Exercise 2 | Wiskunde op Tilburg University

Determine all the stationary points of

$z(x,y)=\frac{2}{3}y^3-\frac{1}{3}x^3+4x-y^2x$ .

Antwoord 1 correct

Correct

Antwoord 2 optie

$(2,0)$

Antwoord 2 correct

Fout

Antwoord 3 optie

$(2,0)$ ,

$(-2,0)$ ,

$(2+2\sqrt{3},1)$ and

$(2-2\sqrt{3},1)$

Antwoord 3 correct

Fout

Antwoord 4 optie

$(2,0)$ and

$(-2,0)$

Antwoord 4 correct

Fout

Antwoord 1 optie

$(2,0)$ ,

$(-2,0)$ ,

$(\sqrt{2},\sqrt{2})$ and

$(-\sqrt{2},-\sqrt{2})$

Antwoord 1 feedback

Correct:

$z'_x(x,y)=-x^2+4-y^2$
$z'_y(x,y)=2y^2-2yx$

$z'_y(x,y)=0$ gives

$2y(y-x)=0$ . Hence,

$y=0$ or

$y=x$ .

First we use

$y=0$ .

$z'_x(x,y)=0$ then gives

$-x^2+4=0$ . Hence,

$x=2$ or

$x=-2$ .

Then we use

$y=x$ .

$z'_x(x,y)=0$ then gives

$-x^2+4-x^2=0$ , or

$2x^2=4$ . Hence,

$x=\sqrt{2}$ (with

$y=x=\sqrt{2}$ ) or

$x=-\sqrt{2}$ (with

$y=x=-\sqrt{2}$ ).

Hence,

$(2,0)$ ,

$(-2,0)$ ,

$(\sqrt{2},\sqrt{2})$ and

$(-\sqrt{2},-\sqrt{2})$ are the stationary point.

Go on.

Antwoord 2 feedback

Wrong:

$z'_x(x,y)\neq -x^2+4$ .

See Partial derivatives.

Antwoord 3 feedback

Wrong:

$z'_y(x,y)\neq 2y^2-2y$ .

See Partial derivatives.

Antwoord 4 feedback

Wrong:

$z'_y(x,y)$ is not only equal to

$0$ for

$y=0$ , but also for

$y=x$ .

Try again.