Find all the local maxima, local minima, and saddle points of f(x, y) = x^3 − 3y^2 + 3xy − 3y.

Question

Find all the local maxima, local minima, and saddle points of f(x, y) = x³ − 3y² + 3xy − 3y.

Answer 1

f(x,y) = x³ - 3y² +3xy - 3y

First Partial Derivatives:

f_{x =} 3x² + 3y

f_y = -6y + 3x - 3

3x² + 3y = 0 -6y + 3x - 3 = 0

3y = -3x² -6(-x²) + 3x - 3 = 0

y = -x² 6x² + 3x - 3 = 0

Critical Points: y = -1, - 1/4 x = -1, 1/2

Second Partial Derivatives:

(-1, -1) (1/2 , -1/4)

f_xx= 6x -6 3

f_xy = 3 3 3

f_yy = -6 -6 -6

D(-1 , 1) = f_xx * f_yy - f_xy = -6 * -6 - 3 = 36 - 3 = 33

33 > 0 and -6 < 0. Therefore, the critical point (-1, -1) is a local maxima.

D( 1/2, -1/4) = f_xx * f_yy - f_{xy =} 3 * -6 - 3 = -18 - 3 = -21

-21 < 0. Therefore the critical point (1/2, -1/4) is a Saddle Point.

Answer 2

Hi Mark T.,

You have submitted 6 problems at once requiring similar techniques to solve (derivatives in 3-space). That's "spamming" your problem set.

--Cheers, -- Mr. d.