Problem 338 - Calculus Volume 1

Question

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

Step-by-Step Solution

Answer

Max: 625 at \(x = 5\); Min: 0 at \(x = 0\) or \(x = 10\).

Step 1: Express one variable in terms of the other

Since we know that the sum of two numbers is 10, we can express one variable in terms of the other. Here, express \(y\) in terms of \(x\) as follows: \(y = 10 - x\).

Step 2: Rewrite the expression to be optimized

Substitute \(y = 10 - x\) into the expression \(x^2 y^2\) to rewrite it as a function of \(x\):\[f(x) = x^2 (10 - x)^2\]

Step 3: Simplify and expand the function

Expand \((10 - x)^2\) as \(100 - 20x + x^2\), then substitute it into \(f(x)\):\[f(x) = x^2 (100 - 20x + x^2) = 100x^2 - 20x^3 + x^4\]

Step 4: Find the first derivative

To find the critical points, compute the derivative of \(f(x)\):\[f'(x) = \frac{d}{dx}(100x^2 - 20x^3 + x^4) = 200x - 60x^2 + 4x^3\]

Step 5: Set the derivative equal to zero to find critical points

Solve \(f'(x) = 0\) to find the critical points:\[4x^3 - 60x^2 + 200x = 0\]Factor out \(4x\):\[4x(x^2 - 15x + 50) = 0\]This gives us \(x = 0\) or solving \(x^2 - 15x + 50 = 0\).

Step 6: Solve the quadratic equation

To solve \(x^2 - 15x + 50 = 0\) use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 1, b = -15, c = 50\):\[ x = \frac{15 \pm \sqrt{225 - 200}}{2} = \frac{15 \pm \sqrt{25}}{2} = \frac{15 \pm 5}{2}\]Thus, \(x = 10\) or \(x = 5\).

Step 7: Evaluate the function at critical points

Evaluate \(f(x)\) at \(x = 0, 5, 10\):- \(f(0) = 0^2(10)^2 = 0\)- \(f(10) = 10^2(0)^2 = 0\)- \(f(5) = 5^2(10-5)^2 = 25 \times 25 = 625\)

Step 8: Analyze and conclude

From the evaluations, \(f(x)\) reaches a maximum of 625 at \(x = 5, y = 5\) and a minimum of 0 at \(x = 0, y = 10\) or \(x = 10, y = 0\).

Key Concepts

Critical Points

In calculus, critical points play a key role in finding the maxima or minima of functions. A critical point occurs where the derivative of a function is zero or undefined. These points are crucial because they pinpoint potential maxima or minima of the function. By solving for critical points, you can determine where the function reaches its highest or lowest values.

To find critical points, you follow these steps:

• Take the derivative of the function.
• Set the derivative equal to zero.
• Solve for the variable.

Analyzing critical points allows you to make informed decisions about optimization, which is especially useful in problems involving economic efficiency or design constraints.

Quadratic Formula

The quadratic formula is a powerful tool in algebra used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). This formula provides the solutions to the equation by relating the coefficients \(a\), \(b\), and \(c\). It is expressed as:

\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula is particularly useful when factorization is complex or impossible. By substituting the coefficients into the formula, you can find the values of \(x\) where the equation holds true.

In optimization problems, the quadratic formula is often used to find critical points. These points identify where the function changes direction, helping determine maxima or minima.

Derivatives

Derivatives represent the rate at which a function changes. In calculus, taking a derivative is like finding the function's speed at a particular point. Mathematically, the derivative of a function \(f(x)\), denoted as \(f'(x)\) or \(\frac{df}{dx}\), measures how much \(f(x)\) changes as \(x\) changes.

In optimization, derivatives help find critical points by:

• Identifying where the slope of the function is zero.
• Determining if the critical point is a maximum or minimum.

Calculating derivatives and analyzing them is vital for solving many real-world problems like maximizing profit or minimizing cost.

Algebra

Algebra is the branch of mathematics dealing with symbols and the rules for manipulating these symbols. It allows us to express and solve equations by finding unknown values. In optimization, algebraic manipulation is key to setting up functions and equations correctly.

Key algebraic steps for solving optimization problems include:

• Expressing one variable in terms of another.
• Simplifying equations to their basic forms.
• Using algebraic identities to simplify further.

Algebra provides the foundation needed to handle complex mathematical expressions, making it essential in optimization tasks.

Functions

Functions are mathematical objects that relate inputs to outputs. Each function assigns exactly one output to each input from its domain. In optimization, we often aim to find the maximum or minimum values of a function, indicating the best outcome under given constraints.

Key properties of functions:

• Domain: All permissible inputs.
• Range: All possible outputs.
• Behavior: How the function changes as the input changes.

Understanding functions allows you to model situations mathematically and solve problems effectively, whether they involve finding the shortest path, the least cost, or the most efficient design.