1) The document discusses maximizing a function f(uA) of a single variable uA on a bounded domain U. 2) It is found that f(uA) is maximized when uA = (2/5)2/3, giving a maximum value of 3/5. 3) The point (uA, uB) = ((2/5)2/3, 3/5) is therefore the optimal solution.

1. ήʔϜཧ࿦#4*$ԋश
ަ‫͚͓ʹࡁܦ׵‬Δަবղ

2. φογϡަবղ
໰୊
ղ౴

3. ަব໰୊ͷఆٛ
ఆٛɿަব໰୊͸
࣮‫ݱ‬Մೳू߹ɿ
ަবͷෆҰக఺ɿ
ͷ૊ ʹΑͬͯఆٛ͞ΕΔ
Ծఆ
w ͸࣍‫ۭؒ਺࣮ݩ‬ ͷίϯύΫτͰತͳ෦෼ू߹
w
w ͷ఺ ͕ଘࡏͯ͠
U
d = (d1
, d2
)
(U, d)
U ℜ2
d ∈ U
U u = (u1
, u2
) ui
di
(i = 1, 2)
0
u2
u1
(d1
, d2
) = (20, 20)
U
100
100

4. ަবղ͕ຬͨ͢΂͖ͭͷެཧ
ύϨʔτ࠷దੑ
͢΂ͯͷ ʹରͯ͠ Ͱ গͳ͘ͱ΋ͭͷ ʹରͯ͠
Ͱ͋Δ ͷ఺ ͸ଘࡏ͠ͳ͍
ରশੑ
ަব໰୊ ͕ରশͳΒ͹
ˎަব໰୊͕ରশͰ͋Δͱ͸ ͳΒ͹
ਖ਼࣍ม‫͔׵‬Βͷಠཱੑ
ަব໰୊ ͕ަব໰୊ ͔Βޮ༻ͷਖ਼࣍ม‫׵‬ ʹΑͬͯಘΒΕΔͱ͖
ແؔ܎ͳ݁Ռ͔Βͷಠཱੑ
̎ͭͷަব໰୊ ͱ ʹ͓͍ͯ ͱ͢Δ
΋͠ ͳΒ͹ Ͱ͋Δ
i = 1, 2 ti
≥ fi
(U, d) i
ti
fi
(U, d) U t = (t1
, t2
)
(U, d) f1
(U, d) = f2
(U, d)
d1
= d2
(u1
, u2
) ∈ U (u2
, u1
) ∈ U
(U′

, d′

) (U, d) ui′

= αi
ui
+ βi
, αi
0 (i = 1,2)
fi
(U′

, d′

) = αi
fi
(U, d) + βi
, αi
0 (i = 1,2)
(U, d) (T, d) T ⊂ U
f(U, d) ∈ T f(T, d) = f(U, d)

5. ަবղ͕ຬͨ͢΂͖ͭͷެཧ
ఆཧ
ަব໰୊ ͷ଒ ͱ͢Δ
ͭͷެཧΛຬͨ͢ަব໰୊ͷղ ͕Ұҙʹଘࡏ͠
͢΂ͯͷަব໰୊ ʹରͯ͠ ͸
ͷղͰ͋Δ
͜ͷղΛφογϡަবղͱ͍͏
·ͨ Λφογϡੵͱ͍͏
(U, d) B
f : B → ℜ2
(U, d) f(U, d)
max
u∈U:u≥d
(u1
− d1
)(u2
− d2
)
(u1
− d1
)(u2
− d2
)

6. ̎ͭͷ෼ׂՄೳͳࡒ ࡒ ࡒ Λަ‫͢׵‬Δࢢ৔Λߟ͑Δ
͜͜Ͱ͸ ਓͷফඅऀ ফඅऀ ফඅऀ# ͕͓Γ ॳ‫ظ‬อ༗ΛͦΕͧΕ
ิ଍ɿ ͱॻ͍ͨͱ͖ ফඅऀ ͷॳ‫ظ‬อ༗ͷ͜ͱͰ͋Γ ͕ࡒͷྔ ͕ࡒͷྔʹͳΔ
֤ফඅऀͷޮ༻ؔ਺Λ ͱ͢Δ
͜ͷਓͷফඅऀ͕ަব͢Δͱߟ͑ φογϡަবղʹΑΔࡒͷ഑෼Λ‫ٻ‬ΊΑ
ͳ͓ ަবܾ྾࣌͸ॳ‫ظ‬อ༗Λ࣋ͬͨ··ͱ͢Δ
eA
= (eA
1 , eA
2 ) = (1,0) eB
= (eB
1 , eB
2 ) = (0,1)
ei
= (ei
1, ei
2) i ei
1 ei
2
uA
(x1, x2) = x
1
3
1
x
1
3
2
uB
(y1, y2) = y
1
2
1
y
1
2
2
໰୊

