Question

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.

Answer 1

The Lagrangian is

`L(x,y,\lambda)=x+8y+\lambda(x^2+y^2-4)`

It has critical points where the first order derivatives vanish:

`L_x=1+2\lambda x=0\implies\lambda=-\frac1{2x}`

`L_y=8+2\lambda y=0\implies\lambda=-\frac4y`

`L_\lambda=x^2+y^2-4=0`

From the first two equations we get

`-\frac1{2x}=-\frac4y\implies y=8x`

Then

`x^2+y^2=65x^2=4\implies x=\pm\frac2{√(65)}\implies y=\pm(16)/(√(65))`

At these critical points, we have

`f\left(\frac2{√(65)},(16)/(√(65))\right)=2√(65)\approx16.125` (maximum)

`f\left(-\frac2{√(65)},-(16)/(√(65))\right)=-2√(65)\approx-16.125` (minimum)