/*! 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 5 A company manufactures \(x\) flo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A company manufactures \(x\) floor lamps and \(y\) table lamps each day. The profit in dollars for the manufacture and sale of these lamps is $$ P(x, y)=18 x+2 y-0.05 x^{2}-0.03 y^{2}+0.02 x y-100 $$ Find the daily production level of each lamp to maximize the company's profits.

Short Answer

Expert verified
Produce approximately 183 floor lamps and 42 table lamps daily.

Step by step solution

01

Find the First Derivatives

To find the production levels that maximize profit, we start by taking the partial derivatives of the profit function \(P(x, y)\) with respect to \(x\) and \(y\). Calculating these, we find:\[P_x = \frac{\partial P}{\partial x} = 18 - 0.1x + 0.02y\]\[P_y = \frac{\partial P}{\partial y} = 2 - 0.06y + 0.02x\]
02

Set Derivatives to Zero

We set both partial derivatives equal to zero in order to find the critical points where the profit could be maximized.\[ 18 - 0.1x + 0.02y = 0 \]\[ 2 - 0.06y + 0.02x = 0 \]
03

Solve the System of Equations

Solve the system of equations from Step 2. Start by solving one equation for one variable and substitute into the other.1. Solve \(18 = 0.1x - 0.02y\) \[ x = \frac{18 + 0.02y}{0.1} \]2. Substitute \(x\) into the second equation: \[ 2 = 0.06y - 0.02\left(\frac{18 + 0.02y}{0.1}\right) \] Simplifying leads to: \[ 2 = 0.06y - 0.36 - 0.004y \] \[ 2.36 = 0.056y \] Solve for \(y\): \[ y = \frac{2.36}{0.056} \approx 42.14 \]
04

Substitute Back to Find x

Use the value of \(y\) to find \(x\). Substitute \(y = 42.14\) into the expression for \(x\):\[ x = \frac{18 + 0.02(42.14)}{0.1} \]Calculate to find \(x\):\[ x \approx 182.8 \]
05

Check Second Derivatives (Second Derivative Test)

