Maximize the function 

f(x,y)=8xy−3x2−3x−4y+10

subject to the constraint

2x+3y=4

Find the location of the maximum and its value. 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 maximum is located at x=
Answer 1 Question 7
,   y=
Answer 2 Question 7
, and f(x,y)=

To maximize the function \( f(x, y) = 8xy - 3x^2 - 3x - 4y + 10 \) subject to the constraint \( 2x + 3y = 4 \), we can use the method of **Lagrange multipliers**.

### Step 1: Set up the Lagrange function
The Lagrange function is given by:
\[
\mathcal{L}(x, y, \lambda) = 8xy - 3x^2 - 3x - 4y + 10 + \lambda (2x + 3y - 4)
\]
where \( \lambda \) is the Lagrange multiplier.

### Step 2: Take partial derivatives
Next, 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. \(\frac{\partial \mathcal{L}}{\partial x} = 8y - 6x - 3 + 2\lambda = 0\)
2. \(\frac{\partial \mathcal{L}}{\partial y} = 8x - 4 + 3\lambda = 0\)
3. \(\frac{\partial \mathcal{L}}{\partial \lambda} = 2x + 3y - 4 = 0\)

### Step 3: Solve the system of equations
We solve the system of equations given by the partial derivatives.

1. From \(\frac{\partial \mathcal{L}}{\partial x} = 0\):
\[
8y - 6x - 3 + 2\lambda = 0 \quad \text{(Equation 1)}
\]

2. From \(\frac{\partial \mathcal{L}}{\partial y} = 0\):
\[
8x - 4 + 3\lambda = 0 \quad \text{(Equation 2)}
\]

3. From \(\frac{\partial \mathcal{L}}{\partial \lambda} = 0\):
\[
2x + 3y = 4 \quad \text{(Equation 3)}
\]

### Step 4: Express \(\lambda\) from the equations and substitute
Let's solve Equation 2 for \(\lambda\):
\[
3\lambda = 4 - 8x \quad \Rightarrow \quad \lambda = \frac{4 - 8x}{3} \quad \text{(Equation 4)}
\]

Now, substitute \(\lambda\) from Equation 4 into Equation 1:
\[
8y - 6x - 3 + 2\left(\frac{4 - 8x}{3}\right) = 0
\]

Simplify:
\[
8y - 6x - 3 + \frac{8 - 16x}{3} = 0
\]

Multiply through by 3 to eliminate the fraction:
\[
24y - 18x - 9 + 8 - 16x = 0
\]

Simplify further:
\[
24y - 34x - 1 = 0 \quad \Rightarrow \quad 24y = 34x + 1 \quad \Rightarrow \quad y = \frac{34x + 1}{24}
\]

### Step 5: Substitute \( y \) into the constraint equation
Substitute \( y = \frac{34x + 1}{24} \) into the constraint \( 2x + 3y = 4 \):
\[
2x + 3\left(\frac{34x + 1}{24}\right) = 4
\]

Multiply through by 24 to clear the fraction:
\[
48x + 3(34x + 1) = 96
\]

Expand and simplify:
\[
48x + 102x + 3 = 96 \quad \Rightarrow \quad 150x = 93 \quad \Rightarrow \quad x = \frac{93}{150} = 0.62
\]

### Step 6: Solve for \( y \)
Now, substitute \( x = 0.62 \) back into the equation for \( y \):
\[
y = \frac{34(0.62) + 1}{24} \approx \frac{21.08 + 1}{24} = \frac{22.08}{24} = 0.92
\]

### Step 7: Calculate \( f(x, y) \) at the maximum point
Finally, substitute \( x = 0.62 \) and \( y = 0.92 \) back into the original function:
\[
f(0.62, 0.92) = 8(0.62)(0.92) - 3(0.62)^2 - 3(0.62) - 4(0.92) + 10
\]

I'll calculate the exact values.The maximum value of the function \( f(x, y) \) occurs at:

- \( x = 0.620 \)
- \( y = 0.920 \)

The maximum value of the function is \( f(0.62, 0.92) = 7.870 \).

Would you like to dive into any specific details, or have any further questions?

Here are some related questions:
1. How do you apply Lagrange multipliers to more complex constraints?
2. What if the constraint was non-linear, how would the method change?
3. How do Lagrange multipliers work in higher dimensions?
4. Could there be multiple maxima or minima for this problem?
5. How does this method compare to solving with substitution directly?
6. What are the necessary conditions for a maximum or minimum using Lagrange multipliers?
7. Can you solve a similar problem with a different constraint, like \( x + y = 5 \)?
8. How does the Hessian matrix help in determining maxima or minima?

**Tip:** Always double-check the partial derivatives and calculations, as small algebraic errors can lead to incorrect results.

Learn how to maximize the function f(x, y) = 8xy - 3x^2 - 3x - 4y + 10 under the constraint 2x + 3y = 4. Detailed step-by-step solution using Lagrange multipliers and partial derivatives. Find the optimal values of x and y and the maximum value of the function. Suitable for advanced undergraduate students and professionals.

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