Exercise 1 | Wiskunde op Tilburg University

Determine all the extrema of

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

Antwoord 1 correct

Correct

Antwoord 2 optie

$z(0,0)=0$ is a minimum
$z(-5,0)=0$ is a minimum

Antwoord 2 correct

Fout

Antwoord 3 optie

$z(0,0)=0$ is a minimum

Antwoord 3 correct

Fout

Antwoord 4 optie

$z(0,0)=0$ is a minimum
$z(-5,0)=0$ ,is a minimum
$(-2\frac{1}{2},\sqrt{\dfrac{1}{12}})=-36.15$ is a maximum
$z(-2\frac{1}{2},-\sqrt{\dfrac{1}{12}})=36.15$ is a maximum

Antwoord 4 correct

Fout

Antwoord 1 optie

There are no extrema.

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

$x^2+5x=0$ with solutions

$x=0$ and

$x=-5$ .

If we plug

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

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

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

Hence,

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

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

$z''_{xx}(x,y)=2y$ ,

$z''_{xy}=2x+5$ and

$z''_{yy}=-6y$ . Hence,

$C(x,y)=-12y^2-(2x+5)^2$ .

Since

$C(0,0)=-25<0$ and

$C(-5,0)=-25<0$ , there are no extrema.

Go on.

Antwoord 2 feedback

Wrong: If

$C(c,d)<0$ , then stationary point

$(c,d)$ is a saddle point.

See Second-order condition extremum.

Antwoord 3 feedback

Wrong: If

$C(c,d)<0$ , then stationary point

$(c,d)$ is a saddle point.

See Second-order condition extremum.

Antwoord 4 feedback

Wrong:

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

Try again.