Exercise 4

Correct: $L(x,y,\lambda)=-xy+2-\lambda(x^2+y-27)$. We differentiate with respect to the variables $x$, $y$ and $\lambda$:

• $L'_x(x,y,\lambda)=-y-2\lambda x$,
  
• $L'_y(x,y,\lambda)=-x-\lambda$,
  
• $L'_{\lambda}(x,y,\lambda)=-x^2-y+27$.

We put the first-order derivatives equal to zero and solve the system: $L'_y(x,y,\lambda)=-x-\lambda=0$ gives $x=-\lambda$. We plug this into $L'_x(x,y,\lambda)=-y-2\lambda x=0$ and that gives $y=2\lambda^2$. We plug that into $L'_{\lambda}(x,y,\lambda)=-x^2-y+27=0$, which results in $\lambda^2=9$, and hence, $\lambda=3$ or $\lambda=-3$. Since $x\geq 0$ and $x=-\lambda$ it must hold that $\lambda=-3$, $x=3$ and hence, $y=18$.$z(3,18)=-52$

The boundaries give: $z(0,27)=2$ and $z(\sqrt{27},0)=2$. Hence, $z(3,18)=-52$ is a minimum and the corresponding shadow price is $\lambda=-3$.

Go on.

Wrong: $x\geq 0$ and $x=-\lambda$.

Try again.

Wrong: $y \neq 2\lambda$.

Try again.

Wrong: $y \neq 2\lambda$.

Try again.