Question

 Find and classify the critical points of the following functions.

(a) f (x, y) = x³ + 3xy² − 15x + y³ − 15y

(b) f (x, y) = (x² + y² )ex²−y²

(c) f (x, y) = ln(ax² + by²  + 1) with a, b > 0

(d) f (x, y) = x³ − 3x/1 + y²

Answer 1

(a) The partial derivatives of f (x, y) are

f_x (x, y) = 3x² + 3y² − 15

f_y (x, y) = 6xy + 3 − 15 .

Critical point are solutions to the equation

3x² + 3y² − 15 = 0

6xy + 3y² − 15 = 0 .

We subtract the second equation from the first to obtain

3y² − 6xy = 0 ⇔ 3x (x − 2y) = 0

We have two cases two consider.

• If x = 0, then the first equation gives

3y² − 15 = 0 ⇒ y = ± √5 ,

leading to the two critical points( 0,√5 ) and (0,− √5)

 • If x = 2y, then the first equation become

 15y² = 15 ⇒ y = ±1

 leading to two more critical points (1, 2) and −1,−2

To determine the type of the critical points we need the second derivatives

f_xx (x, y) = 6x

 f_xy (x, y) = 6y

 f_yy (x, y) = 6x + 6y

Thus we have

 (b) The partial derivatives of f (x, y) a

Critical point are solutions to the equations

This system is equinalent to

In the first equation we see that x² + y² + 1 ≠ 0 and so x = 0. From the second equation we obtain

y = 0 or x² +y² = 1 . This leads to the three solutions (0, 0), (0, 1

To determine the type of the critical points we need the second derivaties

thus we have

   The partial derivativies of f(x,y) are

Critical point are solutions to the equations

Because ax² + by² + 1≠ 0, the only critical point is (0, 0). To determine the type of the critical point we need the second derivatives

Thus we have

(d) The partial derivatives of f (x, y) are

Critical point are solutions to the equations

We can multiply both equations by (1 + y² ) and (1 + y² )² respectively to obtain

Note however that since the first equation specifies x = ±1, it follows that we have y = 0, because the other cases in the second equation cannot happen. Thus we have the two critical points (1,0) and (−1,0). To determine the type of the critical points we need the second derivaties