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

Maxima and minima are very important topics in mathematics in which we can find the maximum or minimum value of a function using its derivatives.

Answer: The function f(x, y) has local maxima at (-1, -1) and saddle point at (1/2, -1/4).

Let's understand the solution in detail.

Explanation:

Given function: f(x, y) = x³ − 3y² + 3xy − 3y

Now, we partially differentiate the function f(x, y) with respect to x to find f_x.

⇒ f_x = 3x² - 0 + 3y - 0 (using rules of derivatives)

⇒ f_x = 3x² + 3y 

Similarly, we partially differentiate the function f(x, y) with respect to y to find f_y.

⇒ f_x = -6y + 3x - 3 

Now, we equate the first derivatives (f_x and f_y) to zero.

⇒ 3x² + 3y = 0 ---- (1)

⇒ -6y + 3x - 3 = 0  ---- (2)

Now, we solve equation (1) and equation (2).

Now, rearranging equation (1), we get 3x² = -3y, which is equal to y = -x².

Now, we substitute the above value of y in equation (2).

⇒ -6(-x²) + 3x - 3 = 0

⇒ 6x² + 3x - 3 = 0

Now, solving the above quadratic equation, we get x = -1, 1/2.

Using y = -x², we find the corresponding values of y, to get the critical points (-1, -1) and (1/2, -1/4).

Now, we find the double derivatives of the function: f_xx, f_yy, f_xy.

⇒ f_xx = 6x

⇒ f_xy = 3

⇒ f_yy = -6

Now, we find the discriminant D = f_xx × f_yy - f_xy for both the critical points to find the nature of the optima.

Hence, D(-1, -1) = 6(-1) × -6 - 3 = 36 - 3 = 33.

Hence, 33 > 0 and -6 < 0, so the point (-1, -1) is a maxima.

Now, D(1/2, -1/4) = 6(1/2) × (-6) - 3 = -18 - 3 = -21

Hence, -21 < 0, so the point (1/2, -1/4) is a saddle point.

Hence, the function f(x, y) has local maxima at (-1, -1) and saddle point at (1/2, -1/4).