Do we get the extrema from that graph?

[mathmari]

Hey!

I am looking at the following exercise:

1. Detremine the extrema of the function $f(x,y)=x^2y$ subject to $3x+2y=9$.
2. Prove also the second order condition. What kind is the extremum?
3. Is this an extremum of the whole function $f(x,y)$?
4. Draw the contour lines of $f(x,y)$ and the constraint.

I have done the following:

1. We can solve the constraint for one variable: \begin{equation*}3x+2y=9 \Rightarrow 3x=9-2y \Rightarrow x=\frac{9-2y}{3} \Rightarrow x=3-\frac{2}{3}y\end{equation*}
   
   Therefore, we get a function of one variable:
   \begin{equation*}h(y)=f\left (3-\frac{2}{3}y, y\right )=\left (3-\frac{2}{3}y\right )^2\cdot y=\left (9-2\cdot 3\cdot \frac{2}{3}y+\frac{4}{9}y^2\right )\cdot y=\left (9-4y+\frac{4}{9}y^2\right )\cdot y=9y-4y^2+\frac{4}{9}y^3\end{equation*}
   
   So, we are looking for the extrema of $h(y)$ using the first derivative:
   \begin{equation*}h'(y)=9-2\cdot 4y+3\cdot \frac{4}{9}y^2=9-8y+\frac{4}{3}y^2\end{equation*}
   
   The critical points of $h(y)=f\left (3-\frac{2}{3}y, y\right )$ are $y=\frac{3}{2}$ and $y=\frac{9}{2}$.
   
   Therefore, the critical points of $f(x,y)$ are the following:
   $\left (3-\frac{2}{3}\cdot \frac{3}{2}, \frac{3}{2}\right )=\left (3-1, \frac{3}{2}\right )=\left (2, \frac{3}{2}\right )$ and $\left (3-\frac{2}{3}\cdot \frac{9}{2}, \frac{9}{2}\right )=\left (3-3, \frac{9}{2}\right )=\left (0, \frac{9}{2}\right )$. The second derivative of $h(y)$ is $h''(y)=-8+\frac{8}{3}y$.
   
   We have that $h''\left (\frac{3}{2}\right )=-8+\frac{8}{3}\cdot \frac{3}{2}=-8+4=-4<0$ and $h''\left (\frac{9}{2}\right )=-8+\frac{8}{3}\cdot \frac{9}{2}=-8+12=4>0$.
   
   That means that $h$ has a maximum at $y=\frac{3}{2}$ and a minimum at $y=\frac{9}{2}$.
   
   From that it implies that $f(x,y)$ has a local maximum at $\left (2, \frac{3}{2}\right )$ and a local minimum $\left (0, \frac{9}{2}\right )$, right? (Wondering)
   
2. The second order condition is the following, or not? (Wondering)
   
   $(x_0, y_0)$ is a
   • local maximum, if \begin{equation*}F_{xx}(x_0, y_0)<0 \ \text{ und } \ F_{xx}(x_0, y_0)F_{yy}(x_0, y_0)-\left (F_{xy}(x_0, y_0)\right )^2>0\end{equation*}
   • local minimum, if\begin{equation*}F_{xx}(x_0, y_0)>0 \ \text{ und } \ F_{xx}(x_0, y_0)F_{yy}(x_0, y_0)-\left (F_{xy}(x_0, y_0)\right )^2>0\end{equation*}
   • saddle point, if \begin{equation*}F_{xx}(x_0, y_0)F_{yy}(x_0, y_0)-\left (F_{xy}(x_0, y_0)\right )^2<0\end{equation*}
   • local maximum, local minimum or saddle point, if \begin{equation*}F_{xx}(x_0, y_0)F_{yy}(x_0, y_0)-\left (F_{xy}(x_0, y_0)\right )^2=0\end{equation*}
     \end{enumerate}
   We have the following: \begin{equation*}f_x=2xy , f_y=x^2, f_{xx}=2y , f_{yy}=0 , f_{xy}=f_{yx}=2x\end{equation*} At the point $\left (2, \frac{3}{2}\right )$ we get the following:
   \begin{equation*}f_{xx}\left (2, \frac{3}{2}\right )=2\cdot \frac{3}{2}=3 , f_{yy}\left (2, \frac{3}{2}\right )=0 , f_{xy}\left (2, \frac{3}{2}\right )=f_{yx}\left (2, \frac{3}{2}\right )=2\cdot 2=4\end{equation*}
   
