# Exercise 2

Determine all the extrema of $z(x,y)=x^2-4x+2xy+5y+y^2-\frac{1}{3}y^3+25$.
$z(5,-3)=3$ is a minimum
• $z(5,-3)=3$ is a minimum
• $z(-1,-3)=39$ is a minimum
$z(5,3)=75$ is a maximum
$z(0,1)=30\frac{2}{3}$ is a maximum
Determine all the extrema of $z(x,y)=x^2-4x+2xy+5y+y^2-\frac{1}{3}y^3+25$.
Antwoord 1 correct
Correct
Antwoord 2 optie
• $z(5,-3)=3$ is a minimum
• $z(-1,-3)=39$ is a minimum
Antwoord 2 correct
Fout
Antwoord 3 optie
$z(5,3)=75$ is a maximum
Antwoord 3 correct
Fout
Antwoord 4 optie
$z(0,1)=30\frac{2}{3}$ is a maximum
Antwoord 4 correct
Fout
Antwoord 1 optie
$z(5,-3)=3$ is a minimum
Antwoord 1 feedback
Correct:
• $z'_x(x,y)=2x-4+2y$
• $z'_y(x,y)=2x+5+2y-y^2$
From $z'_x(x,y)=0$ follows $x=2-y$. Plugging this into $z'_y(x,y)=0$ gives $2(2-y)+5+2y-y^2=0$, or $y^2=9$. Hence, $y=3$ (with $x=2-y=-1$) or $y=-3$ (with $x=2-y=5$).
• $z''_{xx}(x,y)=2$
• $z''_{yy}(x,y)=2-2y$
• $z''_{xy}(x,y)=2$
Hence, $C(x,y)=2(2-2y)-2^2=-4y$. Since $C(-1,3)=-12<0$ it holds that $(-1,3)$ is a saddle point. Since $C(5,-3)=12>0$ and $z''_{xx}(5,-3)=3>0$ it holds that $z(5,-3)=3$ is a minimum.

Go on.
Antwoord 2 feedback
Wrong: $(-1,-3)$ is not a stationary point.

See Stationary point.
Antwoord 3 feedback
Wrong: $(5,3)$ is not a stationary point.

See Stationary point.
Antwoord 4 feedback
Wrong: The fact that $z''_{yy}(0,1)=0$ does not tell you anything about the stationary points of the function.

See Second-order condition extremum