Find all the local maximizers and local - minimizers (if any) of the function f (x, y) = e³x 3x + 4y² - 1 (without any constraints). Determine whether each local maximizer you find is a global maximizer and whether each local minimizer is a global minimizer.
Find all the local maximizers and local - minimizers (if any) of the function f (x, y) = e³x 3x + 4y² - 1 (without any constraints). Determine whether each local maximizer you find is a global maximizer and whether each local minimizer is a global minimizer.
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 Local and Global Extremes of a Multivariable Function**
**Problem Statement:**
Find all the local maximizers and local minimizers (if any) of the function \( f(x, y) = e^{3x} - 3x + 4y^2 - 1 \) (without any constraints). Determine whether each local maximizer you find is a global maximizer and whether each local minimizer is a global minimizer.
**Approach:**
To solve this problem, follow these steps:
1. **Find the First Partial Derivatives:**
- Compute \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).
2. **Set the Partial Derivatives to Zero:**
- Solve the equations \(\frac{\partial f}{\partial x} = 0\) and \(\frac{\partial f}{\partial y} = 0\) to find potential critical points.
3. **Determine the Nature of Critical Points:**
- Compute the second partial derivatives: \(\frac{\partial^2 f}{\partial x^2}\), \(\frac{\partial^2 f}{\partial y^2}\), and the mixed partial derivative \(\frac{\partial^2 f}{\partial x \partial y}\).
- Use the second derivative test or the Hessian determinant to classify each critical point as a local maximizer, local minimizer, or saddle point.
4. **Check for Global Extremes:**
- Compare the values of the function at the critical points to determine if the local maxima and minima are indeed global.
**Explanation:**
- **First Partial Derivatives:**
\[ \frac{\partial f}{\partial x} = 3e^{3x} - 3 \]
\[ \frac{\partial f}{\partial y} = 8y \]
- **Set to Zero and Solve:**
\[ 3e^{3x} - 3 = 0 \implies e^{3x} = 1 \implies 3x = 0 \implies x = 0 \]
\[ 8y = 0 \implies y = 0 \]
So, the critical point is \((x, y) = (0, 0)\).
- **Second Partial Derivatives:**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2c07220d-ca17-4b4a-9c07-9176cc38cf32%2F555cb225-b2fa-494b-b72b-e0499c211e66%2Fx9nhhjp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Finding Local and Global Extremes of a Multivariable Function**
**Problem Statement:**
Find all the local maximizers and local minimizers (if any) of the function \( f(x, y) = e^{3x} - 3x + 4y^2 - 1 \) (without any constraints). Determine whether each local maximizer you find is a global maximizer and whether each local minimizer is a global minimizer.
**Approach:**
To solve this problem, follow these steps:
1. **Find the First Partial Derivatives:**
- Compute \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).
2. **Set the Partial Derivatives to Zero:**
- Solve the equations \(\frac{\partial f}{\partial x} = 0\) and \(\frac{\partial f}{\partial y} = 0\) to find potential critical points.
3. **Determine the Nature of Critical Points:**
- Compute the second partial derivatives: \(\frac{\partial^2 f}{\partial x^2}\), \(\frac{\partial^2 f}{\partial y^2}\), and the mixed partial derivative \(\frac{\partial^2 f}{\partial x \partial y}\).
- Use the second derivative test or the Hessian determinant to classify each critical point as a local maximizer, local minimizer, or saddle point.
4. **Check for Global Extremes:**
- Compare the values of the function at the critical points to determine if the local maxima and minima are indeed global.
**Explanation:**
- **First Partial Derivatives:**
\[ \frac{\partial f}{\partial x} = 3e^{3x} - 3 \]
\[ \frac{\partial f}{\partial y} = 8y \]
- **Set to Zero and Solve:**
\[ 3e^{3x} - 3 = 0 \implies e^{3x} = 1 \implies 3x = 0 \implies x = 0 \]
\[ 8y = 0 \implies y = 0 \]
So, the critical point is \((x, y) = (0, 0)\).
- **Second Partial Derivatives:**
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