Question

Let \( U = \{(x, y) \in \mathbb{R}^2 : x + y \leq 2\} \). Define \( f: U \to \mathbb{R} \) by \[ f(x, y) = (x - 1)^4 + (y - 2)^4. \] The minimum value of \( f \) over \( U \) is: (answer in integer).

• A. 0
• B. \( \frac{1}{16} \)
• C. \( \frac{17}{81} \)
• D. \( \frac{1}{8} \)

Hint:

When dealing with optimization problems in geometry, it is often helpful to check the boundary points and critical points to find the minimum or maximum values of the function.

Answer 1

The Correct Option is D

Step 1: Understand the region \( U \).
The region \( U \) is defined by the inequality \( x + y \leq 2 \). This is a half-plane bounded by the line \( x + y = 2 \).

Step 2: Minimize \( f(x, y) = (x - 1)^4 + (y - 2)^4 \).
To minimize \( f(x, y) \), we look for the point in \( U \) where the function \( f(x, y) \) takes its minimum value. Notice that the function \( f(x, y) \) is a sum of two fourth powers, which are minimized when \( x = 1 \) and \( y = 2 \), i.e., the point closest to the origin of the function's definition.

Step 3: Check if \( (1, 2) \) lies in the region \( U \).
Substituting \( x = 1 \) and \( y = 2 \) into the inequality \( x + y \leq 2 \), we get: \[ 1 + 2 = 3 \quad \text{which is greater than} \quad 2. \] Thus, the point \( (1, 2) \) is not inside the region \( U \).

Step 4: Find the closest point on the boundary.
The point on the boundary closest to \( (1, 2) \) is the point where the line \( x + y = 2 \) intersects the line joining \( (1, 2) \) and the origin. By geometry, the closest point on the boundary is \( (1, 1) \).

Step 5: Compute \( f(1, 1) \).
Substitute \( (x, y) = (1, 1) \) into \( f(x, y) \): \[ f(1, 1) = (1 - 1)^4 + (1 - 2)^4 = 0 + 1 = 1. \] Thus, the minimum value of \( f(x, y) \) over \( U \) is \( \boxed{1} \).