Calculus 3 help?

Question

Find all critical points (if any) of the given function f(x,y) and determine the  whether they are local extreme or saddle points.

a)  f(x,y) = xye^(-x^2-y^2)

b) (x+y)(xy-1)

Answer 1

Let f(x, y) have continuous second partial derivatives on an open region containing a point (a, b) for which fx (a, b) = 0 and fy (a, b) = 0.

To test for relative extrema of f(x, y), consider the quantity d = fxx (x, y) fyy (x, y)  - [ fxy (x, y) ]².

1. If d > 0 and fxx (x, y) > 0, then f(x, y) has a local minimum at (a, b).

2. If d > 0 and fxx (x, y) < 0, then f(x, y) has a local maximum at (a, b).

3. If d < 0, then (a, b, f(a, b)) is a saddle point.

4. The test is inconclusive if d = 0.

a). The function f(x, y) = xye^{(- x2 - y2)}.

Step - 1 : First find the first order partial derivatives.

f_x(x, y) = ye^{(- x2 - y2)}[1 - 2x²].

f_y(x, y) = xe^{(- x2 - y2)}[1 - 2y²].

Step - 2 :Solve the following equations  f_x = 0  and f_y = 0 simultaneously.

f_x(x, y) = ye^{(- x2 - y2)}[1 - 2x²] = 0  and  f_y(x, y) = xe^{(- x2 - y2)}[1 - 2y²] = 0

ye^{(- x2 - y2)}[1 - 2x²] = 0  and  xe^{(- x2 - y2)}[1 - 2y²] = 0

Since exponentials are non zero, this reduces to

y[1 - 2x²] = 0  and  x[1 - 2y²] = 0

Solve : y[1 - 2x²] = 0.

Apply zero product property.

y = 0  and  1 - 2x² = 0

y = 0  and  x = ±1/√2.

Solve : x[1 - 2y²] = 0

x = 0  and  1 - 2y² = 0

x = 0  and  y = ±1/√2.

Therefore, the critical points are (0, 0), (±1/√2, - 1/√2), and (±1/√2, 1/√2).

Step - 3 : Now find the second order partial derivatives.

f_xx(x, y) = 2xy(2x² - 3)e^{(- x2 - y2)}.

f_yy(x, y) = 2xy(2y² - 3)e^{(- x2 - y2)}.

f_xy(x, y) = (1 - 2x²)(1 - 2y²)e^{(- x2 - y2)}.

If either x = 0 or y = 0, then d = fxx (x, y) fyy (x, y)  - [ fxy (x, y) ]²

= [2xy(2x² - 3)e^{(- x2 - y2)}][2xy(2y² - 3)e^{(- x2 - y2)}.] - [(1 - 2x²)(1 - 2y²)e^{(- x2 - y2)} ]².

= [0 * 0] - [1]²

= - 1 < 0.

At (0, 0), d < 0, this is a saddle point.

At (±1/√2, - 1/√2), d = 4e^-2 = 0.5413 > 0, and f_xx(x, y) = 2e^-1 = 0.7357 > 0, hence, we have local minimum here.

At (±1/√2, 1/√2), d = 4e^-2 = 0.5413 > 0, and f_xx(x, y) = - 2e^-1 = - 0.7357 < 0, hence, we have local maximum here.