MHB Finding Extrema under Constraints

[mathmari]

Hey!

I want to find the critical points of the function $f(x_1, x_2)=x_1x_2$ under the constraint $2x_1+x_2=b$.

Using the method of Lagrange multipliers I got the following:

\begin{equation*}L(x_1,x_2,\lambda )=x_1x_2-\lambda \cdot \left (2x_1+x_2-b\right )\end{equation*}

\begin{align*}&L_{x_1}(x_1,x_2,\lambda)=0 \Rightarrow x_2-2\lambda=0 \\ & L_{x_2}(x_1,x_2,\lambda)=0 \Rightarrow x_1-\lambda=0 \\ & L_{\lambda}(x_1,x_2,\lambda)=0 \Rightarrow -\left (2x_1+x_2-b\right )=0\end{align*}

Solving this system we get the solution: $${x_1}_0=\frac{b}{4}, \ {x_2}_0=\frac{b}{2}$$

So, the critical point is $\left (\frac{b}{4}, \frac{b}{2}\right )$, i.e., at this point there might be an extrema. To check what extrema (if we have) it is we do the following:
\begin{align*}&f_{x_1} =x_2 \\ & f_{x_2}=x_1 \\ & f_{x_1x_1}=0 \\ & f_{x_1x_2}=1 \\ & f_{x_2x_2}=0\end{align*}

Then: \begin{equation*}f_{x_1x_2}\left (\frac{b}{4}, \frac{b}{2}\right )=1>0 \ \text{ and } \ f_{x_1x_1}\left (\frac{b}{4}, \frac{b}{2}\right )f_{x_2x_2}\left (\frac{b}{4}, \frac{b}{2}\right )-\left (f_{x_1x_2}\left (\frac{b}{4}, \frac{b}{2}\right )\right )^2=0\cdot 0-1=-1<0\end{equation*}

Therefore, $\left (\frac{b}{4}, \frac{b}{2}\right )$ is a saddle point.

Is this correct?

Because at Wolfram there are some maxima. (Wondering)
Using the reduction method we have the following:
$2x_1+x_2=b \Rightarrow x_2=b-2x_1$.

\begin{align*}&f(x_1, x_2)=x_1x_2 \\ & h(x_1)=f(x_1, b-2x_1)=x_1\cdot (b-2x_1)=bx_1-2x_1^2\end{align*}

The first derivative is \begin{equation*}\frac{dh}{dx_1}=b-4x_1\end{equation*}

The roots of the first derivative are \begin{equation*}\frac{dh}{dx_1}=0 \Rightarrow b-4x_1=0 \Rightarrow x_1=\frac{b}{4}\end{equation*}

So, the critical point of $h(x_1)$ is $x_1=\frac{b}{4}$.

Therefore, the critical point for $f(x_1, x_2)$ is \begin{equation*}(x_1, x_2)=\left (\frac{b}{4}, b-2x_1\right )=\left (\frac{b}{4}, b-2\cdot \frac{b}{4}\right )=\left (\frac{b}{4}, b- \frac{b}{2}\right )=\left (\frac{b}{4}, \frac{b}{2}\right )\end{equation*}

The second deivative is \begin{equation*}\frac{d^2h}{dx_1^2}=-4<0\end{equation*}

So, at $x_1=\frac{b}{4}$ the function $h(x_1)$ has a maximum.

Therefore, the function $f(x_1, x_2)$ has a maximum at \begin{equation*} \left (\frac{b}{4}, \frac{b}{2}\right )\end{equation*}

The maximum is equal to \begin{equation*}f\left (\frac{b}{4}, \frac{b}{2}\right )=\frac{b}{4}\cdot \frac{b}{2}=\frac{b^2}{8}\end{equation*} Why do we get, using that method, a maximum, and by the other one a saddle point? (Wondering)

 

[I like Serena]

Hey mathmari! (Smile)

When we do the second partial derivative test, we're only looking at $f(x,y)=xy$, which has indeed only a saddle point.
However, when we take the boundary condition into account, that changes, and we find a maximum.

 

[mathmari]

When we do the second partial derivative test, we're only looking at $f(x,y)=xy$, which has indeed only a saddle point.
However, when we take the boundary condition into account, that changes, and we find a maximum.

So, using the second partial derivative test we are not taking into considertaion the boundary of the domain?

Does this mean that when we have a constraint with inequalities we have to check always also the boundary?

(Wondering)

 

[I like Serena]

So, using the second partial derivative test we are not taking into considertaion the boundary of the domain?

Correct. Note that we're only using the second derivatives of $f$, and we do not use $2x_1+x_2=b$.

Does this mean that when we have a constraint with inequalities we have to check always also the boundary?

Yes! (Nod)

 

[mathmari]

Correct. Note that we're only using the second derivatives of $f$, and we do not use $2x_1+x_2=b$.Yes! (Nod)

Ah ok. I see! Thank you very much! (Mmm)