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Chapter 10: Problem 11

Find three positive real numbers whose sum is 94 and whose product is a maximum. Enter the three numbers separated by commas: ____________.

Short Answer

Expert verified
31, 31, 32

Step by step solution

01

Introduce the constraint

From the sum equation, we have: \[z = 94 - x - y\] Now we can express the product P in terms of x and y only.
02

Rewrite the product formula

Rewrite the product P using the constraint equation: \[P(x, y) = x \cdot y \cdot (94 - x - y)\]
03

Find the partial derivatives

To optimize the product P(x, y), we will find the gradient, which consists of two partial derivatives: \[\frac{\partial P}{\partial x} = y(94 - x - y) - 2xy\] \[\frac{\partial P}{\partial y} = x(94 - x - y) - 2yx\] We need to find the critical points of the product function P(x, y), so we must solve for x and y when the gradient is equal to zero: \[ \nabla P(x, y) = \begin{cases} \frac{\partial P}{\partial x} = 0 \\ \frac{\partial P}{\partial y} = 0 \end{cases}\]
04

Solve the system of equations

We will now solve the system of equations obtained in step 3 (this may require some algebraic manipulations and substitutions): \[\begin{cases} y(94 - x - y) - 2xy = 0 \\ x(94 - x - y) - 2yx = 0 \end{cases}\] Simplifying and solving for x and y, we find the critical points: \[x = y = 31\]
05

Find the third number

Now that we have the values of x and y, we can find z using the constraint equation we introduced in step 1: \[z = 94 - x - y = 94 - 31 - 31 = 32\]
06

Verify the solution

To check if this solution satisfies the problem statement, let's verify that the numbers are positive real numbers, and their sum is 94: \[x = 31, y = 31, z = 32\] \[x + y + z = 31 + 31 + 32 = 94\] The three numbers are positive real numbers, and they indeed sum up to 94. Answer: \(x = 31\), \(y = 31\), \(z = 32\)

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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 powerful tools used in calculus to find the rate of change of a function with respect to one variable while keeping other variables constant. When dealing with functions of multiple variables, such as the product function in our exercise, partial derivatives help us identify how changes in one variable affect the whole function.

In mathematical optimization problems, partial derivatives are utilized to find maxima or minima of functions that depend on multiple variables. By setting the partial derivatives equal to zero, as we did in this exercise, we can identify critical points, where the function may achieve its highest value. These derivatives effectively break down the multidimensional problem into simpler, more understandable parts, allowing for effective and insightful analysis. They are essential in finding the optimal solution when dealing with multivariate functions.
Critical Points
Critical points are locations in the domain of a function where the derivative is zero or undefined. These points often correspond to maxima, minima, or saddle points of the function, making them key in optimization problems.

In our exercise, we derive two equations from the partial derivatives that lead us to the critical points. By solving the system of equations, we find specific values for the variables that indicate potential points of maximum product.

Once found, critical points must be tested to determine their nature鈥攚hether they are maxima, minima, or saddle points. This testing is essential to ensure that the solution meets the problem's criteria: in this case, maximizing the product of the three numbers. Critical points provide a way to homing in on the solution in multivariate calculus problems.
Constraint Equation
A constraint equation is a condition that the solution must satisfy. In our exercise, the sum of three numbers should be 94, which gives us our constraint equation: \[z = 94 - x - y\].

Using constraint equations simplifies multivariable problems by reducing the number of variables in the main function. Here, it allowed us to express the product function with only two variables instead of three. This reduction makes it feasible to manage and solve complex optimization problems.

Constraints are vital in real-world scenarios where resources or conditions must be met. By incorporating constraints, we ensure that the optimization solution is not only mathematically sound but also practically applicable. They define the boundaries within which optimal solutions must reside.
Product Maximization
Product maximization involves finding values for the variables that lead to the highest possible product under given constraints. This type of problem is common in economics and logistics, where maximizing output or efficiency is the goal.

In this exercise, we aim to maximize the product of three numbers with a fixed sum. By utilizing calculus techniques, including partial derivatives and critical points, we efficiently find the values of these numbers. The solution demonstrates the interplay between mathematical theory and practical problem-solving.

Successful product maximization hinges on correctly applying mathematical principles to analyze and solve the given problem. It illustrates how calculus serves as a powerful tool in translating theoretical concepts into beneficial real-world outcomes. In this particular exercise, it allowed us to pinpoint the specific numbers that maximized the product, given the sum constraint, showcasing the effectiveness of calculus in optimization tasks.

