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Find the critical points and critical values of the function z=x^3+6xy-y^3-1

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Answer:

∂z/∂x = 3x^2 + 6y

∂z/∂y = 6x - 3y^2

Now, we need to solve the system of equations:

3x^2 + 6y = 0

6x - 3y^2 = 0

From the first equation, we get:

y = -x^2/2

Substituting this into the second equation, we get:

6x - 3(-x^2/2)^2 = 0

Simplifying, we get:

6x - 3x^4/4 = 0

Multiplying by 4 and rearranging, we get:

3x^4 - 24x = 0

Factoring out 3x, we get:

3x(x^3 - 8) = 0

Therefore, the critical values of x are x = 0 and x = 2.

For x = 0, we have y = 0 (from y = -x^2/2). So, one critical point is (0, 0).

For x = 2, we have y = -2. So, the other critical point is (2, -2).

To find the critical values of the function, we need to evaluate the function at each critical point:

z(0, 0) = 0^3 + 6(0)(0) - 0^3 - 1 = -1

z(2, -2) = 2^3 + 6(2)(-2) - (-2)^3 - 1 = -13

Therefore, the critical values of the function are -1 and -13.

User Fernando Cardenas
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