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Desafío de Optimización con Derivadas

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Desafío de Optimización con Derivadas

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Problemas de optimización por derivadas

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Andres Alday
Andres Alday
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Desafío de Optimización con Derivadas
 

Desafío de Optimización con DerivadasOnline version

Problemas de optimización por derivadas

by Andres Alday
1

En A(x)= x(10−x) para x en [0,10], ¿dónde se obtiene el máximo de A?

2

R(p)= p(200−p). ¿A qué precio p se maximiza R?

3

Minimiza C(x)= x^2+4x+10 con x≥0. ¿Qué valor de x da el mínimo?

4

Con restricciones x∈[0,4], maximiza y=√x. ¿Qué x da el máximo?

5

Costo promedio A(x)=0.5x+20+100/x. ¿Qué x minimiza A(x)?

6

Para un perímetro fijo de 40, ¿cuál lado x maximiza el área de un rectángulo?

7

P(x)=−x^3+9x^2−24x+6. ¿Qué valores críticos dan solución de optimalidad?

8

f(x)=x^4−8x^3+18x^2. ¿Qué valor de x da mínimo local en x≥0?

9

Maximiza ln(x)−x con x>0. ¿Qué x da el máximo?

10

Una función f(x)=x/(1+x^2). ¿Qué x maximizing f en x>0?

Explicación

Derivada A'(x)=10−2x; A'(5)=0 y es máximo en ese intervalo.

R'(p)=200−2p; igual a 0 en p=100. Es máximo en ese punto.

La parábola es convexa; el mínimo en x≥0 ocurre en la frontera x=0.

y' = 1/(2√x) >0; aumenta hasta el extremo x=4.

A'(x)=0.5−100/x^2; igual a 0 en x^2=200; A''>0 para x>0.

A(x)=x(40−2x); A'(x)=40−4x; igual a 0 en x=10.

P'(x)=−3x^2+18x−24=−3(x−2)(x−4); crítricos en 2 y 4.

f'(x)=4x(x−3)^2; mínimo en x=0, x=3 es punto estacionario con pendiente horizontal pero no cambia signo.

Derivada 1/x−1 = 0 en x=1; es máximo global en x>0.

f'(x)=(1+x^2−2x^2)/(1+x^2)^2=(1−x^2)/(1+x^2)^2; igual a 0 en x=1; máximo en 1.

 
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