   So, we get \begin{equation*}f_{xx}\left (2, \frac{3}{2}\right )f_{yy}\left (2, \frac{3}{2}\right )-\left (f_{xy}\left (2, \frac{3}{2}\right )\right )^2=3\cdot 0-4^2=-16<0\end{equation*}
   
   That means that at $\left (2, \frac{3}{2}\right )$ we have a saddle point. At $\left (0, \frac{9}{2}\right )$ we have the following:
   \begin{equation*}f_{xx}\left (0, \frac{9}{2}\right )=2\cdot \frac{9}{2}=9 , f_{yy}\left (0, \frac{9}{2}\right )=0 , f_{xy}\left (0, \frac{9}{2}\right )=f_{yx}\left (0, \frac{9}{2}\right )=2\cdot 0=0\end{equation*}
   
   We have that \begin{equation*}f_{xx}\left (0, \frac{9}{2}\right )f_{yy}\left (0, \frac{9}{2}\right )-\left (f_{xy}\left (0, \frac{9}{2}\right )\right )^2=9\cdot 0-0^2=0\end{equation*}
   
   So, $\left (0, \frac{9}{2}\right )$ is either a local maximum or a local minimum or a saddle point. We don't get the same result as in the previous question about the kinf of the extrema because at the second order condition we don't use the constraint $3x+2y=9$, right? (Wondering)
3. That means that the function has only extrema under the constraint $3x+2y=9$. The extrema that we found are not extrema of the whole function $f(x,y)$, right? (Wondering)
   
4. We see the contour lines $f(x,y)=c$ for $c\in \{1,2,3,4,5\}$ and the constraint in the following graph:
   
   [DESMOS=-5,5,-4,10]x^2y=1;x^2y=2;x^2y=3;x^2y=4;x^2y=5;3x+2y=9[/DESMOS]
   
   What do we get from that graph? Can we get from there the extrema? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

Hey!

I am looking at the following exercise:

1. Detremine the extrema of the function $f(x,y)=x^2y$ subject to $3x+2y=9$.
2. Prove also the second order condition. What kind is the extremum?
3. Is this an extremum of the whole function $f(x,y)$?
4. Draw the contour lines of $f(x,y)$ and the constraint.

That means that $h$ has a maximum at $y=\frac{3}{2}$ and a minimum at $y=\frac{9}{2}$.

From that it implies that $f(x,y)$ has a local maximum at $\left (2, \frac{3}{2}\right )$ and a local minimum $\left (0, \frac{9}{2}\right )$, right? (Wondering)

Yes.
Additionally we have:
$$f\left (2, \frac{3}{2}\right ) = 6$$
and
$$f\left (0, \frac{9}{2}\right ) = 0$$

[*] The second order condition is the following, or not? (Wondering)

$(x_0, y_0)$ is a

• local maximum, if \begin{equation*}F_{xx}(x_0, y_0)<0 \ \text{ und } \ F_{xx}(x_0, y_0)F_{yy}(x_0, y_0)-\left (F_{xy}(x_0, y_0)\right )^2>0\end{equation*}
• local minimum, if\begin{equation*}F_{xx}(x_0, y_0)>0 \ \text{ und } \ F_{xx}(x_0, y_0)F_{yy}(x_0, y_0)-\left (F_{xy}(x_0, y_0)\right )^2>0\end{equation*}
• saddle point, if \begin{equation*}F_{xx}(x_0, y_0)F_{yy}(x_0, y_0)-\left (F_{xy}(x_0, y_0)\right )^2<0\end{equation*}
• local maximum, local minimum or saddle point, if \begin{equation*}F_{xx}(x_0, y_0)F_{yy}(x_0, y_0)-\left (F_{xy}(x_0, y_0)\right )^2=0\end{equation*}
  \end{enumerate}

We can only apply it at points where $f_x = f_y = 0$.
I'm afraid that's not the case in $(2,\frac 32)$. (Worried)

That means that at $\left (2, \frac{3}{2}\right )$ we have a saddle point.

