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:
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.
- $z'_x(x,y)=-x^2+4-y^2$
- $z'_y(x,y)=2y^2-2yx$
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
Antwoord 3 feedback
Antwoord 4 feedback
Wrong: $z'_y(x,y)$ is not only equal to $0$ for $y=0$, but also for $y=x$.
Try again.
Try again.