Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=xy
 subject to the constraint 4x^2+y^2=8
.

What is the maximum value of f
, and at how many points does it occur?

Maximum value = ?

At how many points does this maximum value occur = ?

What is the minimum value, and at how many points does it occur?

Minimum value = ?

At how many points does this minimum value occur = ?

We are tasked with finding the maximum and minimum values of the function \( f(x, y) = xy \) subject to the constraint \( 4x^2 + y^2 = 8 \) using the method of Lagrange multipliers.

### Step 1: Setting up the problem using Lagrange multipliers
The method of Lagrange multipliers involves introducing a new variable, \( \lambda \) (the Lagrange multiplier), and solving the system of equations formed by:
1. \( \nabla f(x, y) = \lambda \nabla g(x, y) \)
2. The constraint equation \( g(x, y) = 0 \), where \( g(x, y) = 4x^2 + y^2 - 8 \).

So, our function \( f(x, y) = xy \) and the constraint is \( g(x, y) = 4x^2 + y^2 - 8 \).

#### Gradient of \( f(x, y) \):
\[
\nabla f(x, y) = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right) f(x, y) = (y, x)
\]

#### Gradient of \( g(x, y) \):
\[
\nabla g(x, y) = \left( \frac{\partial}{\partial x}, \frac{\partial}{\partial y} \right) g(x, y) = (8x, 2y)
\]

### Step 2: Setting up the system of equations
Using \( \nabla f(x, y) = \lambda \nabla g(x, y) \), we get the system:
\[
y = \lambda (8x) \tag{1}
\]
\[
x = \lambda (2y) \tag{2}
\]

Additionally, we have the constraint:
\[
4x^2 + y^2 = 8 \tag{3}
\]

### Step 3: Solving the system

#### From equation (2):
\[
x = \lambda (2y) \implies \lambda = \frac{x}{2y} \quad \text{(assuming \( y \neq 0 \))}
\]

#### Substitute \( \lambda \) into equation (1):
\[
y = \frac{x}{2y} (8x) = \frac{8x^2}{2y} \implies y^2 = 4x^2
\]
Thus, \( y = 2x \) or \( y = -2x \).

#### Case 1: \( y = 2x \)

Substitute \( y = 2x \) into the constraint equation \( 4x^2 + y^2 = 8 \):
\[
4x^2 + (2x)^2 = 8 \implies 4x^2 + 4x^2 = 8 \implies 8x^2 = 8 \implies x^2 = 1 \implies x = \pm 1
\]
When \( x = 1 \), \( y = 2 \); and when \( x = -1 \), \( y = -2 \).

#### Case 2: \( y = -2x \)

Substitute \( y = -2x \) into the constraint equation:
\[
4x^2 + (-2x)^2 = 8 \implies 4x^2 + 4x^2 = 8 \implies 8x^2 = 8 \implies x^2 = 1 \implies x = \pm 1
\]
When \( x = 1 \), \( y = -2 \); and when \( x = -1 \), \( y = 2 \).

### Step 4: Evaluating \( f(x, y) = xy \) at the critical points

We now have four critical points: \( (1, 2), (-1, -2), (1, -2), (-1, 2) \).

- At \( (1, 2) \), \( f(1, 2) = 1 \times 2 = 2 \).
- At \( (-1, -2) \), \( f(-1, -2) = (-1) \times (-2) = 2 \).
- At \( (1, -2) \), \( f(1, -2) = 1 \times (-2) = -2 \).
- At \( (-1, 2) \), \( f(-1, 2) = (-1) \times 2 = -2 \).

### Step 5: Conclusion

- **Maximum value**: The maximum value of \( f(x, y) = xy \) is \( \boxed{2} \), and it occurs at **2 points**: \( (1, 2) \) and \( (-1, -2) \).
  
- **Minimum value**: The minimum value of \( f(x, y) = xy \) is \( \boxed{-2} \), and it occurs at **2 points**: \( (1, -2) \) and \( (-1, 2) \).

---

Do you need further details or explanations?

Here are some related questions to expand your understanding:

1. How does the method of Lagrange multipliers handle problems with multiple constraints?
2. What are the conditions for a critical point to be a local maximum or minimum in constrained optimization?
3. Can Lagrange multipliers be applied to three-variable functions with two constraints?
4. How do we interpret the Lagrange multiplier \( \lambda \) geometrically?
5. How can we solve constrained optimization problems numerically if they cannot be solved algebraically?

**Tip**: Always verify the solution by checking if the critical points satisfy both the objective function and the constraint.

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=xy subject to the constraint 4x^2+y^2=8. What is the maximum value of f, and at how many points does it occur? Maximum value = ? At how many points does this maximum value occur = ? What is the minimum value, and at how many points does it occur? Minimum value = ? At how many points does this minimum value occur = ?

Learn how to use the method of Lagrange multipliers to find the maximum and minimum values of f(x, y) = xy under the constraint 4x^2 + y^2 = 8. This optimization problem involves multivariable calculus techniques and is suitable for students in advanced calculus or optimization courses. The maximum value is 2, occurring at two points, while the minimum value is -2, also occurring at two points.

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