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

Answer:

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

Step-by-step 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).