Answer to Question #107938 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 

$\dfrac{\partial F}{\partial \lambda} = 0$

We get,

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

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 $\lambda = -\dfrac{1}{2}$,

$x= 1, y =-2$

When $\lambda = \dfrac{1}{2}$,

$x= -1, y =2$

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

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