Exercise 4 | Wiskunde op Tilburg University

Determine the shadow price corresponding to the solution of the following constrained extremum problem.

Minimize $z(x,y)=-xy+2$
Subject to $x^2+y=27$
Where $x,y\geq 0$

Antwoord 1 correct

Correct

Antwoord 2 optie

$\lambda=3$

Antwoord 2 correct

Fout

Antwoord 3 optie

$\lambda=-1+\frac{1}{2}\sqrt{112}$

Antwoord 3 correct

Fout

Antwoord 4 optie

$\lambda=-1-\frac{1}{2}\sqrt{112}$

Antwoord 4 correct

Fout

Antwoord 1 optie

$\lambda=-3$

Antwoord 1 feedback

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.

Antwoord 2 feedback

Wrong:

$x\geq 0$ and

$x=-\lambda$ .

Try again.

Antwoord 3 feedback

Wrong:

$y \neq 2\lambda$ .

Try again.

Antwoord 4 feedback

Wrong:

$y \neq 2\lambda$ .

Try again.