Solve the linear programming problem by the method of corners.

Question

Maximize: P=6x+8y

Subject to:

x+3y>60

9x+5y<320

x>10

y>0

Answer 1

All conditions should have equalities, not inequalities.

We can look

The four lines

x + 3y = 60

9x + 5y = 320

x = 10

y = 0

have six intersections

1)

x + 3y = 60

9x + 5y = 320

9x + 27y = 540

22y = 220

y = 10

x + 3(10) = 60

x = 30

(30, 10)

Does this meet the other two conditions:

x >= 10 Y

y >= 0 Y

Y

2)

x + 3y = 60

x = 10

10 + 3y = 60

y = 50/3

(10, 50/3)

Does this meet the other two inequalities

9x + 5y <= 320

9x + 5y = 9(10) + 5(50/3) = 173 1/3 Y

y >= 0 Y

Y

3)

x + 3y = 60

y = 0

x + 3*0 = 60

x = 60

(60, 0)

Does this meet the other 2 conditions?

9x + 5y <= 320

9x + 5y = 9(60) + 5(0) = 540 N

x = 10 Y

N

4)

9x + 5y = 320

x = 10

9(10) + 5y = 320

90 + 5y = 320

y = 46

(10, 46)

x + 3y >= 60

10+3(46) = 148 Y

y >= 0 Y

Y

5)

9x + 5y = 320

y = 0

9x = 320

x = 320/9 or35 5/9

(320/9, 0)

x + 3y >= 60

x + 3y = 35 5/9 + 0 = 35 5/9 N

x >= 10 Y

N

6)

x = 10

y = 0

(10, 0)

Does this meet the conditions:

x + 3y >= 60 10 + 3(0) = 10 N

9x + 5y <= 320 9(10) + 5(0) = 90 Y

N

Our corners are:

(30, 10), (10, 50/3), (10, 46)

We now calculate the function to maximize

P = 6x + 8y

For (30, 10), P = 6(30) + 8(10) = 260

For (10, 50/3), P = 6(10) + 8(50/3) = 193 1/3

For (10, 46), P = 6(10) + 8(46) = 428

As 428 is the largest value, the maximum is 428 at (10, 46)