Answer to Question #250513 in Calculus for Susan

Question

Find the extreme values of the function f(x,y) = 2x^2 + 3y^2 - 4x - 5

on the set D = {(x,y) : x^2 + y^2 < or = 16 }

Answer 1

Solution

Critical Points and Relative Extrema:

∂f/∂x = 4x – 4 = 0 =>  x = 1

∂f/∂y = 6y = 0 =>  y = 0

f_xx = 4 > 0 and f_yy = 6 > 0, f_xy = 0  => J = f_xxf_yy - f_xy² >0  => point x = 1, y = 0 is relative minimum

Another extreme values are on the bound of D – circle x² + y² = 16

Substituting from equation of circle  y² = 16 - x² into the function f(x,y):

f(x) = 2x² + 3(16 – x²) - 4x – 5  =>  f(x) = 43 – 4x – x² 

f’(x) = -4-2x = 0 => x = -2 => $y\ =\ \pm\sqrt{12}=\pm2\sqrt3$  

So function has a minimum at (x,y) = (1,0) where f(x,y) = -7 and two maximums at $(x,y)\ =\ (-2,\ 2\sqrt3)$  where f(x,y) = 47 and at $(x,y)\ =\ (-2,\ -2\sqrt3)$  where f(x,y) = 47 too.