If `m` is the minimum value of `f(x , y)=x^2-4x+y^2+6y` when `xa n dy` are subjected to the restrictions `0lt=xlt=1a n d0lt=ylt=1,` then the value of `|m|` is________

Options

A. `-1`
B. `-2`
C. `-3`
D. `-5`

Answer

To find the minimum value of the function \( f(x, y) = x^2 - 4x + y^2 + 6y \) under the constraints \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \), we can follow these steps:

### Step 1: Rewrite the function using completing the square.
We will complete the square for both \( x \) and \( y \).

1. For the \( x \) terms:
\[ x^2 - 4x = (x - 2)^2 - 4 \]

2. For the \( y \) terms:
\[ y^2 + 6y = (y + 3)^2 - 9 \]

Now, substituting these back into the function:
\[ f(x, y) = (x - 2)^2 - 4 + (y + 3)^2 - 9 \]
\[ f(x, y) = (x - 2)^2 + (y + 3)^2 - 13 \]

### Step 2: Analyze the function.
The function \( f(x, y) \) can be expressed as:
\[ f(x, y) = (x - 2)^2 + (y + 3)^2 - 13 \]
The term \( (x - 2)^2 + (y + 3)^2 \) represents the distance squared from the point \( (2, -3) \) in the coordinate plane.

### Step 3: Determine the minimum value within the constraints.
We need to evaluate \( f(x, y) \) at the corners of the rectangle defined by the constraints \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \).

1. **At \( (0, 0) \)**:
\[ f(0, 0) = (0 - 2)^2 + (0 + 3)^2 - 13 = 4 + 9 - 13 = 0 \]

2. **At \( (0, 1) \)**:
\[ f(0, 1) = (0 - 2)^2 + (1 + 3)^2 - 13 = 4 + 16 - 13 = 7 \]

3. **At \( (1, 0) \)**:
\[ f(1, 0) = (1 - 2)^2 + (0 + 3)^2 - 13 = 1 + 9 - 13 = -3 \]

4. **At \( (1, 1) \)**:
\[ f(1, 1) = (1 - 2)^2 + (1 + 3)^2 - 13 = 1 + 16 - 13 = 4 \]

### Step 4: Find the minimum value.
From the evaluations:
- \( f(0, 0) = 0 \)
- \( f(0, 1) = 7 \)
- \( f(1, 0) = -3 \)
- \( f(1, 1) = 4 \)

The minimum value is \( -3 \).

### Step 5: Find the absolute value of the minimum.
The problem asks for \( |m| \):
\[ |m| = |-3| = 3 \]

### Final Answer:
The value of \( |m| \) is \( \boxed{3} \).

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