Question

Find the relative maximum and minimum values and the saddle
points. f(x,y)=200y²+x²-x²(y)

Answer 1

To find the relative maximum and minimum values and saddle points of the equation f(x, y) = 200y² + x² - x²(y), use the second partial derivative test. Set the first partial derivatives to zero to find the critical points, solve the system of equations to find the critical points, and use the second derivative test to determine their nature. The critical points (0,√(0²/16)) and (0,-√(0²/16)) are relative minima, while the critical point (x,0) where x ≠ 0 is a saddle point.

Step-by-step explanation:

To identify the relative maximum and minimum values and saddle points of the function f(x, y) = 200y² + x² - x²(y), we can use the second partial derivative test.

1. First, find the critical points by setting the first partial derivatives to zero:

• ∂f/∂x = 2x - 2xy = 0
• ∂f/∂y = 400y - x² = 0

Next, solve the system of equations to find the critical points:

• x = 0
• y = ±√(x²/400) = ±√(x²/400) = ±√(x²/16)

Finally, use the second derivative test to determine whether the critical points are relative maximum, minimum, or saddle points.

• Calculate the second partial derivatives:
  • ∂²f/∂x² = 2
  • ∂²f/∂y² = 400
  • ∂²f/∂x∂y = -2x

Based on the second partial derivatives, we can conclude that:

• The critical point (0,√(0²/16)) is a relative minimum.
• The critical point (0,-√(0²/16)) is a relative minimum.
• The critical point (x,0) where x ≠ 0 is a saddle point.