Homework Chapter 7
Question A
7-6-6
Question B
Max TT= X1 + X2
Max TT = 1.25 X1 + 1.70 X2
St = 1.5 X1 + 3 X2 < 300
4.2 X1 + 3X2 < 597
X1
< 165
X2 > 60
X1
> 0
X2 > 0
PROFIT
P1
P2
P3
MIN FA
X1
1.25
1.5
4.2
X1
X1
NNX1
NNX2
X2
1.70
3
X2
3
X2
P1
RHS
< 300
> 60
< 597
< 165
> 0 NNX1
> 0 NNX1
1.5X1 + 3X2 = 300
1.5X1 + 3X2/3 = 300/3
15X1 = 100
X1 = 100
P2
P3
X2 = 60
4.2 X1 + 3X2 = 597
4.2 X1 / 4.2 + 3X2 = 597/4.2
3X2 = 142.14
X2 = 142.14
1
4.2 X1 + 3X2 = 597
4.2 X1 + 3X2 / 3 = 597/3
4.2X1 = 199
X1 = 199
___________________
P1
X1
X2
0
100
200
0
___________________
P2 X2 = 60
___________________
P3
X1
X2
0
199
142.14
0
___________________
Standing orders X1 = 165 = MIN FA
NNX1
0
NNX2
0
Max TT : 1.25 X1 + 1.70 X2
SLOPE = C1 =
C2
1.25 = 1.2500 = 125 = This goes on the graph…
1.70 = 1.7000 = 170
You can not start to solve for the optimal corner point (solutions for X! & X2) until you
have your graph completed. When you do find the optimal corner point on your graph then to solve for the values of X1 & X2 you frist must write down the two Binding
consrraints (ie the two equations that Cross at your selected Optimal corner point.
Eg.
Dough: 1.5X1 + 3X2 = 300
Min X2:
X2 -= 60
Then continue to solve for X1 and X2.
This is what you were doing below but you made a couple algebra errors.
2
1. When you divide1.5 X1 by 1.5 You must also then divide all terms in the equations
by 1.5
ie,. 1.5 X1 / 1.5 + 3X2 / 1.5 = 300 / 1.5
giving: X1 + 2 X2 = 200
Then you can substitute in the Min X2 equation: X2 = 60 and solve for X!.
..
1.5 X1 + 3X2 = 300
1.5 X1/ 1.5 + 3X2 = 300 + 1.5 = 301.5
X1 + 3(60) = 301.5
X1 + 180/180 = 301.5 – 180
X1 = 121.50
New Max TT: 1.25 X1 + 1.70 X2
1.25 (121.50) + 1.70 (60)
151.88
+ 102 = TT = 253.88
The above blue answers will be used on the graph…..
Question I
New TT = 2.50 X1 + 1.70 X2=
2.50 (121.50) + 1.70 (60) =
303.75 + 102 = 405.75
Slope C1/C2 = 250 / 170
No change on the graph…..
Question J
New TT : 1.25 X1 + 3.00 X2 =
1.25(121.50) + 3.00 (60) =
151.88 + 180
3
Max profit = 331.88
Slope: C1/C2 = 125 / 300 this goes on the graph…….
4