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UPSC 2017 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

Using graphical method, find the maximum value of $2x+y$ subject to

Technique

Graphical LPP — plot half-planes, identify feasible polygon, evaluate the linear objective at each corner (fundamental theorem of LP: optimum at a vertex).

Solution

For a linear objective over a (bounded) convex polygon, the optimum occurs at a vertex (corner point) of the feasible region. We find the feasible polygon, list its vertices, and evaluate $Z=2x+y$ at each.

Step 1 — Boundary lines

• $L_1:\ 4x+3y=12$ — intercepts $(3,0)$ and $(0,4)$.
• $L_2:\ 4x+y=8$ — intercepts $(2,0)$ and $(0,8)$.
• $L_3:\ 4x-y=8$ — intercepts $(2,0)$ and $(0,-8)$.
• Axes $x=0,\ y=0$.

Step 2 — Identify binding constraints near the optimum

The objective $Z=2x+y$ increases to the upper-right, so the optimum lies on the “north-east” boundary, formed by $L_1$ and $L_2$ (with $x,y\ge0$). Note $L_3:\,4x-y\le8$ is slack in this region. Find the intersection of $L_1$ and $L_2$:

So $L_1\cap L_2=\left(\tfrac32,\,2\right)$. Check feasibility: $4x-y=6-2=4\le8$ ✓ (so $L_3$ is indeed non-binding here).

Step 3 — Corner points of the feasible region

The feasible region (intersection of all half-planes with $x,y\ge0$) is the polygon with vertices

(Why these: along the x-axis the tightest bound is $x\le2$ from $L_2,L_3$; the top-right edge is $L_2$ from $A$ up to $B$, then $L_1$ from $B$ to $C$; the left edge is the y-axis up to $(0,4)$, since $L_1$ gives the tightest $y$-intercept.)

Step 4 — Evaluate $Z=2x+y$ at the vertices

The maximum is at $B=\left(\tfrac32,2\right)$:

Answer