ResoluciΓ³n de la Segunda PrΓ‘ctica de EconomΓa MatemΓ‘tica IV 1 1. Hallar la extremal π(π₯) = β«0 (π₯ 2 + π₯Μ 2 )ππ‘ a) Funcio
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ResoluciΓ³n de la Segunda PrΓ‘ctica de EconomΓa MatemΓ‘tica IV 1
1. Hallar la extremal π(π₯) = β«0 (π₯ 2 + π₯Μ 2 )ππ‘ a) Funcional
b) EcuaciΓ³n de Euler
Sujeto a: π₯(0) = 0 π₯(1) = 1
π = π₯ 2 + π₯Μ 2 ππ₯ = 2π₯ ππ₯ = 2π₯Μ πππ₯ = 2π₯Μ ππ‘ πππ₯ ππ₯ = ππ‘ 2π₯ = 2π₯Μ π₯ = π₯Μ π₯ β π₯Μ = 0
c) Hallando raΓces caracterΓsticas π 2 + (0)π β 1 = 0 π2 = 1 π = 1 ; π = β1 d) SoluciΓ³n π₯ = π΄0 π βπ‘ + π΄1 π π‘ e) Reemplazando las condiciones π₯(0) = π΄0 π (0) + π΄1 π (0) = 0 π΄0 + π΄1 = 0 π΄0 = βπ΄1 π₯(1) = π΄0 π (β1) + π΄1 π (1) = 1 f) Hallando Constantes βπ΄1 π (β1) + π΄1 π (1) = 1 π΄1 (π (β1) + π) = 1 1 π΄1 = (β1) π +π π΄1 = 0.324 π΄0 = βπ΄1 , π΄0 = β0.32 g) Hallando la extremal π₯ = π΄0 π βπ‘ + π΄1 π π‘ π₯ = β0.32π βπ‘ + 0.32π π‘
2. Hallar la extremal
2
π½(π) = β«0 (π₯ 2 + 2π₯Μ π₯ + π₯Μ 2 ) ππ‘ π₯(0) = 1 , π₯(2) = 4
a) El Funcional El ejercicio es no lineal
π = π₯ 2 + 2π₯Μ π₯ + π₯Μ 2 π = (π₯ + π₯)Μ2 ππ₯Μ = 2(π₯ + π₯)Μ(π₯) ππ₯Μ = 2π₯ ππ₯Μ > 0 El ejercicio es de MinimizaciΓ³n
b) Por lo tanto el Optimo serΓ‘ π₯(π‘) = π΄0 π‘ + π΄1 c) Reemplazando las condiciones π₯(0) = π΄0 (0) + π΄1 = 1 π΄1 = 1 π₯(2) = π΄0 (2) + 1 = 4 π΄0 =
3 2
d) Hallando la extremal 3 π₯(π‘) = π‘ + 1 2
3. Optimizar
3
π(π¦) = β«1 (π₯ 2 (1 β π₯Μ )2 ) ππ‘
a) El Funcional
π₯(1) = 0 , π₯(3) = 2
π = π₯ 2 (1 β π₯Μ )2 π = (π₯ β π₯π₯Μ )2 ππ₯Μ = 2(π₯ β π₯π₯Μ )(βπ₯) = β2π₯ 2 + 2π₯ 2 π₯Μ ππ₯ΜΜ = 2π₯ 2
ππ₯Μ = 2π₯ ππ₯Μ > 0 El ejercicio es de MinimizaciΓ³n b) Por lo tanto el Optimo serΓ‘ π₯(π‘) = π΄0 π‘ + π΄1 c) Reemplazando las condiciones π₯(1) = π΄0 (1) + π΄1 = 0 π΄1 = βπ΄0 π₯(3) = π΄0 (3) β π΄0 = 2 π΄0 = 1 d) Hallando la extremal π₯(π‘) = π΄0 π‘ + π΄1 π₯(π‘) = π‘ β 1
e) Hallando el MΓnimo 3
π(π¦) = β«(π₯ β π₯π₯Μ )2 ππ‘ 1 3
π(π¦) = β«((π‘ β 1) β (π‘ β 1)(1))2 ππ‘ 1 3
π(π¦) = β«((π‘ β 1) β (π‘ β 1))2 ππ‘ 1 3
π(π¦) = β« 0 ππ‘ 1
π(π¦) = 0 β΄ El Γ³ptimo mΓnimo valor es Cero. 1
4. Hallar el valor optimo a) El funcional
π(π₯) = β«0 (2π₯ + π₯Μ 2 ) ππ‘
π = ππ + πΜ π
ππ₯ = 2 ππ₯Μ = 2π₯Μ πππ₯ = 2π₯Μ ππ‘ b) EcuaciΓ³n de Euler
ππ =
π
ππ π
π
2 = 2π₯Μ π₯Μ = 1
c) Integrando β« π₯Μ = β« 1
β
π₯Μ = π‘ + π΄0
β« π₯Μ = β« π‘ + π΄0 π=
ππ + ππ¨π + π¨π π
d) Hallando Constantes (0)2 + (0)π΄0 + π΄1 = 0 β π¨π = π 2 1 π π₯(1) = + (1)π΄0 + (0) = π β π¨π = π΅ β 2 π π₯(0) =
π₯(0) = 0 , π₯(1) = π
Hallando la condiciΓ³n de transversalidad
ππ = π³ππππ , π» = ππππ β΄ Entonces ππ₯Μ = 0 , ππ₯Μ = 2π₯Μ 2π₯Μ = 0
π₯(1) =
π₯Μ = 0 π₯Μ = π‘ + π΄0 = 0 π΄0 = 0 β π‘ 1 β β1=π 2 1 β =π 2
π‘2 + (1)(βπ‘) + (0) 2
π¨π = π΅ β
π π π π β π¨π = β β = β π π π π
e) Hallando el extremal
π‘2 + π‘(βπ‘) + 0 2 ππ π(π) = β ππ π
π₯(π‘) =
f) Hallando del optimo π(π) =
ππ π
β ππ
πΜ (π) =
ππ β π
ππ = π β ππ β πΜ (π) = βπ
1
π(π₯) = β«(2π₯ + π₯Μ 2 ) ππ‘ 0 1
π(π₯) = β« (2 ( 0 1
π‘2 β π‘ 2 ) + (βπ‘)2 ) ππ‘ 2
π(π₯) = β«(π‘ 2 β 2π‘ 2 ) + (βπ‘)2 ) ππ‘ 0 1
π(π₯) = β«(β(π‘)2 ) + (βπ‘)2 ) ππ‘ 0 1
π(π₯) = β« 0 ππ‘ 0
π(π₯) = 0 β΄ El Γ³ptimo mΓnimo valor es Cero.