Chapter 4, Section 7: Optimization Problems

Welcome to the exciting world of optimization! In this section, we'll be exploring how to use calculus to solve real-world problems that involve finding the best possible solution – whether that's maximizing profit, minimizing cost, or finding the shortest distance. Get ready to put your derivative skills to work!

Key Concepts:

  • Objective Function: This is the function you want to maximize or minimize. It represents the quantity you're trying to optimize (e.g., profit, area, cost). Express it mathematically as $f(x)$ where x is the variable.
  • Constraint: This is an equation that limits the possible values of the variables in your objective function. It defines the relationship between the variables (e.g., the total amount of material available).
  • Critical Points: These are the points where the derivative of the objective function is either zero or undefined ($f'(x) = 0$ or $f'(x)$ does not exist). Critical points are potential locations for maximum or minimum values.
  • Endpoints: If your problem has a defined interval, be sure to check the endpoints of the interval as potential locations for maximum or minimum values.

Problem-Solving Strategy:

  1. Understand the Problem: Read the problem carefully and identify what you are trying to maximize or minimize (the objective function). Also, identify any constraints. Draw a diagram if possible.
  2. Develop a Mathematical Model: Express the objective function and the constraint as mathematical equations. For example, if you are trying to maximize the area of a rectangle with a fixed perimeter, your objective function might be $A = lw$ (where $l$ is length and $w$ is width), and your constraint might be $2l + 2w = P$ (where $P$ is the fixed perimeter).
  3. Reduce to a Single Variable: Use the constraint equation to eliminate one of the variables in the objective function. In our rectangle example, you could solve the constraint for $l$: $l = \frac{P}{2} - w$. Substitute this into the objective function: $A = (\frac{P}{2} - w)w = \frac{P}{2}w - w^2$. Now, $A$ is a function of only one variable, $w$.
  4. Find Critical Points: Find the derivative of the objective function with respect to the remaining variable. Set the derivative equal to zero and solve for the variable. Also, check where the derivative is undefined. So, in our example, $\frac{dA}{dw} = \frac{P}{2} - 2w$. Setting this to zero, we get $w = \frac{P}{4}$.
  5. Determine Maximum or Minimum: Use the first or second derivative test to determine whether each critical point corresponds to a maximum, minimum, or neither. You may also need to consider the endpoints of the interval (if any). In our example, $\frac{d^2A}{dw^2} = -2$, which is negative, so $w = \frac{P}{4}$ corresponds to a maximum.
  6. Answer the Question: Make sure you answer the question that was originally asked. For example, if the question asks for the dimensions of the rectangle, make sure you find both the length and the width. If $w = \frac{P}{4}$, then $l = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}$. So, the rectangle with maximum area for a given perimeter is a square.

Example:

Find two positive numbers whose product is 100 and whose sum is a minimum.

Let the two numbers be $x$ and $y$. We want to minimize $S = x + y$ (the sum) subject to the constraint $xy = 100$ (the product is 100). Solving the constraint for $y$, we get $y = \frac{100}{x}$. Substituting this into the sum, we get $S = x + \frac{100}{x}$.

Now, we find the derivative: $S'(x) = 1 - \frac{100}{x^2}$. Setting this equal to zero, we get $x^2 = 100$, so $x = 10$ (since $x$ must be positive). Then, $y = \frac{100}{10} = 10$. The second derivative is $S''(x) = \frac{200}{x^3}$, which is positive for $x = 10$, so this is a minimum.

Therefore, the two numbers are both 10, and their minimum sum is 20.

Remember, practice makes perfect! Work through several optimization problems to solidify your understanding. Good luck!