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Chapter 2: Problem 6

Find two positive numbers \(x\) and \(y\) that maximize \(Q=x^{2} y\) if \(x+y=2\).

Short Answer

Expert verified
The numbers are \( x = \frac{4}{3} \) and \( y = \frac{2}{3} \).

Step by step solution

01

Define the constraint equation

Given the constraint that the sum of the two positive numbers is 2, write the equation as: \[ x + y = 2 \]
02

Express one variable in terms of the other

From the constraint equation \( x + y = 2 \), solve for \( y \) in terms of \( x \): \[ y = 2 - x \]
03

Substitute into the function to be maximized

Substitute \( y = 2 - x \) into the function \( Q = x^2 y \): \[ Q = x^2 (2 - x) \]
04

Simplify the function

Simplify the function \( Q \): \[ Q = 2x^2 - x^3 \]
05

Find the derivative of the function

To find the maximum value, take the derivative of \( Q \) with respect to \( x \) and set it to zero: \[ Q' = \frac{d}{dx} (2x^2 - x^3) = 4x - 3x^2 \]
06

Solve for the critical points

Solve the equation \( 4x - 3x^2 = 0 \): \[ x(4 - 3x) = 0 \] The solutions are: \[ x = 0 \] and \[ x = \frac{4}{3} \]
07

Determine the feasible solution

Since we need positive numbers and \( x + y = 2 \), check which solution satisfies the constraint: \( x = 0 \) is not valid because it makes \( y = 2 \). So, \( x = \frac{4}{3} \) makes \( y = 2 - \frac{4}{3} = \frac{2}{3} \)
08

Verify the solution

Check that the values satisfy the original constraint and the function is maximized: \( x + y = \frac{4}{3} + \frac{2}{3} = 2 \). Thus, \( x = \frac{4}{3} \) and \( y = \frac{2}{3} \) are the solutions.

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Key Concepts

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

Constraint Equation
In optimization problems, constraint equations define the specific conditions that the variables must satisfy.

For this problem, we are given that the sum of two positive numbers, let's call them x and y, must be 2. This forms our constraint equation: \ \[ x + y = 2 \]

This equation is crucial because it restricts the values of x and y, allowing us to find specific solutions. By expressing one variable in terms of another using this equation, we simplify the problem and make it easier to work with.
Derivatives
Derivatives play a key role in finding the maximum or minimum values of functions.

In this problem, we want to maximize the function Q, defined as \( Q = x^2 y \). After substituting \( y = 2 - x \) into Q, we get: \ \[ Q = x^2 (2 - x) \]

Therefore, our function Q becomes: \ \[ Q = 2x^2 - x^3 \]

To find the maximum value, we need to calculate the derivative of Q with respect to x and then set it to zero: \ \[ Q' = \frac{d}{dx} (2x^2 - x^3) = 4x - 3x^2 \]

The derivative gives us the rate of change of the function. Setting the derivative to zero helps us identify 'critical points'.
Critical Points
Critical points are where the derivative of a function equals zero or does not exist. These points are where potential maxima or minima occur.

Once we have the derivative \( Q' = 4x - 3x^2 \), we set it to zero to find the critical points: \ \[ 4x - 3x^2 = 0 \]

Solving this equation gives us: \ \[ x(4 - 3x) = 0 \]

So our solutions are: \ \[ x = 0 \] and \( x = \frac{4}{3} \).

Critical points are essential in determining where the function can achieve its maximum or minimum value, given the constraints.
Positive Numbers
In the context of this problem, we are asked to find positive numbers x and y that meet our conditions.

From our critical points, we have: \ \[ x = 0 \ and \ x = \frac{4}{3} \]

Since x=0 is not a positive number, we discard it.

That leaves us with \( x = \frac{4}{3} \). Using our constraint equation \( x + y = 2 \), we find y: \ \[ y = 2 - \frac{4}{3} = \frac{2}{3} \]

Both \( x = \frac{4}{3} \) and \( y = \frac{2}{3} \) are positive, satisfying all conditions. This ensures that the provided solution is valid.

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Most popular questions from this chapter

Postal requirements specify that parcels must have length plus girth of at most 84 inches. Consider the problem of finding the dimensions of the square- ended rectangular package of greatest volume that is mailable. (a) Draw a square-ended rectangular box. Label each edge of the square end with the letter \(x\) and label the remaining dimension of the box with the letter \(h\). (b) Express the length plus the girth in terms of \(x\) and \(h\). (c) Determine the objective and constraint equations. (d) Express the quantity to be maximized as a function of \(x\). (e) Find the optimal values of \(x\) and \(h\).

The monthly demand equation for an electric utility company is estimated to be $$ p=60-\left(10^{-5}\right) x \text { , } $$ where \(p\) is measured in dollars and \(x\) is measured in thousands of kilowatt- hours. The utility has fixed costs of 7 million dollars per month and variable costs of $$\$ 30$$ per 1000 kilowatt-hours of electricity generated, so the cost function is $$ C(x)=7 \cdot 10^{6}+30 x \text { . } $$ (a) Find the value of \(x\) and the corresponding price for 1000 kilowatt-hours that maximize the utility's profit. (b) Suppose that rising fuel costs increase the utility's variable costs from $$\$ 30$$ to $$\$ 40$$, so its new cost function is $$ C_{1}(x)=7 \cdot 10^{6}+40 x $$ Should the utility pass all this increase of $$\$ 10$$ per thousand kilowatt- hours on to consumers? Explain your answer.

Coffee consumption in the United States is greater on a per capita basis than anywhere else in the world. However, due to price fluctuations of coffee beans and worries over the health effects of caffeine, coffee consumption has varied considerably over the years. According to data published in The Wall Street Journal, the number of cups \(f(x)\) consumed daily per adult in year \(x\) (with 1955 corresponding to \(x=0)\) is given by the mathematical model $$ f(x)=2.77+0.0848 x-0.00832 x^{2}+0.000144 x^{3} $$ (a) Graph \(y=f(x)\) to show daily coffee consumption from 1955 through 1994 . (b) Use \(f^{\prime}(x)\) to determine the year in which coffee consumption was least during this period. What was the daily coffee consumption at that time? (c) Use \(f^{\prime}(x)\) to determine the year in which coffee consumption was greatest during this period. What was the daily coffee consumption at that time? (d) Use \(f^{\prime \prime}(x)\) to determine the year in which coffee consumption was decreasing at the greatest rate.

Each of the graphs of the functions has one relative maximum and one relative minimum point. Plot these two points and check the concavity there. Using only this information, sketch the graph. $$ f(x)=\frac{1}{3} x^{3}+2 x^{2}-5 x+\frac{8}{3} $$

Sketch the following curves, indicating all relative extreme points and inflection points. $$ y=x^{4}-4 / 3 x^{3} $$
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