Answer to Question #108148 in Calculus for annie

Question

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x−2y subject to the constraint x^2+y^2=5, if such values exist.

maximum = ?

minimum = ?

Answer 1

Given $f(x,y) = x - 2y$ .

Let $g(x,y) = x^{2} + y^{2} - 5$ and

let $F = f(x,y)+\lambda g(x,y)$ , where $\lambda$ is the Lagrangian multiplier.

$F = x - 2y + \lambda(x^{2} + y^{2} -5)$

To find the maximum and minimum values, we have to solve the system

$\dfrac{\partial F}{\partial x} = 0$,$\dfrac{\partial F}{\partial y} = 0$ and $\dfrac{\partial F}{\partial \lambda} = 0$ .

$1 + 2x\lambda = 0\\ -2 + 2y\lambda = 0\\ x^{2} + y^{2} = 5.$

From the first and the second equation we get,

$x = -\dfrac{1}{2\lambda}\\ y = \dfrac{1}{\lambda}\\$

Using these values in the third equation we get,

$\dfrac{1}{4\lambda^{2}} + \dfrac{1}{\lambda^{2}} = 5\\ \lambda^{2} = \dfrac{1}{4}\\ \lambda = \pm\dfrac{1}{2}\\$

When $\lambda = -\dfrac{1}{2}$ ,

$x = 1$ and $y = -2$

When $\lambda = \dfrac{1}{2}$ ,

$x = -1$ and $y = 2$

The maximum and minimum values of f(x,y) are respectively,

$f(1,-2) = 1+4 = 5\\ f(-1,2) = -1-4 = -5\\$