7. ̎ͭͷ෼ׂՄೳͳࡒ ࡒ ࡒ Λަ‫͢׵‬Δࢢ৔Λߟ͑Δ
͜͜Ͱ͸ ਓͷফඅऀ ফඅऀ ফඅऀ# ͕͓Γ ॳ‫ظ‬อ༗ΛͦΕͧΕ
ิ଍ɿ ͱॻ͍ͨͱ͖ ফඅऀ ͷॳ‫ظ‬อ༗ͷ͜ͱͰ͋Γ ͕ࡒͷྔ ͕ࡒͷྔʹͳΔ
֤ফඅऀͷޮ༻ؔ਺Λ ͱ͢Δ
͜ͷਓͷফඅऀ͕ަব͢Δͱߟ͑ φογϡަবղʹΑΔࡒͷ഑෼Λ‫ٻ‬ΊΑ
ͳ͓ ަবܾ྾࣌͸ॳ‫ظ‬อ༗Λ࣋ͬͨ··ͱ͢Δ
ަবܾ྾࣌ͷޮ༻͸ҎԼʹͳΔɿ
eA
= (eA
1 , eA
2 ) = (1,0) eB
= (eB
1 , eB
2 ) = (0,1)
ei
= (ei
1, ei
2) i ei
1 ei
2
uA
(x1, x2) = x
1
3
1
x
1
3
2
uB
(y1, y2) = y
1
2
1
y
1
2
2
dA
= uA
(1,0) = 1
1
30
1
3 = 0 dB
= uB
(0,1) = 0
1
21
1
2 = 0
ղ౴

8. ֤ফඅऀͷޮ༻ؔ਺Λ ͱ͢Δ
࣮‫ݱ‬Մೳू߹Λߟ͑ΔύϨʔτ࠷దͷ৚݅ΑΓ ֤ফඅऀͷ‫ݶ‬ք୅ସ཰͕౳͘͠ͳΔ͜ͱ͔Β
·ͨ ॳ‫ظ‬อ༗Λաෆ଍ͳ͘ফඅ͢Δ͜ͱ͕࠷దͳͨΊ ҎԼͷࣜΛຬͨ͢
Ώ͑ʹ ͜Ε͸ ύϨʔτ࠷దͳ഑෼Ͱ͸ ࡒͱࡒͷফඅྔ͕Ұக͢Δ͜ͱΛҙຯ
·ͨ ΋੒Γཱͭ
ͱ͓͚͹ ΑΓ
Ͱ͋Δ͜ͱʹ஫ҙͯ͠
uA
(x1, x2) = x
1
3
1
x
1
3
2
uB
(y1, y2) = y
1
2
1
y
1
2
2
∂uA
/∂x1
∂uA/∂x2
=
∂uB
/∂y1
∂uB/∂y2
⇔
x1
x2
=
y1
y2
x1 + y1 = 1 x2 + y2 = 1
x1
x2
=
1 − x1
1 − x2
⇔ x1 = x2
y1 = y2
x1 = x2 = t uA
= x
1
3
1
x
1
3
2
= t
1
3t
1
3 = t
2
3 uB
= y
1
2
1
y
1
2
2
= (1 − x1)
1
2(1 − x2)
1
2 = (1 − x1) = (1 − t) uB
= 1 − uA
3
2
0 ≤ t ≤ 1 0 ≤ uA
≤ 1 0 ≤ uB
≤ 1
ղ౴

9. ͸ύϨʔτ࠷దͷ෦෼Ͱ͋Γ ࡒͷഁ‫غ‬Λߟྀͯ͠ ࣮‫ݱ‬Մೳू߹͸
ަবͷෆҰக఺
ΑΓ ্ʹತͰ‫ݮ‬গ
uB
= 1 − uA
3
2
U = {(uA
, uB
)|uB
≤ 1 − uA
3
2
, 0 ≤ uA
, 0 ≤ uB
}
(dA
, dB
) = (0, 0)
duB
duA
= −
3
2
uA
1
2
0
d2
uB
duA2
= −
3
4
uA− 1
2
0
ղ౴
0
uB
uA
1
1

