Exercise 3

Correct: $L(x,y,\lambda)=xy-\lambda(5x+y-25)$. We differentiate with respect to the variables $x$, $y$ and $\lambda$ and put the derivatives equal to zero:

• $L'_x(x,y,\lambda)=y-5\lambda=0$,
  
• $L'_y(x,y,\lambda)=x-\lambda=0$,
  
• $L'_{\lambda}(x,y,\lambda)=-5x-y+25=0$.

From $L'_x(x,y,\lambda)=y-5\lambda=0$ it follows that $y=5\lambda$ and from $L'_y(x,y,\lambda)=x-\lambda=0$ it follows that $x=\lambda$. We plug this into $L'_{\lambda}(x,y,\lambda)=-5x-y+25=0$ and that gives $\lambda=2\frac{1}{2}$. Hence, $x=2\frac{1}{2}$ and $y=12\frac{1}{2}$. $z(2\frac{1}{2},12\frac{1}{2})=31\frac{1}{4}$. The boundaries result in $z(0,25)=0$ and $z(5,0)=0$. Consequently, $z(2\frac{1}{2},12\frac{1}{2})=31\frac{1}{4}$ is the maximum with shadow price $\lambda=2\frac{1}{2}$.

Go on.

Wrong: The Lagrange function is $L(x,y,\lambda)=xy-\lambda(5x+y-25)$.

Try again.

Wrong: $y=5\lambda$.

Try again.

Wrong: $y=5\lambda$.

Try again.