# Exercise 1

Determine all the extrema of $z(x,y)=x^2y+5xy-y^3$.
There are no extrema.
• $z(0,0)=0$ is a minimum
• $z(-5,0)=0$ is a minimum
$z(0,0)=0$ is a minimum
• $z(0,0)=0$ 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
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.