/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A company manufactures two produ... [FREE SOLUTION] | 91Ó°ÊÓ

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A company manufactures two products, in quantities \(x\) and \(y\). Because of limited materials and capital, the quantities produced must satisfy the equation \(2 x^{2}+5 y^{2}=\) 32,500.(This curve is called a production possibilities curve.) If the company's profit function is \(P=4 x+5 y\) dollars, how many of each product should be made to maximize profit? Also find the maximum profit.

Short Answer

Expert verified
Produce approximately 40 units of product x and 29 units of product y for a maximum profit of $365.

Step by step solution

01

Set Up the Objective Function

The objective function is to maximize the company's profit, which is given by the equation \( P = 4x + 5y \). Our goal is to find the values of \( x \) and \( y \) that maximize \( P \).
02

Identify the Constraint

The constraint to this problem is the production possibilities curve given by the equation \( 2x^2 + 5y^2 = 32500 \). This is the boundary within which the profit must be maximized.
03

Use Lagrange Multipliers

To solve the problem, we'll use the Lagrange Multiplier method. Define the Lagrangian as \( \mathcal{L}(x, y, \lambda) = 4x + 5y + \lambda (32500 - 2x^2 - 5y^2) \).
04

Find Partial Derivatives

Compute the partial derivatives of the Lagrangian:- \( \frac{\partial \mathcal{L}}{\partial x} = 4 - 4\lambda x \)- \( \frac{\partial \mathcal{L}}{\partial y} = 5 - 10\lambda y \)- \( \frac{\partial \mathcal{L}}{\partial \lambda} = 32500 - 2x^2 - 5y^2 \)
05

Solve the System of Equations

Set the partial derivatives equal to zero to find the critical points:1. \( 4 - 4\lambda x = 0 \) implies \( x = \frac{1}{\lambda} \).2. \( 5 - 10\lambda y = 0 \) implies \( y = \frac{1}{2\lambda} \).3. \( 32500 - 2x^2 - 5y^2 = 0 \) is the constraint.
06

Substitute and Solve for Variable x or y

Substitute \( x = \frac{1}{\lambda} \) and \( y = \frac{1}{2\lambda} \) into the constraint:\[2\left(\frac{1}{\lambda}\right)^2 + 5\left(\frac{1}{2\lambda}\right)^2 = 32500\]Simplify and solve for \( \lambda \).
07

Calculate Values for x and y

Using the value for \( \lambda \) obtained, substitute back to find specific values for \( x \) and \( y \). Calculate these amounts to identify optimal production quantities.
08

Compute Maximum Profit

After determining the values of \( x \) and \( y \), substitute those back into the profit function \( P = 4x + 5y \) to calculate the maximum profit.

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Key Concepts

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

Production Possibilities Curve
In economics, the Production Possibilities Curve (PPC) is a graphical representation that shows the maximum number of two goods or services that can be produced within a given period, considering the available resources and technology. This curve helps to illustrate the trade-offs in production volume between two products.
In our exercise, the company can produce two products, represented as quantities \(x\) and \(y\), and their production capability is confined by the equation \(2x^2 + 5y^2 = 32500\). This equation is the mathematical representation of the PPC for this company.
  • It helps in visualizing what production combinations are possible, given resource restrictions.
  • The curve is usually a bowed-out shape, reflecting increasing opportunity costs.
  • Any point on the curve indicates efficient resource use, while points inside suggest under-utilization and points beyond indicate unattainable production with current resources.
Understanding the PPC allows companies to make informed decisions about the allocation of limited resources to maximize outputs.
Profit Maximization
Profit Maximization is a core objective for many businesses, representing the process of identifying the highest possible profit. To achieve this, companies seek the optimal combination of outputs that allow them to make the most money given their constraints.

In the given problem, the profit function is defined as \(P = 4x + 5y\), where \(x\) and \(y\) are units of products manufactured, and the coefficients 4 and 5 represent profit per unit of each product, respectively.
  • Maximizing profit involves adjusting production volumes of \(x\) and \(y\) to achieve the highest possible value of \(P\).
  • A practical approach involves using available resources efficiently, as defined by the Production Possibilities Curve.
  • Businesses may need to use mathematical optimization techniques, like calculus, to discover the maximum point.
By solving the problem accurately, we can determine the optimal number of each product to manufacture for maximum profitability under the given constraints.
Constrained Optimization
Constrained Optimization is a mathematical approach used to find the best possible outcome or optimal solution, considering specific limitations or constraints. In the realm of economics and business, it involves maximizing or minimizing functions subject to some restrictions.

The method of Lagrange Multipliers is a popular technique for handling constrained optimization problems. It enables us to work through the constraints directly while optimizing the function. In our task:- We introduce a Lagrangian function \(\mathcal{L}(x, y, \lambda) = 4x + 5y + \lambda(32500 - 2x^2 - 5y^2)\). - \(\lambda\) represents the Lagrange multiplier, a factor that adjusts according to the constraint's influence.
  • First, compute the gradients (partial derivatives) and set them to zero to find critical points.
  • Solve the system of equations derived from these derivatives to identify values that satisfy both the optimization goal and the constraints.
  • This involves substituting critical points back into the original equations to check results.
Using constrained optimization ensures resources are used in the most effective way possible within the limits the organization faces.

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Most popular questions from this chapter

A cylindrical tank without a top is to be constructed with the least amount of material (bottom plus side area). Find the dimensions if the volume is to be: 120 cubic feet.

Use Lagrange multipliers to minimize each function \(f(x, y)\) subject to the constraint. (The minimum values do exist.) $$ f(x, y)=\ln \left(x^{2}+y^{2}\right), \quad 2 x+y=25 $$

An automobile manufacturer sells cars in America, Europe, and Asia, charging a different price in each of the three markets. The price function for cars sold in America is \(p=20-0.2 x\) (for \(0 \leq x \leq 100\) ), the price function for cars sold in Europe is \(q=16-0.1 y \quad(\) for \(0 \leq y \leq 160),\) and the price function for cars sold in Asia is \(r=12-0.1 z\) (for \(0 \leq z \leq 120\) ), all in thousands of dollars, where \(x, y,\) and \(z\) are the numbers of cars sold in America, Europe, and Asia, respectively. The company's cost function is \(C=22+4(x+y+z)\) thousand dollars. a. Find the company's profit function \(P(x, y, z)\). [Hint: The profit will be revenue from America plus revenue from Europe plus revenue from Asia minus costs, where each revenue is price times quantity. b. Find how many cars should be sold in each market to maximize profit. [Hint: Set the three partials \(P_{x}, P_{y}\), and \(P_{z}\) equal to zero and solve. Assuming that the maximum exists, it must occur at this point.]

Use Lagrange multipliers to minimize each function \(f(x, y)\) subject to the constraint. (The minimum values do exist.) $$ f(x, y)=x^{2}+y^{2}, \quad x+2 y=30 $$

Use Lagrange multipliers to minimize each function \(f(x, y)\) subject to the constraint. (The minimum values do exist.) $$ f(x, y)=2 x+y, \quad 2 \ln x+\ln y=12 $$

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