Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=2x−5y subject to the constraint x2+3y2=111, if such values exist.

We are tasked with finding the maximum and minimum values of the function \( f(x, y) = 2x - 5y \) subject to the constraint \( x^2 + 3y^2 = 111 \). To solve this problem, we will use the method of **Lagrange multipliers**.

### Step 1: Set up the Lagrange multiplier equation
We introduce the Lagrange multiplier \( \lambda \) and define the **Lagrangian** function \( \mathcal{L}(x, y, \lambda) \) as follows:

\[
\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda \cdot g(x, y)
\]
where \( g(x, y) = x^2 + 3y^2 - 111 \) is the constraint.

Thus, the Lagrangian becomes:

\[
\mathcal{L}(x, y, \lambda) = (2x - 5y) - \lambda (x^2 + 3y^2 - 111)
\]

### Step 2: Find the partial derivatives
Now, we take the partial derivatives of \( \mathcal{L}(x, y, \lambda) \) with respect to \( x \), \( y \), and \( \lambda \) and set them equal to zero:

1. \( \frac{\partial \mathcal{L}}{\partial x} = 2 - \lambda \cdot 2x = 0 \)
2. \( \frac{\partial \mathcal{L}}{\partial y} = -5 - \lambda \cdot 6y = 0 \)
3. \( \frac{\partial \mathcal{L}}{\partial \lambda} = -(x^2 + 3y^2 - 111) = 0 \) (This is just the constraint equation.)

### Step 3: Solve the system of equations

1. From \( \frac{\partial \mathcal{L}}{\partial x} = 0 \):

\[
2 = \lambda \cdot 2x \quad \Rightarrow \quad \lambda = \frac{1}{x}
\]

2. From \( \frac{\partial \mathcal{L}}{\partial y} = 0 \):

\[
-5 = \lambda \cdot 6y \quad \Rightarrow \quad \lambda = \frac{-5}{6y}
\]

Now we equate the two expressions for \( \lambda \):

\[
\frac{1}{x} = \frac{-5}{6y} \quad \Rightarrow \quad 6y = -5x \quad \Rightarrow \quad y = \frac{-5x}{6}
\]

### Step 4: Substitute into the constraint equation
Substitute \( y = \frac{-5x}{6} \) into the constraint \( x^2 + 3y^2 = 111 \):

\[
x^2 + 3\left( \frac{-5x}{6} \right)^2 = 111
\]

Simplifying:

\[
x^2 + 3 \cdot \frac{25x^2}{36} = 111 \quad \Rightarrow \quad x^2 + \frac{25x^2}{12} = 111
\]

Multiply through by 12 to eliminate the denominator:

\[
12x^2 + 25x^2 = 1332 \quad \Rightarrow \quad 37x^2 = 1332 \quad \Rightarrow \quad x^2 = \frac{1332}{37} = 36
\]

Thus, \( x = \pm 6 \).

### Step 5: Find \( y \)
Using \( y = \frac{-5x}{6} \), we find:

- If \( x = 6 \), then \( y = \frac{-5(6)}{6} = -5 \).
- If \( x = -6 \), then \( y = \frac{-5(-6)}{6} = 5 \).

### Step 6: Compute the function values
Now, we evaluate \( f(x, y) = 2x - 5y \) for the two points:

- At \( (x, y) = (6, -5) \):
  \[
  f(6, -5) = 2(6) - 5(-5) = 12 + 25 = 37
  \]
- At \( (x, y) = (-6, 5) \):
  \[
  f(-6, 5) = 2(-6) - 5(5) = -12 - 25 = -37
  \]

### Step 7: Conclusion
The maximum value of \( f(x, y) \) is **37**, and the minimum value is **-37**.

---

Would you like more details or have any further questions?

Here are 5 related questions you might explore:
1. How does the Lagrange multiplier method work in three dimensions?
2. Can you solve optimization problems with inequality constraints using Lagrange multipliers?
3. What happens when no maximum or minimum exists under certain constraints?
4. How do Lagrange multipliers relate to the geometry of the constraint curve?
5. How do we solve similar problems with multiple constraints?

**Tip**: Always check if the critical points satisfy the constraint equation before concluding your solution!

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=2x−5y subject to the constraint x^2+3y^2=111, if such values exist.

Explore how to use the method of Lagrange multipliers to find the maximum and minimum values of f(x, y) = 2x - 5y, subject to the constraint x^2 + 3y^2 = 111. This detailed solution covers multivariable calculus concepts and optimization techniques.

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