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Optimization

Using derivatives to find the best possible outcome.

Optimization is arguably the most useful application of differential calculus. It answers the questions we care about most in business, engineering, and physics: "What is the maximum profit?", "What is the minimum cost?", or "What is the strongest shape?"

At its core, optimization is about finding the highest point (maximum) or lowest point (minimum) on a graph.

[Image of global maximum vs local maximum graph]

1. The Logic: Slope at the Top

Imagine you are hiking up a hill. As you climb, the slope is positive. When you reach the very peak, for a split second, the ground is perfectly flat. The slope is zero. Then, as you go down, the slope becomes negative.

Calculus tells us that Critical Points (potential maximums or minimums) occur where the derivative is zero.

To optimize f(x):
1. Find f'(x).
2. Set f'(x) = 0.
3. Solve for x.

2. The Standard Procedure

Real-world optimization problems usually involve two equations:

  • Primary Equation: The thing you want to maximize or minimize (e.g., Area, Volume, Cost).
  • Secondary Equation (Constraint): The limitation you must stick to (e.g., "I only have 50 meters of fencing").

3. Classic Example: The Farmer's Fence

Problem: A farmer has 100 meters of fencing and wants to enclose a rectangular field to get the maximum possible area. What dimensions should he use?

[Image of rectangular fence diagram with dimensions x and y]

Step A: Write the Equations

We want to maximize Area (A), but we are constrained by Perimeter (P).

Primary: A = x · y
Constraint: 2x + 2y = 100

Step B: Reduce to One Variable

Use the constraint to solve for y.

2x + 2y = 100 → x + y = 50
y = 50 - x

Substitute this into the Area equation:

A(x) = x(50 - x)
A(x) = 50x - x²

Step C: Take the Derivative

Now we find the peak of this Area function by taking the derivative and setting it to zero.

A'(x) = 50 - 2x
Set A'(x) = 0
50 - 2x = 0
2x = 50
x = 25

Step D: Solve

If x = 25, plug that back into the constraint (y = 50 - x) to find y.

y = 50 - 25 = 25
Dimensions: 25m x 25m
Max Area: 625 m²

Conclusion: A square is the most efficient rectangle!

4. Conclusion

This same process is used to minimize fuel consumption in rockets, maximize volume in packaging design, and optimize financial portfolios. If you can write an equation for it, calculus can optimize it.