Exercise 3

Correct:

• $z'_x(x,y)=2x-4y+y^2$
• $z'_y(x,y)=-4x+2xy+3y^2$

$z'_x(x,y)=0$ gives $x=2y-\frac{1}{2}y^2$. Hence,
$$\begin{align*} z'_y(x,y)& =-4x+2xy+3y^2\\ & = -4(2y-\frac{1}{2}y^2)+2(2y-\frac{1}{2}y^2)y+3y^2\\ & = -y^3+9y^2-8y\\ & = -y(y^2-9y+8). \end{align*}$$
Consequently, $z'_y(x,y)=0$ if $y=0$ (with $x=0$), $y=1$ (with $x=1\frac{1}{2}$) or $y=8$ (with $x=-16$).

• $z''_{xx}=2$
• $z''_{yy}=2x+6y^2$
• $z''_{xy}=-4+2y$

Hence, $C(x,y)=4x+12y^2-(2y-4)^2$.

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.

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.

Wrong: $(0,0)$ is not the only stationary point.

Try again.

Wrong: The correct answer is among them.

Try again.