10. φογϡੵ࠷େԽ໰୊Λߟ͑Δ
TU
ύϨʔτ࠷దੑ͔Β Λຬͨ͢ͷͰ
ͱ͓͚͹ Ұ֊৚݅ΑΓ
͜ͷͱ͖
max
uA
,uB
(uA
− dA
)(uB
− dB
) = max
uA
,uB
uA
uB
uB
≤ 1 − uA
3
2
, 0 ≤ uA
, 0 ≤ uB
uB
= 1 − uA
3
2
max
uA
uA
(1 − uA
3
2
)
f(uA
) = uA
(1 − uA
3
2
) = uA
− uA
5
2
f′

(uA
) = 1 −
5
2
uA
3
2
= 0 ⇔ uA
=
(
2
5)
2
3
uB
= 1−
(
2
5)
2
3
3
2
=
3
5
ղ౴
0
uB
uA
1
1
3
5
(
2
5 )
2
3
≈ 0.54

11. ͜ΕΛຬͨ͢഑෼͸ Ͱ
ΑΓ
ʹ஫ҙ͢Ε͹
͜ͷަ‫Ͱࡁܦ׵‬ͷ‫ڝ‬૪‫ߧۉ‬഑෼͸
ͱͳΓ Ұக͠ͳ͍
uA
=
(
2
5 )
2
3
uB
=
3
5
x1 = x2 = t
uA
= t
2
3 =
(
2
5 )
2
3
t =
2
5
x1 + y1 = 1 x2 + y2 = 1
(x1, x2) =
(
2
5
,
2
5)
(y1, y2) =
(
3
5
,
3
5)
(x1, x2) =
(
1
2
,
1
2)
(y1, y2) =
(
1
2
,
1
2)
ղ౴
0
uB
uA
1
1
3
5
(
2
5 )
2
3
≈ 0.54

12. ิ଍ ‫ݶ‬ք୅ସ཰.34
0A
2
1
ফඅऀͷແࠩผ‫ۂ‬ઢ
ޮ༻ਫ४͕ಉ͡ࡒͷ૊
0A
2
1
ࡒΛ୯Ґ૿΍ͨ͠ͱ͖ʹ
ޮ༻ਫ४ΛอͭͨΊʹ
‫ݮ‬Β͢ࡒͷྔʹ୅ସ཰
1
୅ସ཰
0A
2
1
઀ઓͷ܏͖ʹ‫ݶ‬ք୅ସ཰
‫ݶ‬ք୅ସ཰ແࠩผ‫ۂ‬ઢͷ܏͖
−
dx2
dx1
=
∂u/∂x1
∂u/∂x2
dx1
dx2

13. .34ʹࡒͷ‫ݶ‬քޮ༻ࡒͷ‫ݶ‬քޮ༻
ͷಋग़
‫ݶ‬քޮ༻ͱ͸ ୯ҐͷࡒΛ૿΍ͨ͠ͱ͖ͷޮ༻ͷ૿ՃྔͰ͋Δ
૯ޮ༻ͷ૿ՃྔΛ ͱ͓͘ͱ ͚֤ͩࡒ͕มԽͨ͠ͱ͢Δͱ
ࠓ ޮ༻ਫ४ΛมԽͤ͞ͳ͍ͱ͖ ແࠩผ‫ۂ‬ઢ্ͷͱ͖ Ͱ͋Δ͔Β
−
dx2
dx1
=
∂u/∂x1
∂u/∂x2
du dx1 dx2
du =
∂u
∂x1
dx1 +
∂u
∂x2
dx2
du = 0
0 =
∂u
∂x1
dx1 +
∂u
∂x2
dx2 ⇔ −
dx2
dx1
=
∂u/∂x1
∂u/∂x2
ิ଍ ‫ݶ‬ք୅ସ཰.34

14. ิ଍ ΤοδϫʔεɾϘοΫε
0A
2
1
0B
2
1
ফඅऀͷແࠩผ‫ۂ‬ઢ
ޮ༻ਫ४͕ಉ͡ࡒͷ૊
0A
2
1
0B
2
1
ফඅऀ#ͷແࠩผ‫ۂ‬ઢ
ޮ༻ਫ४͕ಉ͡ࡒͷ૊
0A
2
1
0B
2
1
ফඅऀ#ͷޮ༻ਫ४Λ
Լ͛Δ͜ͱͳ͘
ফඅऀͷޮ༻ਫ४Λߴ͘Ͱ͖Δ
ύϨʔτվળͱ͍͏
ӈ্΄Ͳޮ༻͸ߴ͍
ࠨԼ΄Ͳޮ༻͸ߴ͍
ύϨʔτվળͰ͖ͳ͍ঢ়ଶΛ
ύϨʔτޮ཰ͱ͍͍
ແࠩผ‫ۂ‬ઢ͸઀͢Δ‫ݶ‬ք୅ସ཰͕Ұக

