Question

Find the local maximum and minimum values and saddle point(s) of the function. If you have three dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function. f(x, y) = x3 − 12xy + 8y3 Step 1 To find the critical points of f(x, y), we must solve fx = 0 and fy = 0. For f(x, y) = x3 − 12xy + 8y3 we have fx(x, y) = and fy(x, y) = . fx = 0 implies that x2 = 4y and fy = 0 implies that x = .

Answer 1

Answer: Since the instructions ask for a three-dimensional graph to reveal the important aspects of the function, it would be best to visualize the function using appropriate graphing software.

Explanation:

To find the critical points of the function f(x, y) = x^3 - 12xy + 8y^3, we need to solve the partial derivatives fx = 0 and fy = 0. Let's calculate these derivatives:

fx = 3x^2 - 12y

fy = -12x + 24y^2

Setting fx = 0, we have:

3x^2 - 12y = 0

x^2 = 4y ... (Equation 1)

Setting fy = 0, we have:

-12x + 24y^2 = 0

x = 2y^2 ... (Equation 2)

Now, let's solve equations 1 and 2 simultaneously to find the critical points:

Substituting x = 2y^2 from Equation 2 into Equation 1:

(2y^2)^2 = 4y

4y^4 = 4y

y^4 - y = 0

y(y^3 - 1) = 0

So, we have two cases to consider:

Case 1: y = 0

From Equation 2, when y = 0, x = 2(0)^2 = 0.

Therefore, one critical point is (0, 0).

Case 2: y^3 - 1 = 0

Solving y^3 - 1 = 0, we find y = 1.

Substituting y = 1 into Equation 2, we get x = 2(1)^2 = 2.

So, the other critical point is (2, 1).

To determine whether these critical points are local maxima, local minima, or saddle points, we need to analyze the second-order partial derivatives.