The document describes an optimization problem to maximize the function (fA - dA)(fB - dB) subject to the constraint that (fA, fB) belongs to the set R and is greater than or equal to the target (dA, dB). For three cases where the target is (15,10), (15,10) with additional constraint on uA, and (10,10), it shows solving for the optimal value of fA and the corresponding fB that maximizes the function.

2.
R d
(R, d)
B
f : B → R2
f

3.
f
max
f∈R:( f1,f2)≥(d1,d2)
(f1 − d1)(f2 − d2)

4. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0,uB ≥ 0} d = (dA, dB) = (15, 10)
R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0, uA ≤ 90, uB ≥ 0} d = (dA, dB) = (15, 10)
R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0, uB ≥ 0} d = (dA, dB) = (10, 10)

5. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0,uB ≥ 0} d = (dA, dB) = (15, 10)
uA
uB
1500
10
15
100

6. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0,uB ≥ 0} d = (dA, dB) = (15, 10)
uA
uB
100
1500
10
15
u′A
u′B
u′B = 3(uB − 10) = 3uB − 30
u′A = 2(uA − 15) = 2uA − 30 240
240
0
−30

7. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0,uB ≥ 0} d = (dA, dB) = (15, 10)
uA
uB
100
1500
10
15
u′B = 3(uB − 10) = 3uB − 30
u′A = 2(uA − 15) = 2uA − 30
u′A
u′B
240
240
0
−30

8. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0,uB ≥ 0} d = (dA, dB) = (15, 10)
uA
uB
100
1500
10
15
u′B = 3(uB − 10) = 3uB − 30
u′A = 2(uA − 15) = 2uA − 30
(120, 120)
u′A
u′B
240
240−30
0

9. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0,uB ≥ 0} d = (dA, dB) = (15, 10)
uA
uB
100
1500
10
15
u′B = 3(uB − 10) = 3uB − 30
u′A = 2(uA − 15) = 2uA − 30
(120, 120)
u′A
u′B
240
240−30
0
50
75
(75, 50)

10. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0, uA ≤ 90, uB ≥ 0} d = (dA, dB) = (15, 10)
uB
100
1500
10
15
50
75
(75, 50)
uA
90

11. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0, uB ≥ 0} d = (dA, dB) = (10, 10)
uA
uB
100
1500
10
10
u′B = 3(uB − 10) = 3uB − 30
u′A = 2(uA − 10) = 2uA − 20
u′A
u′B
250
250−20
0
−30

12. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0, uB ≥ 0} d = (dA, dB) = (10, 10)
uA
uB
100
1500
10
u′B = 3(uB − 10) = 3uB − 30
u′A = 2(uA − 10) = 2uA − 20
u′A
u′B
250
250−20
0
−30
10

13. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0, uB ≥ 0} d = (dA, dB) = (10, 10)
uA
uB
100
1500
10
u′B = 3(uB − 10) = 3uB − 30
u′A = 2(uA − 10) = 2uA − 20
u′A
u′B
250
250−20
0
−30
(125, 125)
10

14. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0, uB ≥ 0} d = (dA, dB) = (10, 10)
uA
uB
100
1500
10
u′B = 3(uB − 10) = 3uB − 30
u′A = 2(uA − 10) = 2uA − 20
u′A
u′B
−20
0
−30
250
250
(125, 125)
10

15. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0, uB ≥ 0} d = (dA, dB) = (10, 10)
uA
uB
100
1500
10
u′B = 3(uB − 10) = 3uB − 30
u′A = 2(uA − 10) = 2uA − 20
u′A
u′B
−20
0
−30
(
145
2
,
155
2
)
250
250
(125, 125)
10

16. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0,uB ≥ 0} d = (dA, dB) = (15, 10)
max
f∈R:(fA,fB)≥(15,10)
(fA − 15)(fB − 10)
⇒ max
f∈R:(fA,fB)≥(15,10)
(fA − 15)(
300 − 2fA
3
− 10)
⇔ max
f∈R:(fA,fB)≥(15,10)
1
3
(fA − 15)(270 − 2fA) 13575
fA
(fA, fB) = (75, 50)

17. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0, uA ≤ 90, uB ≥ 0} d = (dA, dB) = (15, 10)
max
f∈R:(fA,fB)≥(15,10)
(fA − 15)(fB − 10)
⇒ max
f∈R:(fA,fB)≥(15,10)
(fA − 15)(
300 − 2fA
3
− 10)
⇔ max
f∈R:(fA,fB)≥(15,10)
1
3
(fA − 15)(270 − 2fA) 13575
fA
(fA, fB) = (75, 50)
90

18. R = {(uA, uB)|2uA + 3uB ≤ 300, uA ≥ 0, uB ≥ 0} d = (dA, dB) = (10, 10)
max
f∈R:(fA,fB)≥(10,10)
(fA − 10)(fB − 10)
⇒ max
f∈R:(fA,fB)≥(10,10)
(fA − 10)(
300 − 2fA
3
− 10)
⇔ max
f∈R:(fA,fB)≥(10,10)
1
3
(fA − 10)(270 − 2fA) 135145
2
fA
(fA, fB) = (
145
2
,
155
2
)

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