Welcome back to Professor Baker's Math Class! In this session, we dive into one of the most practical and powerful applications of derivatives: Optimization Problems (Chapter 4.7). Whether you are a farmer trying to maximize field area, an engineer minimizing material costs, or a commuter trying to minimize travel time, the calculus concepts remain the same.

The 6-Step Strategy for Success

As outlined in the class notes, solving these problems requires a systematic approach. Don't just jump into the algebra! Follow this roadmap:

  1. Understand the Problem: Read carefully. Identify the unknowns, the given quantities, and the conditions.
  2. Draw a Diagram: Visualizing the problem is crucial. Label the fixed quantities and the variable quantities.
  3. Introduce Notation: Assign symbols. Let $Q$ be the quantity you want to maximize or minimize (like Area, Cost, or Time).
  4. Express $Q$: Write an equation for your objective quantity in terms of the other symbols.
  5. Reduce to One Variable: This is the most critical step. Use the given constraints to eliminate variables until $Q$ is expressed as a function of a single variable (e.g., $Q = f(x)$).
  6. Optimize: Use the methods from Sections 4.1 and 4.3 (finding critical numbers where $f'(x) = 0$) to find the absolute maximum or minimum.

Classic Examples from Class

1. The Fencing Problem (Geometry)

We looked at a classic scenario where a farmer has $2400$ ft of fencing to enclose a rectangular field bordering a river (where no fence is needed).

  • Objective: Maximize Area ($A = L \cdot W$).
  • Constraint: The total fencing is fixed. $2400 = L + 2W$.
  • Solution Path: By solving the constraint for $L$ ($L = 2400 - 2W$), we can substitute it into the area equation to get $A(W) = (2400 - 2W)W$. From there, we find the derivative $A'(W)$ to locate the maximum dimensions.

2. Minimizing Distance

Another interesting application is finding the point on a curve (like the parabola $y^2 = 2x$) that is closest to a specific coordinate, such as $(1, 4)$.

Pro Tip: When minimizing distance $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, it is often easier to minimize the square of the distance, $D = d^2$. The minimum occurs at the same $x$-value, but the derivative is much cleaner to calculate!

3. The "Row and Run" Problem (Motion)

We also tackled a complex problem involving different speeds. A woman needs to cross a river and reach a point downstream. She can row at 6 km/h and run at 8 km/h.

Here, the objective is to minimize Time. Since $Time = \frac{Distance}{Speed}$, our total time function becomes:

$$T(x) = \frac{\text{Distance}_{row}}{6} + \frac{\text{Distance}_{run}}{8}$$

This problem perfectly illustrates why drawing a diagram (Step 2) is essential to set up the correct distances using the Pythagorean theorem.

Final Thoughts

Optimization allows us to find the "best" solution among all possible solutions. Remember, the calculus is often the easy part—the challenge lies in setting up the equation correctly. Keep practicing those constraints!

Check the attached PDF for the full step-by-step solutions to the rebate (revenue) problem and the cylinder surface area problem.