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UPSC 2016 Maths Optional Paper 2 Q1e — Step-by-Step Solution

10 marks · Section A

Graphical method · Linear Programming · asked 5× in 13 yrs · Read the full method →

Question

Find the maximum value of $5x+2y$ with constraints $x+2y\ge1,\ 2x+y\le1,\ x\ge0$ and $y\ge0$ by graphical method.

Technique

Graphical LP — plot half-planes, identify feasible polygon, test each candidate vertex for feasibility, evaluate objective at feasible vertices.

Solution

A two-variable LP is solved graphically: sketch the feasible region (intersection of the half-planes), find its corner points (vertices), and evaluate the objective at each — the optimum of a bounded LP occurs at a vertex.

Step 1 — Boundary lines and feasible region

The constraint boundaries are

• $x+2y\ge1$: the side of $L_1$ away from the origin (origin gives $0\ge1$, false).
• $2x+y\le1$: the side of $L_2$ toward the origin (origin gives $0\le1$, true).
• $x\ge0,\ y\ge0$: first quadrant.

Intercepts: $L_1$ meets axes at $(1,0)$ and $(0,\tfrac12)$; $L_2$ at $(\tfrac12,0)$ and $(0,1)$.

Step 2 — Corner points of the feasible region

The feasible region is bounded by $L_1$ (below), $L_2$ (above), and the $y$-axis. Candidate vertices come from intersecting active constraints:

(a) $L_1\cap y$-axis ($x=0$): $2y=1\Rightarrow(0,\tfrac12)$. Check $2x+y=\tfrac12\le1$ ✓ — feasible.

(b) $L_2\cap y$-axis ($x=0$): $y=1\Rightarrow(0,1)$. Check $x+2y=2\ge1$ ✓ — feasible.

(c) $L_1\cap L_2$: solve $x+2y=1,\ 2x+y=1$. Subtract: $(x+2y)-(2x+y)=0\Rightarrow y-x=0\Rightarrow x=y$; then $3x=1\Rightarrow x=y=\tfrac13$. Point $(\tfrac13,\tfrac13)$, both $\ge0$ ✓ — feasible.

(d) $L_2\cap x$-axis $(\tfrac12,0)$ and $L_1\cap x$-axis $(1,0)$: check $x+2y\ge1$ at $(\tfrac12,0)$: $\tfrac12\ge1$ false — infeasible; $(1,0)$ gives $2x+y=2\le1$ false — infeasible. So no vertex on the $x$-axis.

Thus the feasible region is the triangle with vertices

Step 3 — Evaluate $z=5x+2y$ at the vertices

The maximum is at $(\tfrac13,\tfrac13)$:

Answer