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.
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.
Try again.
Antwoord 3 feedback
Wrong: When does it hold that $z'_y(x,y)=0$ if $y=0$?
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Try again.
Antwoord 4 feedback