Determine all the extrema of $z(x,y)=x^2-4xy+xy^2+y^3+2$.
Antwoord 1 correct
Correct
Antwoord 2 optie
- $z(-16,8)=1186$ is a maximum
- $z(0,0)=2$ is a maximum
- $z(1\frac{1}{2},1)=\frac{3}{4}$ is a minimum
Antwoord 2 correct
Fout
Antwoord 3 optie
There are no extrema.
Antwoord 3 correct
Fout
Antwoord 4 optie
None of the other answers is correct.
Antwoord 4 correct
Fout
Antwoord 1 optie
$z(1\frac{1}{2},1)=\frac{3}{4}$ is a minimum
Antwoord 1 feedback
Correct:
$$\begin{align*}
z'_y(x,y)& =-4x+2xy+3y^2\\
& = -4(2y-\frac{1}{2}y^2)+2(2y-\frac{1}{2}y^2)y+3y^2\\
& = -y^3+9y^2-8y\\
& = -y(y^2-9y+8).
\end{align*}$$
Consequently, $z'_y(x,y)=0$ if $y=0$ (with $x=0$), $y=1$ (with $x=1\frac{1}{2}$) or $y=8$ (with $x=-16$).
Since $C(0,0)=-16<0$ and $C(-16,8)=-112<0$ both $(0,0)$ and $(-16,8)$ are saddle points.
Since $C(1\frac{1}{2},1)=14>0$ and $z''_{xx}(1\frac{1}{2},1)=2>0$ it holds that $z(1\frac{1}{2},1)=\frac{3}{4}$ is a minimum.
Go on.
- $z'_x(x,y)=2x-4y+y^2$
- $z'_y(x,y)=-4x+2xy+3y^2$
$$\begin{align*}
z'_y(x,y)& =-4x+2xy+3y^2\\
& = -4(2y-\frac{1}{2}y^2)+2(2y-\frac{1}{2}y^2)y+3y^2\\
& = -y^3+9y^2-8y\\
& = -y(y^2-9y+8).
\end{align*}$$
Consequently, $z'_y(x,y)=0$ if $y=0$ (with $x=0$), $y=1$ (with $x=1\frac{1}{2}$) or $y=8$ (with $x=-16$).
- $z''_{xx}=2$
- $z''_{yy}=2x+6y^2$
- $z''_{xy}=-4+2y$
Since $C(0,0)=-16<0$ and $C(-16,8)=-112<0$ both $(0,0)$ and $(-16,8)$ are saddle points.
Since $C(1\frac{1}{2},1)=14>0$ and $z''_{xx}(1\frac{1}{2},1)=2>0$ it holds that $z(1\frac{1}{2},1)=\frac{3}{4}$ is a minimum.
Go on.
Antwoord 2 feedback
Wrong: If $C(c,d)<0$, then that does not mean that the stationary point $(c,d)$ is a maximum.
See Second-order condition extremum.
See Second-order condition extremum.
Antwoord 3 feedback
Wrong: $(0,0)$ is not the only stationary point.
Try again.
Try again.
Antwoord 4 feedback
Wrong: The correct answer is among them.
Try again.
Try again.