To find the absolute maximum and minimum values of the function f(x, y) = xy² on the given domain defined by the constraints x > 0, y > 0, and x² + y² ≤ 3, follow these steps:
- Understanding the Domain: The inequalities x > 0 and y > 0 indicate that we are only considering the first quadrant of the Cartesian plane. The inequality x² + y² ≤ 3 describes a quarter-circle (a circle of radius √3) located in the first quadrant.
- Finding Critical Points: First, we need to compute the partial derivatives of the function f with respect to x and y:
-
∂f/∂x = y²
-
∂f/∂y = 2xy
- Setting the Partial Derivatives to Zero: To find critical points inside the domain, set the partial derivatives equal to zero:
-
y² = 0 → this gives y = 0 (not in our domain as y > 0)
-
2xy = 0 → this gives x = 0 (not in our domain as x > 0)
- Boundary Points: Since there are no critical points within the domain, we need to examine the boundary. The boundary is defined by the equation x² + y² = 3. We can express y in terms of x:
-
y² = 3 – x² → y = √(3 – x²)
- Substituting into f: Substitute this expression of y back into f:
-
f(x) = x(√(3 – x²))² = 3x – x³
- Finding Critical Points on the Boundary: Now, take the derivative of f with respect to x and set it to zero:
-
df/dx = 3 – 3x² = 0 → 3(1 – x²) = 0 → x = 1
- Evaluating f at the Critical Point and Endpoints: Evaluate f at x = 1 and at the endpoints (where x² = 3, hence y = 0 when x = √3 and vice versa):
-
f(1, √(3 – 1²)) = f(1, √2) = 1(√2)² = 2
-
f(√3, 0) = 0
-
f(0, √3) = 0
- Conclusion: The maximum value of f on the boundary is 2 (at the point (1, √2)), and the minimum value is 0 which occurs at the edges of the boundary (√3, 0) and (0, √3).
Thus, the absolute maximum value of f(x, y) on the specified domain is 2, and the absolute minimum value is 0.