Find Critical Point and classify as rel. min, rel, max, or saddle point.

Question

Find Critical Point and classify as rel. min, rel, max, or saddle point.

z=f(x,y)=2y^2-x^2+6x-4

Answer 1

To find critical points, we first determine the partial derivatives of the function f(x,y) = 2y² - x² + 6x - 4:

Partial derivative of f w/ respect to x:

f_x = -2x + 6

Partial derivative of f w/ respect to y:

f_y = 4y

Next, we determine any critical points by finding points (x,y) where both partial derivatives are 0:

-2x + 6 = 0

4y = 0

We can solve this system of equations:

-2x + 6 = 0

6 = 2x (add 2x to both sides)

x = 3 (divide both sides by 2)

and

4y = 0

y = 0 (divide both sides by 4)

So our only critical points is at x = 3 and y = 0: (3,0).

To determine whether this is a rel. min, max or saddle point: let's use the second derivative test. Find the second derivatives of f:

f_xx = -2

f_yy = 4

f_xy = 0

Now we apply the formula: D(x, y) = f_xx ⋅ f_yy - (f_xy)².

D(0,0) = -2 ⋅ 4 - (0)²

D(0,0) = -8

The partial derivative formula tells us that if D is negative, then the critical point is a saddle point.

Thus our solution is:

Critical point(s) f(x,y) = 2y² - x² + 6x - 4:

(3,0), a saddle point