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Chapter 14: Problem 26

Find the largest product the positive numbers \(x, y,\) and \(z\) can have if \(x+y+z^{2}=16.\)

Short Answer

Expert verified
The maximum product is 64 when \( x = 8, y = 8, z = 0 \).

Step by step solution

01

Establish Expression for Product

We want to maximize the product of three positive numbers, so we first define the product as a function to optimize. Let the product be given by \( P = x imes y \). Since we are given that \( x + y + z^2 = 16 \), this constraint will be used in our optimization process.
02

Express Variables in One Function

From the constraint \( x + y + z^2 = 16 \), solve for \( y = 16 - x - z^2 \). Substitute this into the product function: \[ P = x imes (16 - x - z^2) = 16x - x^2 - xz^2. \] We aim to maximize \( P \) with respect to \( x \) and \( z \).
03

Differentiate and Find Critical Points

Since the two variables are still present, we need to employ partial differentiation. Compute the partial derivative of \( P \) with respetct to \( x \): \( \frac{\partial P}{\partial x} = 16 - 2x - z^2 \) and for \( z \): \( \frac{\partial P}{\partial z} = -2xz \). Set these derivatives to zero to locate the critical points: \[ 16 - 2x - z^2 = 0 \] and \[ -2xz = 0. \] Solving results in \( x = 8 - \frac{z^2}{2} \) and \( z = 0 \) or \( x = 0 \).
04

Assess Solution Cases for Maximum

Consider the case where \( x = 0 \). This results in \( y = 0 \) which doesn't help in maximizing because product is zero. Now consider \( z=0 \), then we have \( x =8 \) from \( 16 - 2x = 0 \), and substituting back, \( y=16-x = 8 \). Thus \( P = 8 \times 8 = 64 \).
05

Conclude with the Solution

We have found that when \( x = 8 \), \( y = 8 \), and \( z = 0 \), the constraint \( x + y + z^2 = 16 \) is satisfied and the product is maximized to \( 64 \). Therefore, the largest product the positive numbers \( x, y, z \) can have is 64.

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Key Concepts

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

Partial Derivatives
When aiming to optimize a function involving multiple variables, partial derivatives become essential tools. They help us understand how changes in one variable affect the function while keeping the other variables constant.
Let's say we have a function of two variables, like the product function from our exercise:
  • First, identify the variables involved, in this case, "x" and "z."
  • Next, calculate the partial derivative of the function with respect to each variable. This tells us how the function changes when that specific variable changes.
  • For "x," it would be: \( \frac{\partial P}{\partial x} = 16 - 2x - z^2 \).
  • For "z," it would be: \( \frac{\partial P}{\partial z} = -2xz \).
By finding where these derivatives equal zero, we can identify potential points where the function might reach a maximum or minimum value.
These are called the "critical points."
Critical Points
Critical points are like stop signs on the road to optimization. They signal the locations where potential extrema (maximum or minimum values) may occur.
In our exercise, we found the critical points by setting the partial derivatives to zero:
  • For \( \frac{\partial P}{\partial x} = 0 \), we solve \( 16 - 2x - z^2 = 0 \).
  • For \( \frac{\partial P}{\partial z} = 0 \), we solve \( -2xz = 0 \).
The challenge is to analyze these points thoroughly. In our case, we derived possibilities such as \( x = 8 - \frac{z^2}{2} \) with \( z = 0 \) or \( x = 0 \). It's crucial to understand which of these points provide valid solutions given the constraints.
Critical points only tell us where to look further, not whether these points really maximize or minimize the function. That's where further evaluation is required.
Constraint Optimization
Constraint optimization involves maximizing or minimizing a function while meeting specific conditions, known as constraints.
In our problem, the constraint is given by the equation \( x + y + z^2 = 16 \). Here's how you handle this:
  • First, solve the constraint for one variable, making it dependent on others. For example, \( y = 16 - x - z^2 \).
  • Substitute this expression into your optimization function to transform it into a single-variable or two-variable problem.
  • This helps in reducing the complexity of the problem and makes it easier to apply calculus to find potential solutions.
Once you have a simplified equation, further differentiation helps to locate critical points and potential extrema under the constraint. Constraint optimization is essential when there are additional criteria that your solution must satisfy.
Maximizing Product
Maximizing a product can be challenging, especially when dealing with multiple variables and constraints. The key to solving such problems lies in structuring your work using calculus and algebra effectively.
Here are the general steps which can help in maximizing a product in problems similar to our exercise:
  • Define the product to be maximized, such as \( P = x \times y \).
  • Use given constraints to express one variable in terms of others, reducing the number of variables.
  • Form the new equation by substituting these expressions back into the product function.
  • Differentiate to find the critical points and assess these points to determine a maximum product.
In our specific example, setting \( z = 0 \) led us to find that \( x = 8 \) and \( y = 8 \) provided the largest product under the constraint. So, strategic use of derivative techniques and careful assessment of critical points ensures success in maximizing products.

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Most popular questions from this chapter

Find the limits by rewriting the fractions first. $$\lim _{(x, y) \rightarrow(0,0)} \frac{\sin \left(x^{2}+y^{2}\right)}{x^{2}+y^{2}}$$

Find the limits. $$\lim _{(x, y) \rightarrow(0,0)} \frac{e^{y} \sin x}{x}$$

By considering different paths of approach, show that the functions have no limit as \((x, y) \rightarrow(0,0)\). $$f(x, y)=\frac{x^{4}}{x^{4}+y^{2}}$$

To find the extreme values of a function \(f(x, y)\) on a curve \(x=x(t), y=y(t),\) we treat \(f\) as a function of the single variable \(t\) and use the Chain Rule to find where \(d f / d t\) is zero. As in any other single-variable case, the extreme values of \(f\) are then found among the values at the a. critical points (points where \(d f / d t\) is zero or fails to exist), and b. endpoints of the parameter domain. Find the absolute maximum and minimum values of the following functions on the given curves. Curves: i) The semicircle \(x^{2}+y^{2}=4, \quad y \geq 0\) ii) The quarter circle \(x^{2}+y^{2}=4, \quad x \geq 0, \quad y \geq 0\) Use the parametric equations \(x=2 \cos t, y=2 \sin t\).

When we try to fit a line \(y=m x+b\) to a set of numerical data points \(\left(x_{1}, y_{1}\right)\) \(\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right),\) we usually choose the line that minimizes the sum of the squares of the vertical distances from the points to the line. In theory, this means finding the values of \(m\) and \(b\) that minimize the value of the function $$w=\left(m x_{1}+b-y_{1}\right)^{2}+\cdots+\left(m x_{n}+b-y_{n}\right)^{2}. \quad (1)$$ (See the accompanying figure.) Show that the values of \(m\) and \(b\) that do this are $$m=\frac{\left(\sum x_{k}\right)\left(\sum y_{k}\right)-n \sum x_{k} y_{k}}{\left(\sum x_{k}\right)^{2}-n \sum x_{k}^{2}},\quad (2)$$ $$b=\frac{1}{n}\left(\sum y_{k}-m \sum x_{k}\right),\quad(3)$$ with all sums running from \(k=1\) to \(k=n\). Many scientific calculators have these formulas built in, enabling you to find \(m\) and \(b\) with only a few keystrokes after you have entered the data. The line \(y=m x+b\) determined by these values of \(m\) and \(b\) is called the least squares line, regression line, or trend line for the data under study. Finding a least squares line lets you 1\. summarize data with a simple expression, 2\. predict values of \(y\) for other, experimentally untried values of \(x\), 3\. handle data analytically. We demonstrated these ideas with a variety of applications in Section 1.4.
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