Question

1. a company that produces customized sports bottles, has developed a profit model that depends on the number (x) of bottles sold per month (in thousands), and the number (y) of advertising packages purchased per month. the monthly profit is given (thousands of euros) by

\\p(x,y) = 20x + 10y - \frac{x^2}{2} - \frac{y^2}{2} + 10\\

the company has a budget constraint relating production and advertising costs, given by

\\8x + 2y = 112\\

3.0 (a) using the method of lagrange multipliers, find the values of (x) and (y) that maximize the profit under the budget constraint. present a conclusion within this specific context.

Explanation:

🆕 New Concept Discovered: Lagrange Multipliers
Finding extrema subject to equality constraints

Step 1: Formulate the Lagrangian Function

To maximize the profit function \( P(x, y) \) subject to the constraint \( g(x, y) = c \), we define the constraint function:

$$ g(x, y) = 8x + 2y = 112 \implies 8x + 2y - 112 = 0 $$

The Lagrangian function \( \mathcal{L}(x, y, \lambda) \) is defined as:

$$ \mathcal{L}(x, y, \lambda) = P(x, y) - \lambda (g(x, y) - 112) $$

$$ \mathcal{L}(x, y, \lambda) = 20x + 10y - \frac{x^2}{2} - \frac{y^2}{2} + 10 - \lambda(8x + 2y - 112) $$

Step 2: Set up the System of Equations

We find the first-order partial derivatives of \( \mathcal{L} \) with respect to \( x \), \( y \), and \( \lambda \), and set them equal to zero:

1. With respect to \( x \):

$$ \frac{\partial \mathcal{L}}{\partial x} = 20 - x - 8\lambda = 0 \implies x = 20 - 8\lambda $$

1. With respect to \( y \):

$$ \frac{\partial \mathcal{L}}{\partial y} = 10 - y - 2\lambda = 0 \implies y = 10 - 2\lambda $$

1. With respect to \( \lambda \):

$$ \frac{\partial \mathcal{L}}{\partial \lambda} = -(8x + 2y - 112) = 0 \implies 8x + 2y = 112 $$

Step 3: Solve for the Multiplier and Variables

Substitute the expressions for \( x \) and \( y \) in terms of \( \lambda \) into the constraint equation:

$$ 8(20 - 8\lambda) + 2(10 - 2\lambda) = 112 $$

$$ 160 - 64\lambda + 20 - 4\lambda = 112 $$

$$ 180 - 68\lambda = 112 $$

$$ 68\lambda = 68 \implies \lambda = 1 $$

Now, substitute \( \lambda = 1 \) back to find \( x \) and \( y \):

$$ x = 20 - 8(1) = 12 $$

$$ y = 10 - 2(1) = 8 $$

Step 4: Contextual Conclusion

We interpret the values of \( x \) and \( y \) within the given business context:

• \( x = 12 \) represents \( 12,000 \) customized sports bottles sold per month.
• \( y = 8 \) represents \( 8 \) advertising packages purchased per month.

Let's calculate the maximum monthly profit at this optimal point:

$$ P(12, 8) = 20(12) + 10(8) - \frac{12^2}{2} - \frac{8^2}{2} + 10 $$

$$ P(12, 8) = 240 + 80 - 72 - 32 + 10 = 226 $$

Since the profit is in thousands of euros, the maximum profit is \( 226,000 \) euros.

Answer:

The values that maximize the profit under the budget constraint are:

• \( x = 12 \) (corresponding to selling 12,000 bottles per month)
• \( y = 8 \) (corresponding to purchasing 8 advertising packages per month)

Conclusion:
To maximize its monthly profit under the given budget constraint, the company should produce and sell 12,000 customized sports bottles and purchase 8 advertising packages per month. This optimal allocation yields a maximum monthly profit of 226,000 euros.