Problem 25
Question
In Exercises \(25-28,\) find three positive numbers \(x, y,\) and \(z\) that satisfy the given conditions. The sum is 120 and the product is maximum.
Step-by-Step Solution
Verified Answer
The three numbers are \(x = 40\), \(y = 40\) and \(z = 40\), and their product is \(64000\).
1Step 1: Variables Setting
Let the three numbers be \(x\), \(y\), and \(z\). According to the problem, we know that \(x + y + z = 120\). We want to maximize the product \(P = x*y*z\).
2Step 2: Rewrite Into Two Variable Problem
To make the problem simpler, express \(z\) from the equation \(x + y + z = 120\), yelding \(z = 120 - x - y\). Substituting \(z\) in the product \(P = x*y*z\), we get \(P = x*y*(120 - x - y)\). We have now transformed the problem into a two variable problem.
3Step 3: Optimization
To find the maximum product P, we take the derivative of \(P\) with respect to both \(x\) and \(y\) and set them equal to zero. Thus, we solve \(\frac{\partial P}{\partial x} = 0\) and \(\frac{\partial P}{\partial y} = 0\). After solving these equations, we find that \(x = y = z = 40\).
4Step 4: Verification
These values do indeed produce the desired sum of 120 as \(x + y + z = 40 + 40 + 40 = 120\). Further, since we have a symmetric situation where each of the variables \(x, y, z\) is set to equal the same value, the product \(P = x*y*z = 40*40*40 = 64000\), which is a maximum that can be achieved for any set of three numbers summing to 120 according to AM-GM inequality.
Key Concepts
Derivative in OptimizationAM-GM InequalityMultivariable Functions
Derivative in Optimization
Derivatives are powerful tools in optimization. They help us find the maximum or minimum values of functions, which is crucial in solving optimization problems. In the given exercise, we aim to maximize the product of three numbers, given their sum. The product function is expressed in terms of two variables, after substituting one variable using the sum constraint.
By taking partial derivatives of the function with respect to each variable, we set these derivatives equal to zero to find the critical points. Specifically, we solve \[\frac{\partial P}{\partial x} = 0 \quad \text{and} \quad \frac{\partial P}{\partial y} = 0\]These critical points are candidates for the locations of the maximum product. Simplifying the above, we find that the derivatives tell us where the slope of the function is zero, which is where maxima or minima might occur.
By taking partial derivatives of the function with respect to each variable, we set these derivatives equal to zero to find the critical points. Specifically, we solve \[\frac{\partial P}{\partial x} = 0 \quad \text{and} \quad \frac{\partial P}{\partial y} = 0\]These critical points are candidates for the locations of the maximum product. Simplifying the above, we find that the derivatives tell us where the slope of the function is zero, which is where maxima or minima might occur.
- Why Zero: Setting the derivative to zero helps us find important points of change in a function.
- Critical Points: These points need further checking to determine whether they are maxima or minima.
AM-GM Inequality
The Arithmetic Mean-Geometric Mean (AM-GM) Inequality is a powerful strategy used in optimization problems involving products and sums. The inequality states that for non-negative numbers, the arithmetic mean is always greater than or equal to the geometric mean.
For three numbers, the AM-GM Inequality is given by:\[\frac{x+y+z}{3} \geq \sqrt[3]{xyz}\]The equality holds when all numbers are equal, i.e., for the maximum product, we need \(x = y = z\).
In our problem, it leads us to set all three numbers equal to each other when their sum is constrained. Thus:
For three numbers, the AM-GM Inequality is given by:\[\frac{x+y+z}{3} \geq \sqrt[3]{xyz}\]The equality holds when all numbers are equal, i.e., for the maximum product, we need \(x = y = z\).
In our problem, it leads us to set all three numbers equal to each other when their sum is constrained. Thus:
- Equality Case: Occurs when all numbers are the same, leading to the optimal product result.
- Verification: It serves as a quick check to ensure that we have the maximum product.
Multivariable Functions
Multivariable functions depend on multiple inputs, providing a richer landscape for exploration in calculus. They are crucial when considering optimization where several variables influence the outcome. In this problem, the function representing the product is initially based on three variables, but can be reduced by using the condition that their sum is fixed.
To handle this product-maximization problem, we utilize two variables by substituting one using the sum constraint. The product function becomes:\[ P = x \cdot y \cdot (120 - x - y)\]
Understanding and analyzing multivariable functions involves tools such as:
To handle this product-maximization problem, we utilize two variables by substituting one using the sum constraint. The product function becomes:\[ P = x \cdot y \cdot (120 - x - y)\]
Understanding and analyzing multivariable functions involves tools such as:
- Partial Derivatives: Essential for exploring changes with respect to each variable separately.
- Critical Points: These are where the partial derivatives are zero, indicating potential maximum/minimum values.
Other exercises in this chapter
Problem 25
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Determine whether the planes \(a_{1} x+b_{1} y+c_{1} z=d_{1}\) and \(a_{2} x+b_{2} y+c_{2} z=d_{2}\) are parallel, perpendicular, or neither. The planes are par
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