Optimization
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:
Take the derivative:
\[ f'(x) \]Find critical points:
Solve \(f'(x) = 0\) and check where \(f'(x)\) is undefined.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
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 |
Exercise
Consider the function on the interval \([0,12]\):
\[ f(x)=−2x^3+27x^2−84x+120 \]
Solution
- Take the derivative:
\[ f'(x)=−6x^2+54x−84 \]
- 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 \(x=7\).
- 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 \(x = 7\)
- Evaluate endpoints
\[ f(0) = 120 \] \[ f(2) = -2(8) + 27(4) - 84(2) + 120 = -16 + 108 - 168 + 120 = 44 \] \[ f(7) = -2(343) + 27(49) - 84(7) + 120 = -686 + 1323 - 588 + 120 = 169 \] \[ f(12) = -2(1728) + 27(144) - 84(12) + 120 = -3456 + 3888 - 1008 + 120 = -456 \]