Problem 339 - Calculus Volume 1

Question

For the following exercises, 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 is 8 at \(x=1\), minimum is approximately -0.1 at \(x=10\).

Step 1: Express y in terms of x

Given that \(x + y = 10\), we can express \(y\) in terms of \(x\) using the equation \(y = 10 - x\). This allows us to rewrite the expression \(f(x) = y - \frac{1}{x}\) as \(f(x) = 10 - x - \frac{1}{x}\).

Step 2: Find the critical points of the function

The function to maximize or minimize is \(f(x) = 10 - x - \frac{1}{x}\). To find critical points, take the derivative of \(f(x)\) with respect to \(x\): \(f'(x) = -1 + \frac{1}{x^2}\). Set \(f'(x) = 0\) to find critical points: \(-1 + \frac{1}{x^2} = 0\). This implies \(\frac{1}{x^2} = 1\) or \(x^2 = 1\). Thus, \(x = 1\).

Step 3: Evaluate the endpoints and critical point

Since \(x\) and \(y\) must be non-negative and \(x + y = 10\), the domain for \(x\) is from \(x = 0\) to \(x = 10\). Evaluate \(f(x)\) at the critical point \(x = 1\), and at the endpoints \(x = 0\) and \(x = 10\):- At \(x = 0\), \(f(x)\) is undefined due to division by zero.- At \(x = 1\), \(f(1) = 10 - 1 - 1 = 8\).- At \(x = 10\), \(y = 0\) since \(x+y = 10\), so \(f(10) = 0 - \frac{1}{10} = -\frac{1}{10}\).

Step 4: Determine the maximum and minimum values

Among the calculated function values, the maximum value is \(8\) at \(x = 1\) and the minimum value is approximately \(-0.1\) at \(x = 10\). We disregard \(x = 0\) since \(f(x)\) is undefined there due to division by zero.

Key Concepts

Understanding Critical Points

Critical points are essential in calculus when you're looking to maximize or minimize a function.

These points occur where the derivative of the function equals zero or does not exist.

This is because, at critical points, the slope of the tangent line to the graph is zero, indicating a potential maximum or minimum.

The first step in the original problem is finding these critical points: the derivative of the function was found, and then set equal to zero.

That's how the critical point at \(x = 1\) was identified.

• Critical points can be potential points of maxima or minima.
• They give insights into the function's behavior around these points.

Identifying the critical points allows us to examine exactly where a function can have high or low points.

The Role of the Derivative

The derivative is a powerful tool that provides the rate of change of a function.

Think of it as a way to understand how a function is behaving at any particular point.

In our exercise, the derivative of the function was given by \(f'(x) = -1 + \frac{1}{x^2}\). This tells us how the function changes as \(x\) changes.

When setting the derivative equal to zero, you find points where the function doesn't increase or decrease, identifying critical points. Such derivations help you to discover potential maxima or minima.

• The derivative indicates where the function's slope is zero, which corresponds to a change in increasing or decreasing behavior.
• By solving \(f'(x) = 0\), we pinpoint critical points that need evaluation for optimization.

Understanding the derivative makes it possible to predict the function's behavior.

Function Evaluation for Optimization

Function evaluation involves checking the calculated values of the function at specific points.

After identifying the critical point and understanding the domain of \(x\) (from 0 to 10 in this problem), these values are then plugged back into the function.

In this scenario:

• At the critical point \(x = 1\), the function has a value of 8, the highest.
• At the endpoint \(x = 10\), the function results in approximately \(-0.1\), the lowest.

Evaluating endpoints is crucial when considering optimization, ensuring you don't miss extreme values that could be potential solutions.

By comparing these values, one can determine both maximum and minimum values for optimization purposes.