Use Lagrange multipliers to find the maximum and minimum values of
f(x, y) = x + 8y
subject to the constraint
x^2 + y^2 = 4
if such values exist.

Round your answers to three decimal places.
If there is no global maximum or global minimum, enter NA in the appropriate answer area.

Respuesta :

The Lagrangian is

[tex]L(x,y,\lambda)=x+8y+\lambda(x^2+y^2-4)[/tex]

It has critical points where the first order derivatives vanish:

[tex]L_x=1+2\lambda x=0\implies\lambda=-\dfrac1{2x}[/tex]

[tex]L_y=8+2\lambda y=0\implies\lambda=-\dfrac4y[/tex]

[tex]L_\lambda=x^2+y^2-4=0[/tex]

From the first two equations we get

[tex]-\dfrac1{2x}=-\dfrac4y\implies y=8x[/tex]

Then

[tex]x^2+y^2=65x^2=4\implies x=\pm\dfrac2{\sqrt{65}}\implies y=\pm\dfrac{16}{\sqrt{65}}[/tex]

At these critical points, we have

[tex]f\left(\dfrac2{\sqrt{65}},\dfrac{16}{\sqrt{65}}\right)=2\sqrt{65}\approx16.125[/tex] (maximum)

[tex]f\left(-\dfrac2{\sqrt{65}},-\dfrac{16}{\sqrt{65}}\right)=-2\sqrt{65}\approx-16.125[/tex] (minimum)