answer

Since both coefficients are positive, z is maximum at the top-right corner $(3,2)$ of the rectangular feasible region, giving 7 times 3 plus 5 times 2, which equals 31.

Analysis & Explanation

This is a Linear Programming Problem where the objective function attains its maximum at a corner (vertex) of the feasible region — a direct application of the Corner Point Theorem.

Why (C) is correct:

Both coefficients (7 and 5) are positive, so z grows as x and y grow. The largest allowed values are $x = 3$ and $y = 2$ , giving $z = 7(3) + 5(2) = 31$ .

Why the distractors are wrong:

(A) 10 — value at $(0,2)$ , taking only y at maximum and $x = 0$ .
(B) 21 — value at $(3,0)$ , taking only x at maximum and $y = 0$ .
(D) 29 — arises from a slip such as $7(2)+5(3) = 29$ (swapping the x and y bounds).

Exam trap: With a simple rectangular region you can spot the answer, but always confirm by testing all corner points rather than assuming.

Common Mistakes

1Substituting only one variable at its maximum and setting the other to zero, giving 21 or 10 instead of testing the corner (3,2).
2Swapping the bounds of x and y, e.g. computing $7(2)+5(3)=29$ .
3Assuming the maximum lies inside the region rather than at a vertex — for a linear objective the optimum is always at a corner point.

Interesting Facts

The Corner Point Theorem guarantees that if an optimal solution to an LPP exists, it occurs at a vertex of the feasible region — this is why we only test corners, not interior points.

Linear programming was developed during the 1940s by George Dantzig, who created the Simplex method to solve large-scale optimisation problems for military logistics.

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Frequently Asked Questions

Why do we only check corner points in a linear programming problem?

Because the objective function is linear, its optimum value over a convex feasible region always occurs at a vertex (corner point). The Corner Point Theorem lets us avoid checking infinitely many interior points and simply compare the objective value at each finite vertex.

How is the feasible region formed from these constraints?

The constraints $x \ge 0$ , $y \ge 0$ restrict us to the first quadrant, while $x \le 3$ and $y \le 2$ cap the variables. Together they enclose a rectangle with vertices (0,0), (3,0), (3,2) and (0,2), which is the feasible region.