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A company manufactures two products. The price function for product \(\mathrm{A}\) is \(p=12-\frac{1}{2} x \quad(\) for \(0 \leq x \leq 24),\) and for product \(\mathrm{B}\) is \(q=20-y \quad(\) for \(0 \leq y \leq 20),\) both in thousands of dollars, where \(x\) and \(y\) are the amounts of products A and B, respectively. If the cost function is $$ C(x, y)=9 x+16 y-x y+7 $$ thousands of dollars, find the quantities and the prices of the two products that maximize profit. Also find the maximum profit.

Short Answer

Expert verified
Quantities: A = 10, B = 7; Prices: A = $7k, B = $13k; Max Profit: $22k.

Step by step solution

01

Define the Profit Function

The profit function is given by the total revenue minus the total cost. For product A, the revenue is given by the price times the quantity, so it's \(x(12-\frac{1}{2}x)\). For product B, the revenue is \(y(20-y)\). Therefore, the total revenue is \(R(x, y) = x(12-\frac{1}{2}x) + y(20-y)\). The total cost is given by \(C(x, y) = 9x + 16y - xy + 7\). Thus, the profit function is \(P(x, y) = R(x, y) - C(x, y)\).
02

Form the Profit Function

Substitute the given revenue and cost expressions into the profit function: \[P(x, y) = [x(12-\frac{1}{2}x) + y(20-y)] - (9x + 16y - xy + 7)\]. Simplifying the expression: \[P(x, y) = 12x - \frac{1}{2}x^2 + 20y - y^2 - 9x - 16y + xy - 7\], which simplifies further to \[P(x, y) = -\frac{1}{2}x^2 - y^2 + xy + 3x + 4y - 7\].
03

Calculate the Partial Derivatives

To find the maximum profit, we need to find where the gradient of the profit function \(abla P(x, y) = (\frac{\partial P}{\partial x}, \frac{\partial P}{\partial y})\) is zero. Calculate these partial derivatives: \(\frac{\partial P}{\partial x} = -x + y + 3\) and \(\frac{\partial P}{\partial y} = -2y + x + 4\).
04

Solve the System of Equations

Set the partial derivatives to zero and solve the system: \(-x + y + 3 = 0\) and \(-2y + x + 4 = 0\). Solving the first equation for \(y\): \(y = x - 3\). Substitute into the second equation: \(-2(x - 3) + x + 4 = 0\) resulting in \(-2x + 6 + x + 4 = 0\) simplifying to \(-x + 10 = 0\), thus, \(x = 10\). Substituting \(x = 10\) into \(y = x - 3\) gives \(y = 7\).
05

Determine the Prices

Substitute \(x = 10\) into the price function for product A: \(p = 12 - \frac{1}{2}(10) = 7\). Substitute \(y = 7\) into the price function for product B: \(q = 20 - 7 = 13\). The prices are \(p = 7\) and \(q = 13\) thousand dollars.
06

Calculate the Maximum Profit

Substitute \(x = 10\) and \(y = 7\) into the profit function. First, find the simplified profit function: \[P(x, y) = -\frac{1}{2}(10)^2 - (7)^2 + (10)(7) + 3(10) + 4(7) - 7\]. After calculations: \[P(10, 7) = -50 - 49 + 70 + 30 + 28 - 7 = 22\]. Therefore, the maximum profit is \(22\) thousand dollars.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Price Function
The price function in economics describes how the price of a product varies with the level of output or quantity of goods being sold. In simpler terms, it's a way to figure out how much you can charge for each unit of a product. In our exercise, there are two price functions for products A and B.

For product A, we have: \ \( p = 12 - \frac{1}{2}x \) for \( 0 \leq x \leq 24 \). This is a linear function, where the price decreases by \( \frac{1}{2} \) for each additional unit of A. Essentially, as more units of product A are sold, the price decreases. This could happen due to factors such as increased supply or reduced demand.

For product B, the price function is: \ \( q = 20 - y \) for \( 0 \leq y \leq 20 \). Similar to product A, this linear relationship implies that for each additional unit of B, the price goes down by 1 unit. These decreasing price functions reflect a common economic principle where more supply may lead to lower prices. By analyzing the price functions, companies can determine optimal pricing strategies to maximize revenue and, ultimately, profit.
Cost Function
The cost function represents the total cost incurred by a company to produce a certain quantity of goods. It includes fixed costs that do not change with the level of output and variable costs that fluctuate with production volume.

In this exercise, the cost function is given by: \ \( C(x, y)=9x+16y-xy+7 \). Here we have:
  • \( 9x \): The cost related to producing x units of product A.
  • \( 16y \): The cost associated with producing y units of product B.
  • \( -xy \): This term shows an interaction effect between producing A and B together, which could represent a form of efficiency or synergy.
  • \( +7 \): A constant cost that could represent expenses that are fixed regardless of production levels.


By analyzing the cost function, companies can determine their expenses at various production levels and work out how to minimize costs while maximizing output. Understanding these dynamics is crucial for optimizing profit margins.
Partial Derivatives
Partial derivatives are used to investigate how a function changes as each of its variables change, holding the others constant. In the context of this exercise, we're dealing with a profit function dependent on two products, A and B.

The profit function derived from our problem is: \ \( P(x, y) = -\frac{1}{2}x^2 - y^2 + xy + 3x + 4y - 7 \). To find the values of \(x\) and \(y\) that maximize profit, we calculate the partial derivatives with respect to \(x\) and \(y\).

For \( \frac{\partial P}{\partial x} \), or the change in profit when only \(x\) is changed, we get \( -x + y + 3 \). For \( \frac{\partial P}{\partial y} \), or the change when only \(y\) is varied, we find \( -2y + x + 4 \).
  • Setting these partial derivatives to zero helps identify the critical points which may maximize (or minimize) the profit function.
  • This technique is powerful in optimization problems where several variables play a role in determining the outcome.


By solving the resulting system of equations, we find optimal quantities of products A and B to maximize profit. The solution offers practical insights for making strategic business decisions.

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