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Chapter 12: Problem 1

Solve the optimization problems. Maximize \(P=x y\) with \(x+y=10\).

Short Answer

Expert verified
The maximum value of the function \(P(x, y) = xy\) with the constraint \(x + y = 10\) is 25, which occurs when \(x = 5\) and \(y = 5\).

Step by step solution

01

Solve the constraint for one of the variables

Let's solve the constraint \(x + y = 10\) for y: \[y = 10 - x\]
02

Substitute the expression from Step 1 into the objective function

Now, substituting the expression for y from Step 1 into the objective function \(P(x, y) = xy\), we obtain \(P(x) = x(10 - x)\): \[P(x) = 10x - x^2\]
03

Find the critical points of the function

To find the critical points of P(x), we need to take the first derivative and set it equal to zero: \[\frac{dP}{dx} = \frac{d}{dx}(10x - x^2) = 10 - 2x\] Now, set the derivative equal to zero and solve for x: \[10 - 2x = 0\] \[x = 5\]
04

Determine the value of y

Now that we have found the value of x that maximizes P(x), we can find the corresponding value of y from the expression derived in Step 1: \[y = 10 - x = 10 - 5 = 5\]
05

Determine the maximum value of P(x, y)

Finally, substitute the values of x and y found in the previous steps into the objective function to find the maximum value of P(x, y): \[P(5, 5) = 5 \cdot 5 = 25\] The maximum value of P(x, y) is 25 when x = 5 and y = 5.

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Key Concepts

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

Objective Function
In optimization problems, the objective function is a mathematical expression that represents the quantity you want to maximize or minimize. In this case, the objective function is the equation you want to optimize: \( P = xy \). This problem's goal is to find the values of \( x \) and \( y \) that maximize the product of these two variables. Understanding the objective function is crucial because it defines what optimization means in the given problem.
  • The objective function acts as a guide, showing what needs to be optimized.
  • It can be a simple expression, like in this case \( xy \), or much more complex when dealing with multi-variable problems.
  • Always start by clearly identifying the objective function, as it is the foundation for solving the problem.
Once you identify the objective function, the next step is to consider any constraints that might affect how you can manipulate \( x \) and \( y \).
Critical Points
Critical points are values of \( x \) (or any other variable) that make the derivative of the objective function zero or undefined. These points are crucial because they often indicate the locations of maximum or minimum values of the objective function. In this problem, after setting up the equation \( P(x) = 10x - x^2 \), we took its first derivative \( \frac{dP}{dx} = 10 - 2x \) and set it equal to zero. This gave us the critical point \( x = 5 \).
  • Finding critical points involves taking the derivative of the objective function.
  • Critical points can sometimes be endpoints or places where the derivative does not exist, so always analyze the specific context of the problem.
  • In this exercise, checking the critical point \( x = 5 \) helped determine where the function reached its maximum value under the given constraints.
Experimenting with critical points gives a deeper understanding of the function's behavior at varying criteria, essential for optimization.
Derivatives
Derivatives play a key role in optimization problems, as they help us understand the rate of change of functions and find critical points. In simple terms, derivatives can tell us how a function behaves as its variables change. For the objective function \( P(x) = 10x - x^2 \), its derivative \( \frac{dP}{dx} = 10 - 2x \) shows us how \( P \) changes with \( x \). Setting the derivative to zero helps find the critical points which are potential maxima or minima for the function.
  • The first derivative of a function gives us the slope at any point along the function, which is used to find critical points.
  • By setting the derivative equal to zero, you can identify where these potential maxima or minima occur.
  • Graphical representation of derivatives can also help visualize how changes occur in the objective function.
Understanding derivatives and their calculations is essential for mastering optimization and solving various mathematical problems.
Constraints in Optimization
Constraints in optimization define the boundaries or conditions that the solution to an optimization problem must satisfy. In this problem, the constraint is given by the linear equation \( x + y = 10 \). This means that any solution must satisfy this equation, which limits the possible values of \( x \) and \( y \). Constraints narrow down the solution space and make the problem complex and realistic.
  • Constraints can be equations or inequalities.
  • They can limit the possible solutions by bounding the variables into a feasible region.
  • Identifying and properly managing constraints is crucial for finding the optimal solution.
In this exercise, solving the constraint allowed us to express \( y \) in terms of \( x \) which simplified the two-variable problem to a single variable, making it easier to handle and find the optimal solution.

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Most popular questions from this chapter

Your friend tells you that he has found a continuous function defined on \((-\infty,+\infty)\) with exactly two critical points, each of which is a relative maximum. Can he be right?

A right circular conical vessel is being filled with green industrial waste at a rate of 100 cubic meters per second. How fast is the level rising after \(200 \pi\) cubic meters have been poured in? The cone has a height of \(50 \mathrm{~m}\) and a radius of \(30 \mathrm{~m}\) at its brim. (The volume of a cone of height \(h\) and crosssectional radius \(r\) at its brim is given by \(V=\frac{1}{3} \pi r^{2} h .\).)

You have been hired as a marketing consultant to Big Book Publishing, Inc., and you have been approached to determine the best selling price for the hit calculus text by Whiner and Istanbul entitled Fun with Derivatives. You decide to make life easy and assume that the demand equation for Fun with Derivatives has the linear form \(q=m p+b\), where \(p\) is the price per book, \(q\) is the demand in annual sales, and \(m\) and \(b\) are certain constants you'll have to figure out. a. Your market studies reveal the following sales figures: when the price is set at \(\$ 50.00\) per book, the sales amount to 10,000 per year; when the price is set at \(\$ 80.00\) per book, the sales drop to 1000 per year. Use these data to calculate the demand equation. b. Now estimate the unit price that maximizes annual revenue and predict what Big Book Publishing, Inc.'s annual revenue will be at that price.

The distance of a UFO from an observer is given by \(s=2 t^{3}-3 t^{2}+100\) feet after \(t\) seconds \((t \geq 0)\). Obtain the extrema, points of inflection, and behavior at infinity. Sketch the curve and interpret these features in terms of the movement of the UFO.

By thinking about extrema, show that, if \(f(x)\) is a polynomial, then between every pair of zeros \((x\) -intercepts) of \(f(x)\) there is a zero of \(f^{\prime}(x)\)
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