Identify any extrema of the function by recognizing its given form or its form after completing the square. Verify your results by using the partial derivatives to locate any critical points and test for relative extrema.

g(x, y) = 5 - (x - 3)² - (y + 2)²

Step-by-step explanation:

Given that,

g(x, y) = 5 - (x - 3)² - (y + 2)²

Let find the grad of the function,

The grad of a function is defined as

∇g= ∂g/∂x •i + ∂g/∂y •j + ∂g/∂z •k

∇g = gx•i + gy•j + gz•k

gx = -2(x-3) = -2x+6

gy = -2(y+2) = -2y -4

∇g = (-2x+6) •i + (-2y-4)•j

We have a maximum or a minimum If g conservative, then, ∇g = 0i +0j

Then comparing this to the grad of the function

(-2x+6) •i + (-2y-4)•j = 0i +0j

Then, -2x+6 = 0

2x=6

x = 3

Also, -2y-4=0

-2y=4

y = -2

Then, g(x, y) = 5 - (x - 3)² - (y + 2)²

g(x, y) = 5 - (3 - 3)² - (-2 + 2)²

g(x, y) = 5

So the critical point is (3, -2, 5)

gx =-2x+6

Second derivative of gx with respect to x

gxx=-2

gy=-2y-4

Second derivative of gy with respect to y

gyy=-2

Second derivative of gx with respect to y

gxy =0

d =gxxgyy - (gxy)²

Then, gyy=gxx

d = -2×-2 -0²

d = 4-0=4

Since d>0

Since d is greater than 0, then, it is not a chair points.

Then, since gxx=-2<0, gyy=-2<0

Cause the second derivative in x (or in y) is less than zero, then the point is relatively maximum

So the maximum point is (3, -2, 5).