Let
f(x,y)=x 
2
 +2y 
2
 −4x−1
. (a) Find the critical points of
f. (b) Find the extreme values of
f
on the disk
x 
2
 +y 
2
 <9

Let’s walk through the problem step-by-step to solve both parts (a) and (b).  

Given the function:

\[
f(x, y) = x^2 + 2y^2 - 4x - 1
\]

---

### **(a) Finding the Critical Points**  
Critical points occur where the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\) are **zero** or **undefined**.

1. **Partial derivative with respect to \(x\):**  
   \[
   f_x = \frac{\partial f}{\partial x} = 2x - 4
   \]
   Set \(f_x = 0\):
   \[
   2x - 4 = 0 \quad \Rightarrow \quad x = 2
   \]

2. **Partial derivative with respect to \(y\):**  
   \[
   f_y = \frac{\partial f}{\partial y} = 4y
   \]
   Set \(f_y = 0\):
   \[
   4y = 0 \quad \Rightarrow \quad y = 0
   \]

So, the only **critical point** is:
\[
(2, 0)
\]

---

### **(b) Finding the Extreme Values of \(f(x, y)\) on the Disk \(x^2 + y^2 < 9\)**  
This part requires finding the minimum and maximum values of \(f(x, y)\) on the disk \(x^2 + y^2 < 9\). This involves:

1. **Evaluating the function at the critical point**.
2. **Checking the boundary** of the disk, which is the circle \(x^2 + y^2 = 9\).

---

#### 1. Evaluating \(f(x, y)\) at the critical point \((2, 0)\):  
\[
f(2, 0) = (2)^2 + 2(0)^2 - 4(2) - 1 = 4 - 8 - 1 = -5
\]
So, \(f(2, 0) = -5\).

---

#### 2. Checking the boundary \(x^2 + y^2 = 9\)  
To find the extreme values on the boundary, we substitute \(x = 3\cos \theta\) and \(y = 3\sin \theta\) (a parametrization of the circle of radius 3) into the function:

\[
f(3\cos \theta, 3\sin \theta) = (3\cos \theta)^2 + 2(3\sin \theta)^2 - 4(3\cos \theta) - 1
\]
\[
= 9\cos^2 \theta + 18\sin^2 \theta - 12\cos \theta - 1
\]

Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\), we simplify the expression:

\[
f(3\cos \theta, 3\sin \theta) = 9 + 9\sin^2 \theta - 12\cos \theta - 1
\]
\[
= 8 + 9\sin^2 \theta - 12\cos \theta
\]

Now, we need to find the maximum and minimum of this function with respect to \(\theta\).

---

#### 3. Finding the Extreme Values on the Boundary  
We now maximize and minimize:

\[
g(\theta) = 8 + 9\sin^2 \theta - 12\cos \theta
\]

- **Maximum Value:** This occurs when \(12\cos \theta\) is smallest (i.e., \(\cos \theta = -1\)).  
  Substituting \(\cos \theta = -1\):
  \[
  g(\theta) = 8 + 9\sin^2 \theta + 12 = 20
  \]

- **Minimum Value:** This occurs when \(12\cos \theta\) is largest (i.e., \(\cos \theta = 1\)).  
  Substituting \(\cos \theta = 1\):
  \[
  g(\theta) = 8 + 9\sin^2 \theta - 12 = -4
  \]

---

### **Conclusion: Extreme Values**  
- **At the critical point** \((2, 0)\), the function value is \(f(2, 0) = -5\).
- **On the boundary** \(x^2 + y^2 = 9\), the minimum value is \(-4\) and the maximum value is \(20\).

Thus, the extreme values of \(f(x, y)\) on the disk \(x^2 + y^2 < 9\) are:
- **Maximum:** \(20\)  
- **Minimum:** \(-5\) (at the critical point \((2, 0)\))

---

### Do you have any further questions, or would you like a deeper explanation of any step?  

Here are **5 related questions** you could explore:
1. How do we use Lagrange multipliers to find extrema on constraints?
2. What is the significance of Hessians in classifying critical points?
3. How can we generalize this approach to functions of three variables?
4. What is the difference between local and global extrema?
5. Can the same method apply to non-circular boundaries, such as ellipses?

**Tip:** Always check both the interior and boundary points when optimizing over a region!

Let f(x,y) = x^2 + 2y^2 - 4x - 1. (a) Find the critical points of f. (b) Find the extreme values of f on the disk x^2 + y^2 < 9.

Explore how to find critical points and extreme values of the function f(x,y) = x² + 2y² - 4x - 1 within the disk x² + y² < 9. Learn how to apply partial derivatives, optimization techniques, and boundary parametrization to solve this calculus problem. Suitable for undergraduate students in Calculus II or III.

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