Optimization

Whether you’re minimizing cost, maximizing efficiency, or locating peaks in data—optimization tells us where a function reaches its highest or lowest values. 🎯

Optimization Steps

Optimization involves finding the maximum or minimum value of a function over a given domain.

There are two types of extrema:

• Global (absolute) extrema: the highest or lowest value over the entire domain.
• Local (relative) extrema: the highest or lowest value in a small neighborhood.

Steps:

1. Take the derivative:
   f'(x)

2. Find critical points:
   Solve f'(x) = 0 and check where f'(x) is undefined.

3. Classify using a test:

   • First Derivative Test (sign change in f'(x))
   • Second Derivative Test:
     • If f''(x) > 0, local minimum
     • If f''(x) < 0, local maximum

4. Evaluate endpoints (if on a closed interval) to find global extrema.

First Derivative Test

Check the sign of f'(x) around critical points:

Behavior of f'(x)	Type of Point
Changes + \to -	Local maximum
Changes - \to +	Local minimum
No sign change	Not an extremum

Second Derivative Test

Use the second derivative at a critical point:

f''(x)

f''(x) Value	Conclusion
> 0	Local minimum
< 0	Local maximum
= 0	Inconclusive

NoteExercise

Consider the function on the interval [0,12]:

f(x)=−2x^3+27x^2−84x+120

TipSolution

1. Take the derivative:

f'(x)=−6x^2+54x−84

2. Find critical points:

Set f'(x) = 0

\begin{aligned} −6x^2+54x−84 &= 0 \\ x^2-9x+14 &=0 \ \text{(divide by -6)} \\ (x-2)(x-7) &=0 \end{aligned}

Critical points are x=2 and x=7.

3. Classify using a test:

f''(x)=−12x+54

At x = 2:f''(2)=−12(2)+54=30>0 Local minimum at x = 2

At x = 7:f''(7)=−12(7)+54=−30<0 Local maximum at x = 7

4. Evaluate endpoints

f(0) = 120 f(2) = -2(8) + 27(4) - 84(2) + 120 = -16 + 108 - 168 + 120 = 44 f(7) = -2(343) + 27(49) - 84(7) + 120 = -686 + 1323 - 588 + 120 = 169 f(12) = -2(1728) + 27(144) - 84(12) + 120 = -3456 + 3888 - 1008 + 120 = -456