Lm/300 Week 4

Chapter 3, Problem 8
Information from problem 7.
Formulate LPM to determine the number of basketballs and footballs to produce in order to maximize profit.
X1 -- # of basketballs
X2 -- # of footballs
Maximize Z = 12x1 + 16x2
Subject to: 3x1 + 2x2 ≤ 500 4x1 + 5x2 ≤ 800 X1, x2 ≥ 0
Transform this model to standard form.
Maximize Z = 12x1 + 16x2 + 0s1 +0s2 3x1 + 2x2 + s1 = 500 4x1 +5x2 + x2 = 800 X1, x2, s1, s2 ≥ 0
a). Identify the amount of unused resources (slack) at each of the graphical points.
3x1 + 2x2= 500 4x1 + 5x2 = 800
X1 = 0, x2 = 250 X1 = 0, x2 = 160
X2 = 0, x1 = 500/3 X2 = 0, x1 = 200
Maximize Z = 12x1 + 16x2
At point A (0, 0) Z = 0
At point B (500/3, 0) Z = 12(500/3) + 16(0) Z = 2000
At point D (0,
160) Z = 12(0) + 16(160) Z = 2560
At point C (900/7, 400/7) Z = 12(900/7) +16(400/7) Z = 2457

Point D (0, 160) is an optimal solution. X1 = 0, x2 = 160, Z = 2560
Unused resources:
At point D (0, 160)
3x1 + 2x2 + s1 = 500 4x1 +5x2 + s2 = 800
3(0) + 2(160) + s1 = 500 4(0) + 5(160) +s2 = 800
S1 = 500 -320 s2 = 800 - 800
S1 = 180 s2 = 0

At point C (900/7, 400/7)
3(900/7) +2(400/7) + s1 = 500 4(900/7) + 5(400/7) + s2 = 800
2700/7 +800/7 + s1 = 500 3600/7 +2000/7 + s2 = 800
S1 = 3500/7 -500 s2 = 800 – 5600/7
S1 = 0 s2 = 0

At point B (500/3, 0)
3(500/3) + 2(0) + s1 = 500 4(500/3) +5(0) +s2 = 800
500 + s1 = 500 2000/3 + s2 = 800
S1 = 0 s2 = 400/3

At point A (0, 0)
3(0) +2(0) + s1 = 500 4(0) +5(0) +s2 = 800
S1 = 500 s2 = 800

b).
What would be effect on the optimal solution if the profit for a basketball changed from $12 to $13?
Z = 12x1 + 16x2 becomes Z = 13x1 + 16x2
At point D (0, 160) Z = 2560
At point B (500/3, 0) Z = 2166.67
At point C (900/7, 400/7) Z = 2585.71
Point C would become an optimal solution if profit for basketballs changed from $12 to $13.

What would be the effect if the profit for a football changed from $16 to $15?
Z = 12x1 + 16x2 becomes Z = 12x1 + 15x2
At point D (0, 160) Z = 2400
At point B (500/3, 0) Z = 2000
At point C (900/7, 400/7) Z = 2400
Either point D or C would be an optimal solution if profit for football changed from $16 to $15.

c) What would be the effect on the optimal solution if 500 additional pounds of rubber could be obtained?
3x1 + 2x2= 500 becomes 3x1 + 2x2= 1000
3x1 + 2x2= 1000
X1 = 0, x2 = 500
X2 = 0, x1 = 1000/3

4x1 + 5x2 = 800
X1 = 0, x2 = 160
X2 = 0, x1 = 200
Maximize Z = 12x1 +16x2
A (0,0) Z = 0
D (0,160) Z = 2560
B (200,0) Z = 2400
Point D is an optimal solution.

What would be the effect if 500 additional square feet of leather could be obtained?
4x1 + 5x2 = 800 becomes 4x1 + 5x2 = 1300
4x1 + 5x2 = 1300
X1 = 0, x2 =
260
X2 = 0, x1 = 325

3x1 + 2x2= 500
X1 = 0, x2 = 250
X2 = 0, x1 = 500/3
Maximize Z = 12x1 + 16x2
A (0, 0) Z = 0
D (0, 250) Z = 4000
B (500/3, 0) Z = 2000
Point D is an optimal solution.