Calculate the second partial derivatives to use the second derivative test:\[ P_{xx} = -0.1, \quad P_{yy} = -0.06, \quad P_{xy} = 0.02 \]The determinant of the Hessian matrix is:\[ H = P_{xx}P_{yy} - (P_{xy})^2 = (-0.1)(-0.06) - (0.02)^2 = 0.006 - 0.0004 = 0.0056 \] Since \(H > 0\) and \(P_{xx} < 0\), the critical point is a maximum. Thus, the maximum profit occurs when \(x \approx 182.8\) and \(y \approx 42.14\), which means rounding to whole numbers: approximately 183 floor lamps and 42 table lamps.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Partial Derivatives
Partial derivatives are fundamental tools in optimization, particularly in economics when dealing with functions that depend on multiple variables. A partial derivative measures how a function changes as one of its variables changes, while keeping the other variables constant.
In the context of our lamp example, the profit function \( P(x, y) \) depends on two variables: the number of floor lamps \( x \) and the number of table lamps \( y \). To determine how the profit changes as we vary \( x \) and \( y \) independently, we calculate two partial derivatives:
  • \( P_x = \frac{\partial P}{\partial x} \): Measures the rate of change of profit with respect to the number of floor lamps, keeping the number of table lamps constant.
  • \( P_y = \frac{\partial P}{\partial y} \): Measures the rate of change of profit with respect to the number of table lamps, keeping the number of floor lamps constant.
These derivatives help us identify how sensitive overall profit is to changes in either type of lamp produced.
Critical Points
When optimizing a function like the profit function, critical points are points where the first derivatives (partial derivatives) are zero. This is where the function could potentially have a maximum, a minimum, or a saddle point.
For the lamp production problem, we set the partial derivatives \( P_x \) and \( P_y \) to zero to locate the critical points:
  • \( 18 - 0.1x + 0.02y = 0 \)
  • \( 2 - 0.06y + 0.02x = 0 \)
Solving these equations gives us the values of \( x \) and \( y \) at the critical point. These values are not guaranteed to be optimum, so further analysis, such as the second derivative test, is needed to confirm their nature.
Second Derivative Test
The second derivative test is a method to determine the nature of critical points found by solving the first derivative conditions. It involves calculating the second partial derivatives and evaluating the Hessian matrix.
In the lamp problem, after identifying the critical points, we calculate:
  • \( P_{xx} = \frac{\partial^2 P}{\partial x^2} = -0.1 \)
  • \( P_{yy} = \frac{\partial^2 P}{\partial y^2} = -0.06 \)
  • \( P_{xy} = \frac{\partial^2 P}{\partial x \partial y} = 0.02 \)
The results are used to form the Hessian matrix, and its determinant helps to determine if the critical point is a maximum, minimum, or saddle point.
Hessian Matrix
The Hessian matrix is a square matrix comprising the second partial derivatives of a function with respect to its variables. It's a key tool in multivariable calculus for analyzing critical points.
The Hessian matrix \( H \) for the given profit function is:\[H = \begin{bmatrix}P_{xx} & P_{xy} \P_{xy} & P_{yy} \end{bmatrix} = \begin{bmatrix}-0.1 & 0.02 \0.02 & -0.06 \end{bmatrix}\]To apply the second derivative test, we compute the determinant of this matrix:\[H = P_{xx}P_{yy} - (P_{xy})^2\]For the lamp example, the determinant is \( 0.0056 \), which is positive. If \( H > 0 \) and \( P_{xx} < 0 \), the critical point corresponds to a local maximum. This allows us to conclude that the production of about 183 floor lamps and 42 table lamps will maximize profit.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A continuous function \(f(x, y)\) takes on its absolute extrema on a closed and bounded region either at an interior point or at a boundary point of the region. Find the absolute extrema of $$ f(x, y)=48 x y-32 x^{3}-24 y^{2} $$ on the square region \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\).

Consider an athlete competing in the shot put. What factors influence the length of his or her throw? Construct a model that predicts the distance thrown as a function of the initial velocity and angle of release. What is the optimal angle of release? If the athlete cannot maximize the initial velocity at the angle of release you propose, should he or she be more concerned with satisfying the angle of release or generating a high initial velocity? What are the trade-offs?

One of the difficulties in managing a fishery in a free-enterprise system is that excess capacity may be created through overcapitalization. This happened in 1970 when the capacity of the Peruvian anchoveta fishermen was sufficient to catch and process the maximum annual growth rate in less than 3 months. A disadvantage of restricting access to the fishery by closing the season after a quota is reached is that this excess capacity is idle during much of the season, which creates a politically and economically unsatisfying situation. An alternative is to control the capacity in some manner. Suggest several procedures for controlling the capacity that is developed. What difficulties would be involved in implementing a procedure such as restricting the number of commercial fishing permits issued?

Problems 5 and 6 are related to fishing regulation. One of the difficulties in managing a fishery in a free-enterprise system is that excess capacity may be created through overcapitalization. This happened in 1970 when the capacity of the Peruvian anchoveta fishermen was sufficient to catch and process the maximum annual growth rate in less than 3 months. A disadvantage of restricting access to the fishery by closing the season after a quota is reached is that this excess capacity is idle during much of the season, which creates a politically and economically unsatisfying situation. An alternative is to control the capacity in some manner. Suggest several procedures for controlling the capacity that is developed. What difficulties would be involved in implementing a procedure such as restricting the number of commercial fishing permits issued?

Cow to Market-A cow currently weighs \(800 \mathrm{lb}\) and is gaining \(35 \mathrm{lb}\) per week. It costs \(\$ 6.50\) a week to maintain the cow. The market price today is \(\$ 0.95\) per pound but is falling \(\$ 0.01\) per day. Formulate a mathematical model and find the optimal period to keep the cow until it is sold to maximize profits.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.