# Math Help - Lagrange multipliers

1. ## Lagrange multipliers

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

f(x,y) = 4x + 6y; x^2 + y^2 = 13

I'm not sure how to solve this.

2. Originally Posted by paulrb
Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

f(x,y) = 4x + 6y; x^2 + y^2 = 13

I'm not sure how to solve this.
Do you know how to use Lagrange Multipliers? Basically let us define $f(x,y)=4x+6y$ and $g(x,y)=x^2+y^2-13$. We then need to solve the system of equations given by

$\left\{\begin{array}{cc}f(x,y)_x=\lambda{g(x,y)_x} \\f(x,y)_y=\lambda{g(x,y)_y}\\g(x,y)=0\end{array}\ right\}$

A little work shows the above is equivalent to

$\left\{\begin{array}{cc}4=2x\lambda\\6=2y\lambda\\ x^2+y^2+-13=0\end{array}\right\}$

And unless my calculations are off I calculate the solutions to be

$\left\{\begin{array}{cc}\left\{x,y,\lambda\right\} =\left\{-2,-3-1\right\}\\\left\{x,y,\lambda\right\}=\left\{2,3,1 \right\}\end{array}\right\}$

So now $f(-2,-3)=-26$ and $f(2,3)=26$. From there we can conclude that the point $(-2,-3,-26)$ is a minimum point and $(2,3,26)$ a maximum point.

3. Hello, paulrb!

I must assume you know the basic steps for Lagrange multipliers.

Use Lagrange multipliers to find the maximum and minimum values
of the function subject to the given constraints.

. . $f(x,y) \:= \:4x + 6y \qquad x^2 + y^2 \:=\: 13$

We have: . $F(x,y,\lambda) \;=\;4x + 6y + \lambda(x^2+y^2-13)$

Take partial derivatives and equate to zero.

. . $F_x \;=\;4 + 2x\lambda\:=\:0 \quad\Rightarrow\quad x \:=\:-\frac{2}{\lambda}$ .[1]

. . $F_y \;=\;6 + 2y\lambda \:=\:0 \quad\Rightarrow\quad y \:=\:-\frac{3}{\lambda}$ .[2]

. . $F_{\lambda} \;=\;x^2-y^2-13 \:=\:0 \quad\Rightarrow\quad x^2+y^2 \:=\:13$ .[3]

Substitute [1] and [2] into [3]: . $\left(-\frac{2}{\lambda}\right)^2 + \left(-\frac{3}{\lambda}\right)^2 \:=\:13 \quad\Rightarrow\quad \frac{4}{\lambda^2} + \frac{9}{\lambda^2} \:=\:13
$

. . . . . . $\frac{13}{\lambda^2} \:=\:13 \quad\Rightarrow\quad \lambda^2 \:=\:1\quad\Rightarrow\quad\lambda \:=\:\pm1$

Substitute into [1]: . $x \:=\:-\frac{2}{\pm1} \quad\Rightarrow\quad x \:=\:\pm2$

Substitute into [2]: . $y \:=\:-\frac{3}{\pm1} \quad\Rightarrow\quad y \:=\:\pm3$

. . $\begin{array}{cccccc}f(2,3) &=& 4(2) + 6(3) &=& 26 & \text{maximum} \\ \\[-3mm]
f(\text{-}2,\text{-}3) &=& 4(\text{-}2) + 6(\text{-}3) &=& \text{-}26 & \text{minimum} \end{array}$