Solve the following linear programming problem graphically: Minimize $Z = 13x - 15y$ subject to constraints $x + y \le 7$ , $2x - 3y + 6 \ge 0$ , $x \ge 0$ , $y \ge 0$ .

Q: Solve the following linear programming problem graphically: Minimize $Z = 13x - 15y$ subject to constraints $x + y \le 7$, $2x - 3y + 6 \ge 0$, $x \ge 0$, $y \ge 0$.

The minimum value of Z is **−30**, attained at the corner point **(0, 2)**. **Feasible region:** Bounded by $x \ge 0$, $y \ge 0$, $x+y \le 7$, and $2x-3y+6 \ge 0$, with corner points $(0,0)$, $(7,0)$, $(3,4)$, and $(0,2)$. **Corner point evaluation of $Z = 13x - 15y$:** - $(0,0)$: $Z = 0$ - $(7,0)$: $Z = 91$ - $(3,4)$: $Z = 39-60 = -21$ - $(0,2)$: $Z = 0-30 =$ **−30 (minimum)**

Linear Programming

Linear Programming Problem — Graphical Method

Official Answer

The minimum value of Z is −30, attained at the corner point (0, 2).

Feasible region: Bounded by $x \ge 0$ , $y \ge 0$ , $x+y \le 7$ , and $2x-3y+6 \ge 0$ , with corner points $(0,0)$ , $(7,0)$ , $(3,4)$ , and $(0,2)$ .

Corner point evaluation of $Z = 13x - 15y$ :

$(0,0)$ : $Z = 0$
$(7,0)$ : $Z = 91$
$(3,4)$ : $Z = 39-60 = -21$
$(0,2)$ : $Z = 0-30 =$ −30 (minimum)

linear programming problemgraphical methodfeasible regioncorner point theoremobjective function minimisationbounded region

Marking Scheme

11 mark: correct plotting of both constraint lines and identification of the feasible region (with correct shading direction, verified at the origin).
21 mark: correct identification of all four corner points $(0,0)$ , $(7,0)$ , $(3,4)$ , $(0,2)$ .
31 mark: correct evaluation of Z at each vertex and correct identification of the minimum value $-30$ at $(0,2)$ .

Hint

Plot the boundary lines $x+y=7$ and $2x-3y+6=0$ , shade the feasible region in the first quadrant, and evaluate Z at each corner point.

Quick Oral Answer

The feasible region is a quadrilateral with vertices $(0,0)$ , $(7,0)$ , $(3,4)$ , $(0,2)$ ; evaluating $Z = 13x-15y$ at each gives the minimum value $-30$ at $(0,2)$ .

Analysis & Explanation

This is a standard bounded-region LPP requiring correct plotting of two intersecting half-planes and vertex evaluation.

Concept: By the Corner Point Theorem, the optimal value of a linear objective function over a bounded convex feasible region always occurs at a vertex. Here the feasible region is a quadrilateral bounded by the two given lines and the coordinate axes.

Exam trap: A frequent error is misplotting the inequality $2x-3y+6 \ge 0$ — testing the origin gives $2(0)-3(0)+6 = 6 \ge 0$ , true, so the origin's side is included; students sometimes shade the wrong side. Another common mistake is missing the vertex $(0,2)$ , assuming the feasible region only touches $x+y=7$ and the axes.

Real-world link: Minimisation LPPs of this type model real cost-minimisation problems, such as minimising transportation or production cost subject to resource and demand constraints.

Common Mistakes

1Shading the wrong side of the inequality $2x-3y+6 \ge 0$ , leading to an incorrect feasible region and wrong vertices.
2Missing the vertex $(0,2)$ and only checking $(0,0)$ , $(7,0)$ and $(3,4)$ , leading to an incorrect minimum.
3Sign errors while computing $Z = 13x-15y$ at negative-contributing vertices like $(0,2)$ , turning $-30$ into $+30$ .

Interesting Facts

The Corner Point Theorem underlying this method is part of linear programming theory developed alongside George Dantzig's simplex method in 1947.

Companies like airlines and logistics firms use the same corner-point logic daily, scaled up via the simplex algorithm, to minimise fuel and operating costs subject to scheduling constraints.

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Frequently Asked Questions

How do you decide which side of a constraint line to shade?

Substitute a test point not on the line (usually the origin) into the inequality; if it satisfies the inequality, shade the side containing that point, otherwise shade the opposite side.

Why is it enough to check only the corner points of the feasible region?

By the Corner Point (Extreme Point) Theorem, a linear objective function over a bounded convex region attains its maximum and minimum only at vertices of the region, so checking all vertices is sufficient.