Exercise 1 | Wiskunde op Tilburg University

Determine all the stationary points of

$z(x,y)=x^2y+5xy-y^3$ .

Antwoord 1 correct

Correct

Antwoord 2 optie

$(0,0)$ ,

$(-5,0)$ ,

$(-2\frac{1}{2},\sqrt{\dfrac{1}{12}})$ and

$(-2\frac{1}{2},-\sqrt{\dfrac{1}{12}})$

Antwoord 2 correct

Fout

Antwoord 3 optie

$(0,0)$

Antwoord 3 correct

Fout

Antwoord 4 optie

$(-2\frac{1}{2},0)$

Antwoord 4 correct

Fout

Antwoord 1 optie

$(0,0)$ and

$(-5,0)$

Antwoord 1 feedback

Correct:

$z'_x(x,y)=2xy+5y$ and

$z'_y(x,y)=x^2+5-3y^2$ . Hence,

$z'_x(x,y)=0$ if

$y=0$ or if

$x=-2\frac{1}{2}$ .

If we plug

$y=0$ into

$z'_y(x,y)$ we get the equation

$x^2+5x=0$ . The solutions of this equation are

$x=0$ and

$x=-5$ .

If we plug

$x=-2\frac{1}{2}$ into

$z'_y(x,y)$ we get the equation

$-6\frac{1}{4}-3y^2=0$ and that equation has no solutions.

Hence,

$(x,y)=(0,0)$ and

$(x,y)=(-5,0)$ are the only stationary points.

Go on.

Antwoord 2 feedback

Wrong:

$-6\frac{1}{4}-3y^2=0$ has no solutions.

Try again.

Antwoord 3 feedback

Wrong: When does it hold that

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

$y=0$ ?

Try again.

Antwoord 4 feedback

Wrong:

$z'_y(-2\frac{1}{2},0)\neq 0$ .

See Stationary point.