Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=3x−2y subject to the constraint x2+y2=13, if such values exist.

maximum =
minimum =
*(For either value, enter **DNE** if there is no such value.)*

We are asked to find the maximum and minimum values of the function \( f(x, y) = 3x - 2y \), subject to the constraint \( x^2 + y^2 = 13 \). This is a standard optimization problem that can be solved using the method of **Lagrange multipliers**.

### Step 1: Define the Lagrange Multiplier Equation
We introduce the Lagrange multiplier \( \lambda \) and set up the system of equations by defining the Lagrange function \( \mathcal{L}(x, y, \lambda) \):
\[
\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda (g(x, y) - c)
\]
where \( g(x, y) = x^2 + y^2 \) and the constraint is \( g(x, y) = 13 \).

So, the Lagrange function becomes:
\[
\mathcal{L}(x, y, \lambda) = (3x - 2y) - \lambda (x^2 + y^2 - 13)
\]

### Step 2: Compute the Gradient
Now, take the partial derivatives of \( \mathcal{L}(x, y, \lambda) \) with respect to \( x \), \( y \), and \( \lambda \), and set them equal to zero:
\[
\frac{\partial \mathcal{L}}{\partial x} = 3 - 2\lambda x = 0 \quad \Rightarrow \quad 3 = 2\lambda x \quad \Rightarrow \quad \lambda = \frac{3}{2x}
\]
\[
\frac{\partial \mathcal{L}}{\partial y} = -2 - 2\lambda y = 0 \quad \Rightarrow \quad -2 = 2\lambda y \quad \Rightarrow \quad \lambda = -\frac{1}{y}
\]
\[
\frac{\partial \mathcal{L}}{\partial \lambda} = -(x^2 + y^2 - 13) = 0 \quad \Rightarrow \quad x^2 + y^2 = 13
\]

### Step 3: Solve the System of Equations
Now, we have two expressions for \( \lambda \):
\[
\lambda = \frac{3}{2x} \quad \text{and} \quad \lambda = -\frac{1}{y}
\]
Set them equal to each other:
\[
\frac{3}{2x} = -\frac{1}{y} \quad \Rightarrow \quad 3y = -2x \quad \Rightarrow \quad y = -\frac{2}{3}x
\]

### Step 4: Substitute into the Constraint
Substitute \( y = -\frac{2}{3}x \) into the constraint \( x^2 + y^2 = 13 \):
\[
x^2 + \left(-\frac{2}{3}x\right)^2 = 13 \quad \Rightarrow \quad x^2 + \frac{4}{9}x^2 = 13 \quad \Rightarrow \quad \frac{13}{9}x^2 = 13
\]
\[
x^2 = 9 \quad \Rightarrow \quad x = 3 \quad \text{or} \quad x = -3
\]

### Step 5: Find the Corresponding \( y \)-values
For \( x = 3 \):
\[
y = -\frac{2}{3}(3) = -2
\]
For \( x = -3 \):
\[
y = -\frac{2}{3}(-3) = 2
\]

### Step 6: Evaluate \( f(x, y) \) at the Critical Points
Now, evaluate the function \( f(x, y) = 3x - 2y \) at the points \( (3, -2) \) and \( (-3, 2) \):

- At \( (3, -2) \):
\[
f(3, -2) = 3(3) - 2(-2) = 9 + 4 = 13
\]
- At \( (-3, 2) \):
\[
f(-3, 2) = 3(-3) - 2(2) = -9 - 4 = -13
\]

### Step 7: Conclusion
- The maximum value of \( f(x, y) \) is \( \boxed{13} \).
- The minimum value of \( f(x, y) \) is \( \boxed{-13} \).

Would you like a deeper explanation or details? Here are five related questions:

1. How does the method of Lagrange multipliers work for functions with more than two variables?
2. Can the Lagrange multiplier method be used with inequality constraints?
3. How do you visualize the function and constraint in this problem geometrically?
4. What are alternative methods for constrained optimization problems besides Lagrange multipliers?
5. What happens if there is no solution to the system of equations derived from the Lagrange multiplier method?

**Tip:** When solving with Lagrange multipliers, always verify the constraint is satisfied at the critical points to ensure correctness.

Use Lagrange multipliers to find the maximum and minimum values of f(x, y) = 3x − 2y subject to the constraint x^2 + y^2 = 13, if such values exist.

Find the maximum and minimum values of the function f(x, y) = 3x - 2y subject to the constraint x^2 + y^2 = 13 using Lagrange multipliers. Learn how to set up the Lagrange function, solve the system of equations, and apply critical point analysis in this college-level optimization problem.

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