Use Lagrange multipliers to find the optimal value of the function 

f(x,y)=x2+y2+xy

subject to the constraint

x+3y=7

Find the location of the optimal point, the value of λ
 and the value of f(x,y)
 at this point. Enter non-integer numerical values as decimals to at least 3 decimal places. Note: you must use a . and not , for a decimal point.

The optimal point is located at x=
Answer 1 Question 6
,   y=
Answer 2 Question 6
,  with  λ=
Answer 3 Question 6
, and f(x,y)=
Answer 4 Question 6


To solve this problem using Lagrange multipliers, we proceed as follows:

### Step 1: Define the problem

We have the function:
\[
f(x, y) = x^2 + y^2 + xy
\]
and the constraint:
\[
g(x, y) = x + 3y - 7 = 0
\]

### Step 2: Set up the Lagrange function

The Lagrange function is:
\[
\mathcal{L}(x, y, \lambda) = f(x, y) - \lambda \cdot g(x, y)
\]
Substituting the given functions:
\[
\mathcal{L}(x, y, \lambda) = x^2 + y^2 + xy - \lambda(x + 3y - 7)
\]

### Step 3: Find the partial derivatives and set them to zero

The partial derivatives with respect to \(x\), \(y\), and \(\lambda\) are:
\[
\frac{\partial \mathcal{L}}{\partial x} = 2x + y - \lambda = 0
\]
\[
\frac{\partial \mathcal{L}}{\partial y} = 2y + x - 3\lambda = 0
\]
\[
\frac{\partial \mathcal{L}}{\partial \lambda} = -(x + 3y - 7) = 0
\]

### Step 4: Solve the system of equations

We solve this system of equations:

1. \(2x + y = \lambda\)  (Equation 1)
2. \(x + 2y = 3\lambda\) (Equation 2)
3. \(x + 3y = 7\) (Constraint)

We can solve this system step by step. Let's start by expressing \(\lambda\) from Equation 1 and Equation 2.

From Equation 1:
\[
\lambda = 2x + y \quad \text{(Equation 4)}
\]
From Equation 2:
\[
\lambda = \frac{x + 2y}{3} \quad \text{(Equation 5)}
\]

Equating Equation 4 and Equation 5:
\[
2x + y = \frac{x + 2y}{3}
\]
Multiply everything by 3 to clear the fraction:
\[
6x + 3y = x + 2y
\]
Simplify:
\[
5x + y = 0 \quad \text{(Equation 6)}
\]
Now use Equation 6 in the constraint \(x + 3y = 7\) to find \(x\) and \(y\).

From Equation 6, express \(y\) in terms of \(x\):
\[
y = -5x
\]
Substitute this into the constraint:
\[
x + 3(-5x) = 7
\]
\[
x - 15x = 7
\]
\[
-14x = 7
\]
\[
x = -\frac{1}{2}
\]

Substitute \(x = -\frac{1}{2}\) back into Equation 6:
\[
y = -5(-\frac{1}{2}) = \frac{5}{2}
\]

### Step 5: Calculate \(\lambda\) and \(f(x, y)\)

Substitute \(x = -\frac{1}{2}\) and \(y = \frac{5}{2}\) into Equation 4 to find \(\lambda\):
\[
\lambda = 2(-\frac{1}{2}) + \frac{5}{2} = -1 + \frac{5}{2} = \frac{3}{2}
\]

Finally, calculate \(f(x, y)\) at the optimal point:
\[
f\left(-\frac{1}{2}, \frac{5}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(\frac{5}{2}\right)^2 + \left(-\frac{1}{2}\right)\left(\frac{5}{2}\right)
\]
\[
= \frac{1}{4} + \frac{25}{4} - \frac{5}{4} = \frac{21}{4} = 5.25
\]

### Conclusion

The optimal point is located at:
- \(x = -0.5\)
- \(y = 2.5\)
- \(\lambda = 1.5\)
- \(f(x, y) = 5.25\) 

If you have any questions or need further details, feel free to ask! Here are some related problems to consider:

1. Use Lagrange multipliers to find the maximum of \(f(x, y) = xy\) subject to the constraint \(x + y = 10\).
2. Solve a similar problem with the constraint \(x^2 + y^2 = 25\).
3. Explore the case when \(f(x, y) = x^2 - y^2\) and the constraint is \(x + y = 1\).
4. How does the method of Lagrange multipliers change when there are multiple constraints?
5. Find the optimal value of \(f(x, y) = x^2 + 3y^2\) subject to \(x + 2y = 6\).
6. Investigate how Lagrange multipliers apply to optimization problems in three dimensions.
7. Solve the problem when \(f(x, y, z) = x + y + z\) subject to \(x^2 + y^2 + z^2 = 1\).
8. Use Lagrange multipliers for a problem with an inequality constraint, such as \(x + y \leq 10\).

**Tip:** Remember to check the constraint carefully before solving the system of equations using Lagrange multipliers.

Learn how to optimize the function f(x, y) = x^2 + y^2 + xy using Lagrange multipliers under the constraint x + 3y = 7. Detailed step-by-step solution provided, including the optimal point, Lagrange multiplier value, and function value at the optimal point.

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