Question

Use the method of Lagrange multipliers to find the absolute maximum and minimum values of(x, y) = x² + y²-x-y + 8 on the unit disc, namely, D = {(x, y) | x² + y² S 1).

Answer 1

To find the absolute maximum and minimum values of the function on the unit disc, we can use the method of Lagrange multipliers. The absolute maximum value is 43/9 at (1/3, 2/3) and (-1/3, -2/3), and the absolute minimum value is 11/4 at (-√2/2, √2/2) and (√2/2, -√2/2).

Step-by-step explanation:

To find the absolute maximum and minimum values of the function f(x, y) = x² + y² - x - y + 8 on the unit disc, we can use the method of Lagrange multipliers.

1. First, we need to set up the system of equations using the equation of the function and the constraint equation x² + y² = 1:
   ∂f/∂x = λ∂g/∂x
   ∂f/∂y = λ∂g/∂y
   x² + y² = 1
2. Next, we differentiate the function and the constraint equation with respect to x and y:
   ∂f/∂x = 2x - 1 = λ(2x)
   ∂f/∂y = 2y - 1 = λ(2y)
   x² + y² = 1
3. Solving the system of equations, we find the critical points:
   x = 1/3, y = 2/3; x = -1/3, y = -2/3; x = -√2/2, y = √2/2; x = √2/2, y = -√2/2
4. Finally, we evaluate the function at the critical points and compare the values:
   f(1/3, 2/3) = 43/9; f(-1/3, -2/3) = 43/9; f(-√2/2, √2/2) = 11/4; f(√2/2, -√2/2) = 11/4

The absolute maximum value is 43/9 at (1/3, 2/3) and (-1/3, -2/3), and the absolute minimum value is 11/4 at (-√2/2, √2/2) and (√2/2, -√2/2).