Question

What are the critical points of the function f (x,y) = x³y³ - 3x² - 3y² - 8?

a) There are no critical points
b) (0, 0)
c) (1, 1)
d) (-1, -1)

Answer 1

The critical points of the function f(x,y) = x³y³ - 3x² - 3y² - 8 are (-1,-1).

Step-by-step explanation:

To find the critical points of the function f(x,y) = x³y³ - 3x² - 3y² - 8, we need to find the points where the partial derivatives with respect to x and y are both equal to zero.

For the derivative of f(x,y) with respect to x, we have:

fx(x,y) = 3x²y³ - 6x

Setting fx(x,y) = 0 and solving for x:

• 3x²y³ - 6x = 0
• x(3x²y³ - 6) = 0
• x = 0 or 3x²y³ - 6 = 0

For the derivative of f(x,y) with respect to y, we have:

fy(x,y) = 3x³y² - 6y

Setting fy(x,y) = 0 and solving for y:

• 3x³y² - 6y = 0
• y(3x³y² - 6) = 0
• y = 0 or 3x³y² - 6 = 0

Therefore, we have four possibilities for the critical points:

1. (0,0)
2. (0,√2/√3)
3. (0,-√2/√3)
4. (√2/√3,0)

So, the correct answer is d) (-1,-1).