I'm afraid we don't, since that point is not a critical point of $f$. (Shake)

[*] That means that the function has only extrema under the constraint $3x+2y=9$. The extrema that we found are not extrema of the whole function $f(x,y)$, right? (Wondering)

Actually, if we find the critical points of $f$ without constraints, we'll see that the lines $x=0$ and $y=0$ both consist entirely of critical points. The function has a minimum in all those points with value $0$.

[*] We see the contour lines $f(x,y)=c$ for $c\in \{1,2,3,4,5\}$ and the constraint in the following graph:

What do we get from that graph? Can we get from there the extrema? (Wondering)

We can see that the line touches the contour lines at $(2,\frac 32)$, which is where we have a maximum on the line with value $6$.
And we can see that the line crosses the contour lines in the bottom of a valley at $(0,\frac 92)$, which where we have a minimum with value $0$.

It would help to plot the function of $h$ that you find, since that will show how the extrema relate. (Thinking)

 

[mathmari]

Additionally we have:
$$f\left (2, \frac{3}{2}\right ) = 6$$
and
$$f\left (0, \frac{9}{2}\right ) = 0$$
We can only apply it at points where $f_x = f_y = 0$.
I'm afraid that's not the case in $(2,\frac 32)$. (Worried)
I'm afraid we don't, since that point is not a critical point of $f$. (Shake)

So when we find a critical point for $h$ the respective point for $f$ is not necessarily a critical point? (Wondering)

Actually, if we find the critical points of $f$ without constraints, we'll see that the lines $x=0$ and $y=0$ both consist entirely of critical points. The function has a minimum in all those points with value $0$.

Ahh, these are the roots of $f_x$ and $f_y$. (Thinking)

We can see that the line touches the contour lines at $(2,\frac 32)$, which is where we have a maximum on the line with value $6$.
And we can see that the line crosses the contour lines in the bottom of a valley at $(0,\frac 92)$, which where we have a minimum with value $0$.

I ot stuck right now... How can $(2,\frac 32)$ be a maximum but not a critical point? (Wondering)

It would help to plot the function of $h$ that you find, since that will show how the extrema relate. (Thinking)

The graph of $h$ is the following:

[DESMOS=-5,10,-10,10]9x-4x^2+\frac{4}{9}x^3[/DESMOS]

From that we get that a local minimum at $y=4.5$ and a local maximum $y=1.5$, right? (Wondering)

 

[I like Serena]

So when we find a critical point for $h$ the respective point for $f$ is not necessarily a critical point? (Wondering)

Indeed.

Ahh, these are the roots of $f_x$ and $f_y$. (Thinking)

yep

I ot stuck right now... How can $(2,\frac 32)$ be a maximum but not a critical point? (Wondering)

Suppose we're walking on a mountain. Its only critical point is at the very top where the ground is horizontal yes?
Now suppose we're walking on a road that is on a mountainside. Then at some points that road can be level can't it? Even though the mountainside is not horizontal.

The graph of $h$ is the following:

From that we get that a local minimum at $y=4.5$ and a local maximum $y=1.5$, right? (Wondering)

Yes.
Additionally we can see that neither of them is a global extremum.
Comparing it with the contour graph, we can see that the minimum corresponds to a road that traverses the bottom of a valley.
And the maximum corresponds to a road on a mountainside that reaches some highest point after which it goes down again. (Thinking)

 

[mathmari]

Indeed.

So, every time we apply that method, i.e., solving the constraint for one variable and finding the extrema of the function of one variable that we get, do we have to check at the end if the points that we get are indeed roots of $f_x$ and $f_y$ ? (Wondering)

 

[I like Serena]

So, every time we apply that method, i.e., solving the constraint for one variable and finding the extrema of the function of one variable that we get, do we have to check at the end if the points that we get are indeed roots of $f_x$ and $f_y$ ? (Wondering)

Usually we're only interested in the extrema with the constraint. The extrema without the constraint are usually different and of no particular importance. (Thinking)