Problem 339 - Calculus Volume 1

Question

Consider two nonnegative numbers \(x\) and \(y\) such that \(x+y=10\). Maximize and minimize the quantities. \(y-\frac{1}{x}\)

Step-by-Step Solution

Answer

Maximum value is 8 at \(x=1\); Minimum value is \(-\infty\) as \(x\) approaches 0.

Step 1: Set up the Equation

We need to express the function we want to maximize or minimize, which is given as \(f(x, y) = y - \frac{1}{x}\). Also, we know that \(x + y = 10\). Thus, we can express \(y\) in terms of \(x\): \( y = 10 - x\). Substitute \(y = 10 - x\) into the function \(f(x, y)\).

Step 2: Substitute in the Function

Substitute \(y = 10 - x\) into the function \(f(x, y)\) to get a function of one variable: \(f(x) = (10 - x) - \frac{1}{x}\). So, the function becomes \(f(x) = 10 - x - \frac{1}{x}\).

Step 3: Find the Derivative

To find the critical points, first find the derivative of the function \(f(x) = 10 - x - \frac{1}{x}\). The derivative \(f'(x)\) is: \(f'(x) = -1 + \frac{1}{x^2}\).

Step 4: Set Derivative to Zero

Set the derivative \(f'(x)\) equal to zero to find the critical points: \(-1 + \frac{1}{x^2} = 0\). Solve for \(x\): \(\frac{1}{x^2} = 1\), which simplifies to \(x^2 = 1\). Thus, \(x = 1\) (since \(x\) is nonnegative).

Step 5: Verify the Endpoints

Consider the endpoints for the possible values of \(x\) where \(0 < x \leq 10\). Plug \(x = 1\), \(x = 0^+\), and \(x = 10\) into the function.

Step 6: Calculate Function at Critical Points

Evaluate the function at the critical point: \(f(1) = 10 - 1 - \frac{1}{1} = 8\).

Step 7: Evaluate Function at Endpoints

Evaluate the function at the endpoints (approaching values): \(f(10) = 10 - 10 - \frac{1}{10} = -0.1\). As \(x\) approaches 0 from the right, \(\lim_{x \to 0^+} f(x) = 10 - 0 - \frac{1}{0^+} = -\infty\).

Step 8: Determine Maximum and Minimum Values

The maximum value of \(f(x)\) is 8 when \(x = 1\) and the minimum value is \(-\infty\) as \(x\) approaches 0 from the right.

Key Concepts

Critical Points

Critical points in calculus are the values of a variable where the derivative of a function is zero or undefined. These points can indicate where a function might have a maximum or a minimum, or even a point of inflection.
To find critical points, we first need to take the derivative of the function. For example, consider the function presented in the problem:

• Initial function: \( f(x) = 10 - x - \frac{1}{x} \)
• Derivative: \( f'(x) = -1 + \frac{1}{x^2} \)

Setting \( f'(x) = 0 \) finds the critical points:

• \( -1 + \frac{1}{x^2} = 0 \)
• Solving gives \( x^2 = 1 \)
• Hence, \( x = 1 \), since only positive values are considered

The critical points are potential candidates for local maxima or minima of the function.

Derivative

A derivative in calculus is a tool that measures how a function changes as its input changes. It is a fundamental concept to determine the rate at which things happen.
The derivative of a function helps identify where a function's slope is zero, which is crucial for finding critical points. In this exercise, the derivative is calculated as follows:

• Function: \( f(x) = 10 - x - \frac{1}{x} \)
• Derivative: \( f'(x) = -1 + \frac{1}{x^2} \)

The derivative \( f'(x) \) indicates how the function \( f(x) \) changes at any point \( x \). When \( f'(x) = 0 \), it tells us there might be a critical point at which function behavior can change, indicating potential maxima or minima.
To solve for this derivative and find critical points, we set \( f'(x) \) to zero and solve for \( x \). This helps determine where the function's slope is zero, guiding us to potential peaks or troughs in the function's graph.

Maxima and Minima

Finding the maxima and minima of a function involves identifying the highest and lowest points the function can reach within a particular range.
In the provided exercise, by evaluating critical points and endpoints, we determine these values. After finding the critical point (\( x = 1 \)), it's essential to evaluate the function at this point and at the boundaries. Here's how:

• Calculate \( f(1) = 8 \), suggesting a possible maximum.
• Evaluate endpoints: \( f(10) = -0.1 \).
• Note behavior as \( x \) approaches 0 from the right: \( f(x) \to -\infty \).

Therefore, the maximum value of the function is 8 when \( x = 1 \), and the function approaches negative infinity as \( x \) nears zero. Understanding these points involves analyzing where the function's output reaches its highest and lowest values, guided by calculating derivatives and checking the function's limits.