Problem 8: For the function f (x, y) = 5 ? 6x + 4y + 2x2 + y2 . (a) Find fx and

Question

Problem 8: For the function f (x, y) = 5 ? 6x + 4y + 2x2 + y2 .

(a) Find fx and fy and solve for the critical point(s) of f .

(b) Compute the second partials fxx , fxx=fxy and fyy .

(c) For each critical point, evaluate the second partials, evaluate the quantity D = fxx fyy ? (fxy)^2 ,
and perform the second partials test to determine whether the critical point is a local maximum,
minimum, or saddle point.

(d) Give all maximum and minimum values, or coordinates of saddle points (evaluate f at the
appropriate locations).

Explanation / Answer

f(x,y) = 5 - 6x + 4y + 2x2 + y2

fx = -6 + 4x fy = 4 + 2y

Set them equal to 0. 4x - 6 = 0 4x = 6 x = 3/2

4 + 2y = 0 2y = -4 y = -2.

Thus there is only one critical point and it's (3/2, -2)

fxx = 4 fyy = 2 fxy = fyx = 0

D(3/2, -2) = 4*2 - 0 = 8 > 0

fxx(3/2, -2) = 4 > 0

Then in accordance with the second partial derivative test:

(3/2, -2) is a local min.

At tha point, the value of f(x,y) is:

f(3/2, -2) = 5 - 6*(3/2) + 4*-2 +2*(3/2)^2 + (-2)^2

f(3/2, -2) = 5 - 9 - 8 + (9/2) + 4

f(3/2, -2) = (9/2) - 8

f(3/2, -2) = (9/2) - (16/2) = -7/2

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