# Extrema under constraints

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 critical point $\left (\frac{b}{4}, \frac{b}{2}\right )$.

To check what extrema (if there exists) it is, we do the following: $$f_{x_1} =x_2 , \ f_{x_2}=x_1 , \ f_{x_1x_1}=0 . \ f_{x_1x_2}=1 , \ f_{x_2x_2}=0$$

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.



Then I want to check if there are extrema if we have an other constraint, $\{(x_1, x_2)\in \mathbb{R}^2 \mid x_1\geq 0, x_2\geq 0\}$.

A critical point is \begin{equation*}\nabla f=\begin{pmatrix}0 \\ 0\end{pmatrix}\Rightarrow \begin{pmatrix}x_2 \\ x_1\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix} \Rightarrow x_1=0 \ \text{ und } \ x_2=0\end{equation*}

But this point is again a saddle point, right?

Then I wan to check if there are extrema if we have an other constraint, $\{(x_1, x_2)\in \mathbb{R}^2 \mid x_1\geq 0, x_2\geq 0\}$.

A critival point is \begin{equation*}\nabla f=\begin{pmatrix}0 \\ 0\end{pmatrix}\Rightarrow \begin{pmatrix}x_2 \\ x_1\end{pmatrix}=\begin{pmatrix}0 \\ 0\end{pmatrix} \Rightarrow x_1=0 \ \text{ and } \ x_2=0\end{equation*}

But this point is again a saddle point, right?

• Are you obliged to use Lagrange multipliers ? The problem is simple since $x_2=b-2x_1$ makes $f=x_1(b-2x_1)$. – Claude Leibovici Mar 15 '17 at 11:26
• I thought that I have to use Lagrange multipliers because at the nect question it is asked to interpret the Lagrange multiplier. If we use the method use proposed, would we get an other result? @ClaudeLeibovici – Mary Star Mar 15 '17 at 11:43

$$\begin{array}{ll} \text{maximize} & x y\\ \text{subject to} & 2x + y = b\end{array}$$

From the equality constraint, we have $y = b - 2 x$. Let

$$g (x) := x (b - 2 x)$$

The derivative of $g$ vanishes at $\frac b4$. Hence, the maximizer is $(\bar x, \bar y) := \left(\frac b4, \frac b2\right)$ and the maximum is $\frac{b^2}{8}$. For example, if $b = 4$, we have

• I see!! When we have the other constraint $\{(x_1, x_2)\in \mathbb{R}^2 \mid x_1\geq 0, x_2\geq 0\}$ what do we do then? Do we have to use in that case Lagrange multipliers? – Mary Star Mar 15 '17 at 11:59
• @MaryStar Well, in that case, the domain of $g$ is $[0,\frac b2]$. To find the minimum, evaluate $g$ at the two endpoints and take the minimum of those two values. – Rodrigo de Azevedo Mar 15 '17 at 12:03
• @MaryStar Evaluate the objective function at the boundary of the nonnegative quadrant, i.e., at the nonnegative parts of the axes. The objective function is zero at the boundary. That is the minimum. The maximum is obviously $\infty$. – Rodrigo de Azevedo Mar 15 '17 at 12:10
• @MaryStar That critical point is at the boundary. At the boundary, the objective function is zero. Since the objective function is $x y$, it is nonnegative on the nonnegative quadrant. Thus, the minimum is zero and the minimizer is any point on the boundary. – Rodrigo de Azevedo Mar 15 '17 at 14:04
• @MaryStar Make $t_1^2 := x_1$ and $t_2^2 := x_2$. Minimizing $x_1 x_2$ subject to nonnegativity constraints is the same as minimizing $t_1^2 t_2^2$ without any constraints. Differentiate and see where the derivative vanishes. It vanishes at the boundary of the nonnegative quadrant. – Rodrigo de Azevedo Mar 15 '17 at 14:51