To find the absolute maximum and minimum values of the function f(x, y) = xy^2 + 5 on the set D, we will follow these steps:
Step 1: Determine the critical points of f(x, y) within the given domain D. These are the points where the gradient of f(x, y) is zero or undefined.
First, find the partial derivatives of f(x, y) with respect to x and y:
∂f/∂x = y^2
∂f/∂y = 2xy
Setting each partial derivative equal to zero and solving for x and y, we get:
y^2 = 0 (from ∂f/∂x = 0)
=> y = 0
2xy = 0 (from ∂f/∂y = 0)
=> x = 0 or y = 0
The critical points are (0, 0) and any points on the x-axis or y-axis that lie within the domain D.
Step 2: Evaluate the function f(x, y) at the critical points and boundary points of D.
Let's consider the boundary of D, which is described by the equation x^2 + y^2 = 3. We can parameterize this boundary by letting:
x = √(3)cosθ
y = √(3)sinθ
where θ is an angle between 0 and 2π.
Now substitute these values of x and y into f(x, y):
f(√(3)cosθ, √(3)sinθ) = (√(3)cosθ)(√(3)sinθ)^2 + 5
= 3cosθsin^2θ + 5
Step 3: Identify the maximum and minimum values of f(x, y) by comparing the values obtained from critical points and boundary points.
Let's evaluate f(x, y) at the critical point (0, 0):
f(0, 0) = (0)(0^2) + 5
= 5
Now, evaluate f(x, y) at the boundary points:
f(√(3)cosθ, √(3)sinθ) = 3cosθsin^2θ + 5
To find the maximum and minimum values of f(x, y) on the boundary, we can analyze the function 3cosθsin^2θ. Since this function is periodic with period π, we only need to consider values of θ between 0 and π.
We take the derivative of the function with respect to θ:
d/dθ (3cosθsin^2θ) = 3cosθ[2sinθcosθ] = 6cosθsinθcosθ = 3sin2θcosθ
Setting the derivative equal to zero and solving for θ, we get:
3sin2θcosθ = 0
sin2θcosθ = 0
From this equation, the critical points occur when sin2θ = 0 or cosθ = 0.
When sin2θ = 0:
2θ = 0 or π, leading to θ = 0 or π/2
When cosθ = 0:
θ = π/2 or 3π/2
These critical points correspond to the values of θ for which the function 3cosθsin^2θ is at a maximum or minimum.
Now we evaluate f(x, y) at these critical points using the expression:
f(√(3)cosθ, √(3)sinθ) = 3cosθsin^2θ + 5
For θ = 0:
f(√(3)cos0, √(3)sin0) = 3cos0sin^20 + 5
= 0 + 5
= 5
For θ = π/2:
f(√(3)cos(π/2), √(3)sin(π/2)) = 3cos(π/2)sin^2(π/2) + 5
= 0 + 5
= 5
For θ = π:
f(√(3)cosπ, √(3)sinπ) = 3cosπsin^2π + 5
= 0 + 5
= 5
For θ = 3π/2:
f(√(3)cos(3π/2), √(3)sin(3π/2)) = 3cos(3π/2)sin^2(3π/2) + 5
= 0 + 5
= 5
The maximum value of f(x, y) occurs at the points (0, 0) and on the boundary of D, where f(x, y) = 5. The minimum value of f(x, y) is also 5 on the entire set D.
Therefore, the absolute maximum value of f on the set D is 5, and the absolute minimum value is also 5.