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Worked Examples · Example 12

Q.Solve the following Linear Programming Problem Graphically. Maximize Z=15x+10yZ = 15x + 10y Subject to 4x+6y≤3604x + 6y \leq 360 3x≤1803x \leq 180 5y≤2005y \leq 200 x≥0x \geq 0, and y≥0y \geq 0

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✓ Free question

Plot the feasible region defined by the constraints, identify the corner points, evaluate the objective function at each vertex, and select the maximum. The optimal solution is x=60, y=20\boxed{x = 60, \, y = 20} with maximum value Z=1100\boxed{Z = 1100}.

Why the graphical method works

Linear programming finds the best outcome (maximum or minimum) of a linear objective function subject to linear constraints. The fundamental theorem tells us that if an optimal solution exists, it must occur at a corner point (vertex) of the feasible region. This is because the objective function is a plane, and as we slide it across the feasible region, the last point of contact before leaving the region is always a vertex.

For two-variable problems, we can visualize this geometrically: plot each constraint as a boundary line, shade the feasible side, find where they all overlap, then test the corners.

Step-by-step solution

1. Convert inequalities to boundary equations

Each constraint inequality becomes an equation for its boundary line:

  • 4x+6y=3604x + 6y = 360 (simplifies to 2x+3y=1802x + 3y = 180)
  • 3x=1803x = 180 (vertical line at x=60x = 60)
  • 5y=2005y = 200 (horizontal line at y=40y = 40)
  • x=0x = 0 (the yy-axis)
  • y=0y = 0 (the xx-axis)

2. Find intercepts for each boundary

For 2x+3y=1802x + 3y = 180:

  • When x=0x = 0: y=60y = 60, giving point (0,60)(0, 60)
  • When y=0y = 0: x=90x = 90, giving point (90,0)(90, 0)

For 3x=1803x = 180: vertical line through (60,0)(60, 0) and (60,y)(60, y) for all yy

For 5y=2005y = 200: horizontal line through (0,40)(0, 40) and (x,40)(x, 40) for all xx

3. Determine the feasible region

Test a point (say the origin) in each inequality to find which side is feasible:

  • 4(0)+6(0)=0≤3604(0) + 6(0) = 0 \leq 360 ✓ (origin side is feasible)
  • 3(0)=0≤1803(0) = 0 \leq 180 ✓ (left of the line)
  • 5(0)=0≤2005(0) = 0 \leq 200 ✓ (below the line)
  • x≥0x \geq 0, y≥0y \geq 0 (first quadrant)

The feasible region is the intersection of all these half-planes in the first quadrant.

4. Identify corner points

The vertices of the feasible region occur where boundary lines intersect:

Intersection ofCoordinatesCheck feasibility
x=0x = 0, y=0y = 0(0,0)(0, 0)All constraints satisfied ✓
x=0x = 0, 5y=2005y = 200(0,40)(0, 40)Check: 4(0)+6(40)=240≤3604(0) + 6(40) = 240 \leq 360 ✓
x=0x = 0, 2x+3y=1802x + 3y = 180(0,60)(0, 60)Check: 5(60)=300>2005(60) = 300 > 200 ✗
3x=1803x = 180, y=0y = 0(60,0)(60, 0)All constraints satisfied ✓
3x=1803x = 180, 5y=2005y = 200(60,40)(60, 40)Check: 4(60)+6(40)=480>3604(60) + 6(40) = 480 > 360 ✗
3x=1803x = 180, 2x+3y=1802x + 3y = 180(60,20)(60, 20)2(60)+3(20)=1802(60) + 3(20) = 180 ✓; 5(20)=100≤2005(20) = 100 \leq 200 ✓
5y=2005y = 200, 2x+3y=1802x + 3y = 180(30,40)(30, 40)4(30)+6(40)=3604(30) + 6(40) = 360 ✓; 3(30)=90≤1803(30) = 90 \leq 180 ✓
2x+3y=1802x + 3y = 180, y=0y = 0(90,0)(90, 0)Check: 3(90)=270>1803(90) = 270 > 180 ✗
Watch out

Not every intersection of two boundary lines is a corner of the feasible region. Always verify that the point satisfies all constraints before including it.

The valid corner points are: (0,0)(0, 0), (0,40)(0, 40), (30,40)(30, 40), (60,20)(60, 20), and (60,0)(60, 0).

5. Evaluate the objective function at each corner

Z=15x+10yZ = 15x + 10y

Corner pointZ=15x+10yZ = 15x + 10y
(0,0)(0, 0)15(0)+10(0)=015(0) + 10(0) = 0
(0,40)(0, 40)15(0)+10(40)=40015(0) + 10(40) = 400
(30,40)(30, 40)15(30)+10(40)=85015(30) + 10(40) = 850
(60,0)(60, 0)15(60)+10(0)=90015(60) + 10(0) = 900
(60,20)(60, 20)15(60)+10(20)=900+200=110015(60) + 10(20) = 900 + 200 = 1100

6. Select the maximum

The maximum value of ZZ is 11001100, occurring at the point (60,20)(60, 20).

Figure 8.7 — bounded pentagon feasible region for Maximize Z = 15x + 10y solved by the iso-profit method
Figure 8.7 — bounded pentagon feasible region for Maximize Z = 15x + 10y solved by the iso-profit method

Sliding the iso-profit line up-right across the bounded pentagon, the maximum Z = 1100 is reached at B(60, 20).

Tip

In graphical LP, once you've plotted the feasible region, you can also visualize the objective function as a family of parallel lines 15x+10y=k15x + 10y = k for different values of kk. Slide this line outward (for maximization) until it just touches the last corner of the feasible region.

✓Final answer

The maximum value is Z=1100\boxed{Z = 1100} at x=60, y=20\boxed{x = 60, \, y = 20}.

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