Exercise 5

Correct: $L(x,y,\lambda)=y^3x-\lambda(5x^2+\frac{15}{8}y^4-50)$. We differentiate with respect to the variables $x$, $y$ en $\lambda$:

• $L'_x(x,y,\lambda)=y^3-10\lambda x$,
  
• $L'_y(x,y,\lambda)=3y^2x-7\frac{1}{2}\lambda y^3$,
  
• $L'_{\lambda}(x,y,\lambda)=-5x^2-\frac{15}{8}y^4+50$.

$L'_x(x,y,\lambda)=y^3-10\lambda x=0$ gives $x=\dfrac{y^3}{10\lambda}$. We plug this into $L'_y(x,y,\lambda)=3y^2x-7\frac{1}{2}\lambda y^3=0$ and solving gives $y=5\lambda$ (with $x=12\frac{1}{2}\lambda^2$) or $y=-5\lambda$ (with $x=-12\frac{1}{2}\lambda^2$). We plug this into $L'_{\lambda}(x,y,\lambda)=-5x^2-\frac{15}{8}y^4+50=0$, which gives $\lambda=\frac{2}{5}$, $x=2$, $y=2$. $z(2,2)= 16$. We check the boundaries: $z(\sqrt{10},0)=0$ and $z(0,\sqrt[4]{26\frac{2}{3}})=0$ and hence, $z(2,2)= 16$ is the maximum. The corresponding shadow price is $\lambda=\frac{2}{5}$.

Go on.

Wrong: The shadow price is not equal to the value of $x$ (or $y$) at the maximum.

See Extra explanation: Schadow price.

Wrong: The shadow price is not the maximum value.

See Extra explanation: Schadow price.

Wrong: The correct answer is among them.

Try again.