Question

This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. f(x, y) = 4x + 8y; x2 + y2 = 20

Answer 1

(2,4) and (-2,-4).

Explanation:

We consider the system

`\left \{ {{gra(f(x,y))= λ gra(g(x,y))} \atop {x^(2)+y^(2)=20}} \right.`

where g(x,y)=
`x^(2) +y^(2) = 20`

gra(f(x,y))= (4,8)

gra(g(x,y))=(2x,2y)

So,

`\left \{ {{(4,8)=λ(2x,2y)} \atop {x^(2)+y^(2)=20}} \right.`

We have then that 4=2λx and 8=2λy. Dividing the second equation by 2 at both sides we obtain 4=λy. So, 4=2λx and 4=λy, we equalize both equations:

2λx=λy ⇔ 2x=y.

We replace that y value in the constraint equation:

`x^(2) +(2x)^(2) = 20`

`x^(2) + 4x^(2)= 20`

`5x^(2)= 20`

`x^(2)= 4`

`x= 2` and

`x= -2`

With this two x values and the y equation y= 2x we can obtain the extremes: (2,4) and (-2,-4).