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The corner points of the bounded feasible region of the LPP: Maximize z=x+yz = x + y subject to constraints 2x+5y≤1002x + 5y \leq 100, 8x+5y≤2008x + 5y \leq 200, x≥0x \geq 0, y≥0y \geq 0 are

Solution

✅ Correct Option: 1

The corner points of the feasible region occur where the boundary lines intersect.

The boundary lines from the constraints are:

  • 2x+5y=1002x + 5y = 100
  • 8x+5y=2008x + 5y = 200
  • x=0x = 0 (y-axis)
  • y=0y = 0 (x-axis)

Intersection of x=0x = 0 and y=0y = 0:

Point: (0,0)(0, 0)


Intersection of x=0x = 0 and 2x+5y=1002x + 5y = 100:

5y=1005y = 100

y=20y = 20

Point: (0,20)(0, 20)


Intersection of x=0x = 0 and 8x+5y=2008x + 5y = 200:

5y=2005y = 200

y=40y = 40

Point: (0,40)(0, 40)

Checking 2x+5y≤1002x + 5y \leq 100:

2(0)+5(40)=200≰1002(0) + 5(40) = 200 \not\leq 100

This point is outside the feasible region.


Intersection of y=0y = 0 and 2x+5y=1002x + 5y = 100:

2x=1002x = 100

x=50x = 50

Point: (50,0)(50, 0)

Checking 8x+5y≤2008x + 5y \leq 200:

8(50)+5(0)=400≰2008(50) + 5(0) = 400 \not\leq 200

This point is outside the feasible region.


Intersection of y=0y = 0 and 8x+5y=2008x + 5y = 200:

8x=2008x = 200

x=25x = 25

Point: (25,0)(25, 0)


Intersection of 2x+5y=1002x + 5y = 100 and 8x+5y=2008x + 5y = 200:

8x+5y=2008x + 5y = 200

−(2x+5y=100)-(2x + 5y = 100)

6x=1006x = 100

x=503x = \frac{50}{3}

Substituting into 2x+5y=1002x + 5y = 100:

2(503)+5y=1002\left(\frac{50}{3}\right) + 5y = 100

1003+5y=100\frac{100}{3} + 5y = 100

5y=100−10035y = 100 - \frac{100}{3}

5y=20035y = \frac{200}{3}

y=403y = \frac{40}{3}

Point: (503,403)\left(\frac{50}{3}, \frac{40}{3}\right)


The corner points of the bounded feasible region are:

(0,0)(0,0), (25,0)(25,0), (0,20)(0,20), (503,403)\left(\frac{50}{3}, \frac{40}{3}\right)

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