/*! 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 46 An electronics company's profit ... [FREE SOLUTION] | 91Ó°ÊÓ

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An electronics company's profit \(P(x, y)\) from making \(x\) DVD players and \(y\) CD players per day is given below. a. Find the marginal profit function for DVD players. b. Evaluate your answer to part (a) at \(x=200\) and \(y=300\) and interpret the result. c. Find the marginal profit function for CD players. d. Evaluate your answer to part (c) at \(x=200\) and \(y=100\) and interpret the result. \(P(x, y)=3 x^{2}-4 x y+4 y^{2}+80 x+100 y+200\)

Short Answer

Expert verified
The marginal profit for DVD players at 200, 300 is $80; for CD players at 200, 100 is $100.

Step by step solution

01

Marginal Profit Function for DVD Players

To find the marginal profit function for DVD players, we take the partial derivative of the profit function \(P(x, y)\) with respect to \(x\).\[\frac{\partial P}{\partial x} = \frac{\partial}{\partial x} (3x^2 - 4xy + 4y^2 + 80x + 100y + 200)\]Evaluating the derivative, we get:\[\frac{\partial P}{\partial x} = 6x - 4y + 80\]
02

Evaluate Marginal Profit Function for DVD Players

Substitute \(x = 200\) and \(y = 300\) into the marginal profit function \(\frac{\partial P}{\partial x}\):\[\frac{\partial P}{\partial x} = 6(200) - 4(300) + 80 = 1200 - 1200 + 80 = 80\]The marginal profit for producing one more DVD player when \(200\) DVD players and \(300\) CD players are produced is \(80\) dollars.
03

Marginal Profit Function for CD Players

To find the marginal profit function for CD players, we take the partial derivative of the profit function \(P(x, y)\) with respect to \(y\).\[\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (3x^2 - 4xy + 4y^2 + 80x + 100y + 200)\]Evaluating the derivative, we obtain:\[\frac{\partial P}{\partial y} = -4x + 8y + 100\]
04

Evaluate Marginal Profit Function for CD Players

Substitute \(x = 200\) and \(y = 100\) into the marginal profit function \(\frac{\partial P}{\partial y}\):\[\frac{\partial P}{\partial y} = -4(200) + 8(100) + 100 = -800 + 800 + 100 = 100\]The marginal profit for producing one more CD player when \(200\) DVD players and \(100\) CD players are produced is \(100\) dollars.

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Key Concepts

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

Partial Derivatives
Partial derivatives are a cornerstone in multivariable calculus, especially when dealing with functions like the profit function in this exercise. When we have a function of multiple variables, such as the profit function \( P(x, y) \), which depends on both \( x \) (DVD players) and \( y \) (CD players), partial derivatives allow us to see how the function changes as one variable changes, while keeping the other constant. This is crucial for calculating marginal values in economics and business.

For example, to find the marginal profit function for DVD players, we take the partial derivative of \( P(x, y) \) with respect to \( x \). This involves differentiating \( P(x, y) \) as if \( y \) were a constant. Similarly, to find the marginal profit function for CD players, we take the partial derivative with respect to \( y \).

  • Application: Helps determine the rate of change of profit with respect to each type of product independently.
  • Result: For DVD players, we found the partial derivative \( \frac{\partial P}{\partial x} = 6x - 4y + 80 \). For CD players, it is \( \frac{\partial P}{\partial y} = -4x + 8y + 100 \).
Profit Function
In economics, a profit function represents the total profit a company makes from selling goods or services. It is expressed as a function of quantities of goods, costs of production, and sometimes other economic factors. In our exercise, the profit function \( P(x, y) = 3x^2 - 4xy + 4y^2 + 80x + 100y + 200 \) encapsulates the relationship between producing and selling \( x \) DVD players and \( y \) CD players.

Understanding this function is crucial for businesses to strategize and make informed production decisions. The function usually includes:
  • Revenue components: Gains from selling products.
  • Cost components: Expenses associated with production.
  • Constant term: Fixed costs or baseline profit.
In this case, the function includes terms for both variables \(x\) and \(y\), allowing us to calculate profit based on any combination of DVD and CD players produced daily.
Marginal Analysis
Marginal analysis is a technique often used in economics to assess the impact of a unit change in the decision variable. In our context, it helps determine how much additional profit can be made from producing one more unit of a product.

This exercise uses marginal analysis to find the marginal profit, which is the additional profit from making one more DVD player or CD player. After computing the partial derivatives, we substitute specific values to evaluate at given production levels.

  • DVD Players at \(x = 200\), \(y = 300\): The marginal profit was \(80\) dollars, meaning that each additional DVD player contributes \\(80 of profit under these conditions.
  • CD Players at \(x = 200\), \(y = 100\): The marginal profit was \(100\) dollars, meaning that each additional CD player contributes \\)100 of profit under these conditions.
Thus, marginal analysis provides valuable insights into optimizing production and maximizing profits based on current operational levels.

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Most popular questions from this chapter

Find the total differential of each function. $$ f(x, y, z)=x y+y z+x z $$

Show that for the Cobb-Douglas production function \(P=a L^{b} K^{1-b}\) restricted to the isocost line \(w L+r K=C^{*}\) (where \(L\) and \(K\) are the amounts of labor and capital with prices \(w\) and \(r\), and \(C^{*}\) is a given total cost), the maximum possible production is $$ P^{*}=a\left(\frac{b}{w}\right)^{b}\left(\frac{1-b}{r}\right)^{1-b} C^{*} $$ when the labor and capital are $$ L^{*}=\left(\frac{b}{w}\right) \mathrm{C}^{*} \text { and } K^{*}=\left(\frac{1-b}{r}\right) \mathrm{C}^{*} $$ and \(\lambda\) is $$ \lambda^{*}=-a\left(\frac{b}{w}\right)^{b}\left(\frac{1-b}{r}\right)^{1-b} $$

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Find the total differential of each function. $$ f(x, y, z)=e^{x y z} $$

THE SLOPE OF \(f(x, y)=c\) On page 556 we used the fact that the slope in the \(x\) - \(y\) plane of the curve defined by \(f(x, y)=c\) (for constant \(c\) ) is given by the formula \(-\frac{f_{x}}{f_{v}}\). Verify this formula by justifying the following five steps. a. If \(f(x, y)=c\) can be solved explicitly for a function \(y=F(x)\), then we may write \(f(x, F(x))=c .\) Justify: \(f(x+\Delta x, F(x+\Delta x))-f(x, F(x))=0\). b. Justify: \(f(x+\Delta x, F(x+\Delta x)-F(x)+F(x))\) \(-f(x, F(x))=0\) c. Defining \(\Delta F\) by \(\Delta F=F(x+\Delta x)-F(x)\), we may write the previous equation as $$ f(x+\Delta x, \Delta F+F(x))-f(x, F(x))=0 $$ Then, writing \(F\) for \(F(x)\), this becomes $$ f(x+\Delta x, F+\Delta F)-f(x, F)=0 $$ Justify: \(f_{x} \Delta x+f_{y} \Delta F \approx 0\). d. Justify: \(\frac{\Delta F}{\Delta x} \approx-\frac{f_{x}}{f_{y}}\). e. Justify: \(\frac{d F}{d x}=-\frac{f_{x}}{f_{y}}\). This shows that the slope of \(F(x)\), and therefore the slope of \(f(x, y)=c\), is \(-\frac{f_{x}}{f_{y}}\).

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