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Worked Examples · Example 11

Q.Minimize Z=3x+5yZ = 3x + 5y Subject to constraints: x,y≥0x, y \geq 0 x+3y−3≥0x + 3y - 3 \geq 0 x+y−2≥0x + y - 2 \geq 0

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This is a linear programming problem where we minimise Z=3x+5yZ = 3x + 5y subject to x+3y≥3x + 3y \geq 3, x+y≥2x + y \geq 2 and x,y≥0x, y \geq 0. The feasible region is unbounded, but the minimum occurs at the corner point (32,12)\left(\tfrac32, \tfrac12\right), giving Z=7Z = 7.

We are asked to minimise Z=3x+5yZ = 3x + 5y subject to x,y≥0x, y \geq 0, x+3y≥3x + 3y \geq 3, and x+y≥2x + y \geq 2. Because the objective and all constraints are linear, the minimum — if it exists — occurs at a corner point of the feasible region. The region here turns out to be unbounded, so after evaluating the corners we confirm that the objective cannot be driven any lower by moving outward.

Step 1 – Draw the boundary lines.

The constraint boundaries are

L1:x+3y=3,L2:x+y=2,L_1: x + 3y = 3, \qquad L_2: x + y = 2,

together with the axes x=0x = 0 and y=0y = 0. Both constraints are "≥\geq" inequalities, so the feasible region lies above both lines within the first quadrant and is unbounded.

Step 2 – Find the feasible corner points.

A corner point is where two boundary lines meet and the point satisfies every constraint. Testing each candidate intersection:

IntersectionPointFeasible?
L1∩L2L_1 \cap L_2(32,12)\left(\tfrac32, \tfrac12\right)Yes — satisfies all constraints
L1∩(y=0)L_1 \cap (y=0)(3,0)(3,0)Yes — x+y=3≥2x+y = 3 \geq 2 ✓
L2∩(x=0)L_2 \cap (x=0)(0,2)(0,2)Yes — x+3y=6≥3x+3y = 6 \geq 3 ✓
L1∩(x=0)L_1 \cap (x=0)(0,1)(0,1)No — x+y=1<2x+y = 1 < 2
L2∩(y=0)L_2 \cap (y=0)(2,0)(2,0)No — x+3y=2<3x+3y = 2 < 3

For L1∩L2L_1 \cap L_2, subtract L2L_2 from L1L_1: (x+3y)−(x+y)=3−2⇒2y=1(x+3y)-(x+y) = 3-2 \Rightarrow 2y = 1, so y=12y = \tfrac12 and x=2−12=32x = 2 - \tfrac12 = \tfrac32.

So the feasible region has exactly three corner points: A(3,0)A(3,0), B(32,12)B\left(\tfrac32, \tfrac12\right) and C(0,2)C(0,2).

Watch out

Not every intersection of two boundary lines is a corner of the feasible region — each candidate must be checked against all the constraints. Here (0,1)(0,1) and (2,0)(2,0) each satisfy the two lines that produce them but violate the remaining constraint, so they are infeasible and are discarded.

Step 3 – Evaluate Z=3x+5yZ = 3x + 5y at each feasible corner.

Corner pointZ=3x+5yZ = 3x + 5y
A(3,0)A(3,0)99
B(32,12)B\left(\tfrac32, \tfrac12\right)92+52=7\tfrac92 + \tfrac52 = 7
C(0,2)C(0,2)1010

The smallest corner value is Z=7Z = 7 at B(32,12)B\left(\tfrac32, \tfrac12\right).

Step 4 – Confirm the minimum over the unbounded region. …

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