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Most popular questions from this chapter

In this exercise we consider how to apply the Method of Lagrange Multipliers to optimize functions of three variable subject to two constraints. Suppose we want to optimize \(f=f(x, y, z)\) subject to the constraints \(g(x, y, z)=c\) and \(h(x, y, z)=k\). Also suppose that the two level surfaces \(g(x, y, z)=c\) and \(h(x, y, z)=k\) intersect at a curve \(C\). The optimum point \(P=\left(x_{0}, y_{0}, z_{0}\right)\) will then lie on \(C\). a. Assume that \(C\) can be represented parametrically by a vector-valued function \(\mathbf{r}=\mathbf{r}(t) .\) Let \(\overrightarrow{O P}=\mathbf{r}\left(t_{0}\right) .\) Use the Chain Rule applied to \(f(\mathbf{r}(t)), g(\mathbf{r}(t)),\) and \(h(\mathbf{r}(t)),\) to explain why $$\begin{array}{l} \nabla f\left(x_{0}, y_{0}, z_{0}\right) \cdot \mathbf{r}^{\prime}\left(t_{0}\right)=0 \\ \nabla g\left(x_{0}, y_{0}, z_{0}\right) \cdot \mathbf{r}^{\prime}\left(t_{0}\right)=0, \text { and } \\ \nabla h\left(x_{0}, y_{0}, z_{0}\right) \cdot \mathbf{r}^{\prime}\left(t_{0}\right)=0\end{array}$$ Explain how this shows that \(\nabla f\left(x_{0}, y_{0}, z_{0}\right), \nabla g\left(x_{0}, y_{0}, z_{0}\right),\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) are all orthogonal to \(C\) at \(P\). This shows that \(\nabla f\left(x_{0}, y_{0}, z_{0}\right), \nabla g\left(x_{0}, y_{0}, z_{0}\right),\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) all lie in the same plane. b. Assuming that \(\nabla g\left(x_{0}, y_{0}, z_{0}\right)\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) are nonzero and not parallel, explain why every point in the plane determined by \(\nabla g\left(x_{0}, y_{0}, z_{0}\right)\) and \(\nabla h\left(x_{0}, y_{0}, z_{0}\right)\) has the form \(s \nabla g\left(x_{0}, y_{0}, z_{0}\right)+t \nabla h\left(x_{0}, y_{0}, z_{0}\right)\) for some scalars \(s\) and \(t\). c. Parts (a.) and (b.) show that there must exist scalars \(\lambda\) and \(\mu\) such that $$\nabla f\left(x_{0}, y_{0}, z_{0}\right)=\lambda \nabla g\left(x_{0}, y_{0}, z_{0}\right)+\mu \nabla h\left(x_{0}, y_{0}, z_{0}\right)$$ So to optimize \(f=f(x, y, z)\) subject to the constraints \(g(x, y, z)=c\) and \(h(x, y, z)=k\) we must solve the system of equations $$\nabla f(x, y, z)=\lambda \nabla g(x, y, z)+\mu \nabla h(x, y, z)$$ $$\begin{array}{l}g(x, y, z)=c, \text { and } \\\h(x, y, z)=k .\end{array}$$ for \(x, y, z, \lambda,\) and \(\mu .\) Use this idea to find the maximum and minimum values of \(f(x, y, z)=\) \(x+2 y\) subject to the constraints \(y^{2}+z^{2}=8\) and \(x+y+z=10\).

Let \(z=u^{2}-v^{2}\) and suppose that $$\begin{array}{l}u=e^{x} \cos (y) \\\v=e^{x} \sin (y)\end{array}$$ a. Find the values of \(u\) and \(v\) that correspond to \(x=0\) and \(y=2 \pi / 3\). b. Use the Chain Rule to find the general partial derivatives $$\frac{\partial z}{\partial x} \text { and } \frac{\partial z}{\partial y}$$ and then determine both \(\left.\frac{\partial z}{\partial x}\right|_{\left(0, \frac{2 \pi}{3}\right)}\) and \(\left.\frac{\partial z}{\partial y}\right|_{\left(0, \frac{2 \pi}{3}\right)}\).

The airlines place restrictions on luggage that can be carried onto planes. \- A carry-on bag can weigh no more than 40 lbs. \- The length plus width plus height of a bag cannot exceed 45 inches. \- The bag must fit in an overhead bin. Let \(x, y,\) and \(z\) be the length, width, and height (in inches) of a carry on bag. In this problem we find the dimensions of the bag of largest volume, \(V=x y z,\) that satisfies the second restriction. Assume that we use all 45 inches to get a maximum volume. (Note that this bag of maximum volume might not satisfy the third restriction.) a. Write the volume \(V=V(x, y)\) as a function of just the two variables \(x\) and \(y\) b. Explain why the domain over which \(V\) is defined is the triangular region \(R\) with vertices \((0,0),(45,0),\) and (0,45) . c. Find the critical points, if any, of \(V\) in the interior of the region \(R\). d. Find the maximum value of \(V\) on the boundary of the region \(R\), and then determine the dimensions of a bag with maximum volume on the entire region \(R\). (Note that most carry-on bags sold today measure 22 by 14 by 9 inches with a volume of 2772 cubic inches, so that the bags will fit into the overhead bins.)

Find the first partial derivatives of \(f(x, y)=\sin (x-y)\) at the point \((-4,\) -4) A. \(f_{x}(-4,-4)=\) _________. B. \(f_{y}(-4,-4)=\) _________.

If \(f(x, y, z)=2 z y^{2},\) then the gradient at the point (2,2,4) is \(\nabla f(2,2,4)=\) _____________.
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