Answers:
a) vertices: (400, 500); (0, 100); (1400,0)
b) production levels that yield the maximum profit and maximum profit:
x = 1400, y = 0, P = 18,700.
Explanation
Part A.
1) State the restrictions:
i) The production of y must exceed the production of x by at least 100 units:
⇒ y ≥ x + 100
ii) Production levels are limited by the formula x+2y ≤ 1400.
⇒ y ≤ 700 - x/2
iii) x and y cannot be negative:
⇒ x ≥ 0 and y ≥ 0
2) The feasible region is the area inside the lines defined by:
y = x + 100, y = 700 - x/2, y = 0, and x = 0
3) The vertices are found by solving 3 separated systems of equations:
i) intersection of lines y = x + 100 and y = 700 - x/2
⇒ x + 100 = 700 - x/2
⇒ 3x/2 = 600 ⇒ x= 1200 / 3 = 400
⇒ y = 400 + 100 = 500
⇒ vertex = (400, 500)
ii) intersection of lines y = x + 100 and x = 0
⇒ x = 0 and y = 100
⇒ vertex = (0, 100)
iii) intersection of lines y = 700 - x/2 and y = 0
⇒ y = 0 ⇒ x/2 = 700 ⇒ x = 1400
⇒ vertex = (1400, 0)
Therefore the vertices of the feasible region are:
(400, 500); (0, 100); (1400,0)
Part B.
The production levels that yield the maximum profit, and the maximum profit are found by replacing the vertices in the profit equation. The vertex that yields the maximum profit is the solution
1) vertex (400, 500)
P = 14x + 22y - 900 = 14(400) + 22(500) - 900 = 15,700
2) vertex (0, 100)
P = 14x + 22y - 900 = 14(0) + 22(100) - 900 = 1,300 (clearly less profit)
3) vertex (1400, 0)
P = 14x + 22y - 900 = 14(1400) + 22(0) - 900 = 18,700, which is the highest profit.