15. ফඅऀͷޮ༻ؔ਺Λ ͱ͢Δͱ
Ώ͑ʹ ‫ݶ‬ք୅ସ཰͸
uA
(x1, x2) = x
1
3
1
x
1
3
2
∂uA
(x1, x2)
∂x1
=
1
3
x− 2
3
1
x
1
3
2
∂uA
(x1, x2)
∂x2
=
1
3
x
1
3
1
x−2
3
2
∂uA
(x1, x2)
∂x1
∂uA(x1, x2)
∂x2
=
1
3
x− 2
3
1 x
1
3
2
1
3
x
1
3
1 x−2
3
2
=
x2
x1
ิ଍ ภඍ෼

16. ফඅऀͷॳ‫ظ‬อ༗Λ ޮ༻ؔ਺Λ
ޮ༻࠷େԽ͸‫ݶ‬ք୅ସ཰Ձ֨ൺ
ΑΓ
ಉ༷ʹ ফඅऀ#ͷॳ‫ظ‬อ༗Λ ޮ༻ؔ਺Λ
ޮ༻࠷େԽ͸‫ݶ‬ք୅ସ཰Ձ֨ൺ
ΑΓ
eA
= (eA
1 , eA
2 ) = (1,0) uA
(x1, x2) = x
1
3
1
x
1
3
2
∂uA
(x1, x2)
∂x1
∂uA(x1, x2)
∂x2
=
x2
x1
=
p1
p2
p1x1 + p2x2 = p1 ⋅ 1 x1(p1, p2, eA
) =
1
2
x2(p1, p2, eA
) =
1
2
p1
p2
eB
= (eB
1 , eB
2 ) = (0,1) uB
(y1, y2) = y
1
2
1
y
1
2
2
∂uB
(y1, y2)
∂y1
∂uB(y1, y2)
∂y2
=
y2
y1
=
p1
p2
p1y1 + p2y2 = p2 y1(p1, p2, eB
) =
1
2
p2
p1
y2(p1, p2, eB
) =
1
2
ิ଍ धཁؔ਺ͷಋग़
0A
2
1
ফඅऀͷແࠩผ‫ۂ‬ઢ
ޮ༻ਫ४͕ಉ͡ࡒͷ૊
༧ࢉઢɿp1x1 + p2x2 = p1
༧ࢉ੍໿Λຬͨ͢தͰޮ༻࠷େ

17. ‫ڝ‬૪‫ߧۉ‬഑෼Ͱ͸ ֤ফඅऀ͸ޮ༻࠷େԽ৚݅Λຬͨ͠
͔ͭ௒աधཁ͕ͳ͍഑෼Ͱ͋Δ
ޮ༻࠷େԽ৚݅ΑΓ
ফඅऀͷधཁؔ਺
ফඅऀ#ͷधཁؔ਺
௒աधཁ͕ͳ͍͜ͱ͔Β ࡒͷ৚݅ΑΓ
Ώ͑ʹ ͜ΕΛधཁؔ਺ʹ୅ೖ͢Ε͹‫ڝ‬૪‫ߧۉ‬഑෼
x1(p1, p2, eA
) =
1
2
x2(p1, p2, eA
) =
1
2
p1
p2
y1(p1, p2, eB
) =
1
2
p2
p1
y2(p1, p2, eB
) =
1
2
x1(p1, p2, eA
) + y1(p1, p2, eB
) = 1 ⇔
p2
p1
= 1
(x1, x2) =
(
1
2
,
1
2)
(y1, y2) =
(
1
2
,
1
2)
ิ଍ ‫ڝ‬૪‫ߧۉ‬഑෼
0A
2
1
ফඅऀͷແࠩผ‫ۂ‬ઢ
ޮ༻ਫ४͕ಉ͡ࡒͷ૊
༧ࢉઢɿp1x1 + p2x2 = p1
༧ࢉ੍໿Λຬͨ͢தͰޮ༻࠷େ

18. ήʔϜཧ࿦#4*$ԋश
ަ‫͚͓ʹࡁܦ׵‬Δަবղ