Problem 340 - Calculus Volume 1

Question

Consider two nonnegative numbers \(x\) and \(y\) such that \(x+y=10\). Maximize and minimize the quantities. \(x^{2}-y\)

Step-by-Step Solution

Answer

Maximum is 100 at \(x = 10\), and minimum is -10 at \(x = 0\).

Step 1: Express one variable in terms of the other

Given the equation \(x + y = 10\), we can express \(y\) in terms of \(x\). This gives us \(y = 10 - x\). We will use this expression to eliminate \(y\) from the function we want to maximize/minimize.

Step 2: Substitute into the function

Substitute \(y = 10 - x\) into the expression \(x^2 - y\). This gives: \[ x^2 - y = x^2 - (10 - x) = x^2 + x - 10 \].Now, we have a function of a single variable to maximize or minimize, which is \(f(x) = x^2 + x - 10\).

Step 3: Find critical points

To find the critical points, compute the derivative of \(f(x) = x^2 + x - 10\). The derivative is \(f'(x) = 2x + 1\). Set the derivative equal to zero and solve for \(x\):\[ 2x + 1 = 0 \]\[ 2x = -1 \]\[ x = -\frac{1}{2} \].

Step 4: Evaluate endpoints and critical point

Evaluate the function \(f(x) = x^2 + x - 10\) at the endpoints of the interval and at the critical point. Since \(x\) and \(y\) are nonnegative and \(x + y = 10\), \(x\) must be in the interval \([0, 10]\).- At \(x = 0\), \(f(0) = 0^2 + 0 - 10 = -10\).- At \(x = 10\), \(f(10) = 10^2 + 10 - 10 = 100\).- At \(x = -\frac{1}{2}\), since it's outside the valid range, it does not need evaluation.

Step 5: Determine maximum and minimum values

Based on evaluations: - The maximum value of \(f(x) = x^2 + x - 10\) is \(100\) at \(x = 10\). - The minimum value of \(f(x) = x^2 + x - 10\) is \(-10\) at \(x = 0\).

Key Concepts

Critical Points

Critical points are essential in optimization problems. They help us find where a function may reach its maximum or minimum values. Identifying critical points involves the derivative of the function. By setting the derivative to zero, or finding where it's undefined, we can locate these crucial points.

In our exercise, we found the function representing our condition:

• The function to examine was \[f(x) = x^2 + x - 10\].
• We derived the function to get \[f'(x) = 2x + 1\].
• Setting the derivative to zero provided us with the critical point \[x = -\frac{1}{2}\].

This calculation indicates a potential maximum or minimum. Yet, be cautious. Here, the critical point was outside our valid range \([0, 10]\), so we disregarded it. Always check your domain!

Derivative

The derivative is a powerful tool in calculus. It gives us the function's rate of change, revealing how the function behaves at any given point. Use derivatives to find slopes of tangent lines, investigate increasing or decreasing trends, and, importantly, locate critical points.

For the problem at hand:

• Our function, \[f(x) = x^2 + x - 10\], had a derivative \[f'(x) = 2x + 1\].
• Taking the derivative tells us how the function \(f(x)\) changes as \(x\) varies.
• By solving \[2x + 1 = 0\], we find values of \(x\) where the change switches direction, indicating critical points.

The derivative isn't just about computation. It provides insight into the behavior of functions and aids in solving optimization challenges. It's fundamental when determining where functions attain their extreme values.

Function Evaluation

Function evaluation is the process of calculating the output of a function given specific inputs. In optimization, we evaluate the function at critical points and boundaries to find extremes—the highest or lowest values the function can take.

For our case, we looked at the function \(f(x)\) as follows:

• Evaluating at the endpoints \(x = 0\) and \(x = 10\) gave us \(-10\) and \(100\) respectively.
• The endpoint evaluations are crucial because they can capture maximum or minimum values.

Notice, the critical point \(-\frac{1}{2}\) was discarded because it fell outside the domain \([0, 10]\). Evaluation confirms whether or not obtained critical points provide practical solutions, benefiting overall decision-making in optimization scenarios.