the absolute maximum and minimum values of f(x,y)= x²-2xy+2y on the rectangle defined I in the interval [0,3] and y in the interval [0,2]
the absolute maximum and minimum values of f(x,y)= x²-2xy+2y on the rectangle defined I in the interval [0,3] and y in the interval [0,2]
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Finding Absolute Maximum and Minimum Values of a Function in a Given Rectangle**
Consider the function \( f(x, y) = x^2 - 2xy + 2y \).
We are tasked with finding the absolute maximum and minimum values of this function in the rectangle defined by \( x \) in the interval \([0, 3]\) and \( y \) in the interval \([0, 2]\).
**Steps to Solve:**
1. **Identify the Function and Constraints:**
- Function: \( f(x, y) = x^2 - 2xy + 2y \)
- Constraints: \( 0 \leq x \leq 3 \)
\( 0 \leq y \leq 2 \)
2. **Find the Critical Points:**
- To find these, set the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \) to zero.
- \( \frac{\partial f}{\partial x} = 2x - 2y = 0 \)
- \( \frac{\partial f}{\partial y} = -2x + 2 = 0 \)
- Solve the system of equations:
- \( 2x - 2y = 0 \) ⟹ \( x = y \)
- \( -2x + 2 = 0 \) ⟹ \( x = 1 \)
- Therefore, \( x = 1 \) and \( y = 1 \). So, the critical point is \( (1, 1) \).
3. **Evaluate \( f(x, y) \) at the Critical Point:**
- \( f(1, 1) = 1^2 - 2(1)(1) + 2(1) = 1 - 2 + 2 = 1 \)
4. **Evaluate \( f(x, y) \) at the Boundary Points:**
- Compute \( f(x, y) \) at the corners of the rectangle:
- \( (0, 0): \)
\( f(0, 0) = 0^2 - 2(0)(0) + 2(0) = 0 \)
- \( (0, 2): \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffdb8cee4-3263-42a2-bbf8-01e2cd6c2adc%2Fc87346cc-7a37-4454-a66f-24d6283b11c4%2F8iiu9l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding Absolute Maximum and Minimum Values of a Function in a Given Rectangle**
Consider the function \( f(x, y) = x^2 - 2xy + 2y \).
We are tasked with finding the absolute maximum and minimum values of this function in the rectangle defined by \( x \) in the interval \([0, 3]\) and \( y \) in the interval \([0, 2]\).
**Steps to Solve:**
1. **Identify the Function and Constraints:**
- Function: \( f(x, y) = x^2 - 2xy + 2y \)
- Constraints: \( 0 \leq x \leq 3 \)
\( 0 \leq y \leq 2 \)
2. **Find the Critical Points:**
- To find these, set the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \) to zero.
- \( \frac{\partial f}{\partial x} = 2x - 2y = 0 \)
- \( \frac{\partial f}{\partial y} = -2x + 2 = 0 \)
- Solve the system of equations:
- \( 2x - 2y = 0 \) ⟹ \( x = y \)
- \( -2x + 2 = 0 \) ⟹ \( x = 1 \)
- Therefore, \( x = 1 \) and \( y = 1 \). So, the critical point is \( (1, 1) \).
3. **Evaluate \( f(x, y) \) at the Critical Point:**
- \( f(1, 1) = 1^2 - 2(1)(1) + 2(1) = 1 - 2 + 2 = 1 \)
4. **Evaluate \( f(x, y) \) at the Boundary Points:**
- Compute \( f(x, y) \) at the corners of the rectangle:
- \( (0, 0): \)
\( f(0, 0) = 0^2 - 2(0)(0) + 2(0) = 0 \)
- \( (0, 2): \)
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