Finding Absolute Max and Min Using EVT: Worked Problems

📚 Understanding the Extreme Value Theorem (EVT)

The Extreme Value Theorem (EVT) is a cornerstone of calculus that guarantees the existence of absolute maximum and minimum values for a continuous function on a closed interval. In simpler terms, if you have a nice, unbroken curve (continuous function) on a specific segment of the x-axis (closed interval), there’s definitely a highest and lowest point on that curve within that segment.

📜 History and Background

The EVT, though not explicitly stated in its modern form until later, has its roots in the work of mathematicians like Bernard Bolzano and Karl Weierstrass in the 19th century. Weierstrass formalized the theorem, solidifying its place in real analysis. The theorem relies on the completeness of real numbers and is crucial for optimization problems.

📌 Key Principles of EVT

• 🔗 Continuity: The function $f(x)$ must be continuous on the closed interval $[a, b]$. This means there are no breaks, jumps, or vertical asymptotes within the interval.
• 🔒 Closed Interval: The interval $[a, b]$ must be closed, meaning it includes both endpoints $a$ and $b$. This is crucial because the absolute max or min could occur at the endpoints.
• 🔑 Finding Critical Points: To find the absolute max and min, you need to identify the critical points of $f(x)$ within the interval $(a, b)$. Critical points occur where the derivative $f'(x)$ is either equal to zero or undefined.
• 📊 Evaluation: Evaluate the function $f(x)$ at all critical points and at the endpoints $a$ and $b$.
• 🏆 Comparison: Compare the values obtained in the previous step. The largest value is the absolute maximum, and the smallest value is the absolute minimum of $f(x)$ on the interval $[a, b]$.

💡 Real-World Examples

Example 1: Maximizing Profit

A company's profit $P(x)$ (in thousands of dollars) as a function of the number of units sold $x$ is given by $P(x) = -0.1x^2 + 2x + 5$ on the interval $[0, 15]$. Find the maximum profit.

1. 🔍 Find the derivative: $P'(x) = -0.2x + 2$.
2. 📍 Find critical points: Set $P'(x) = 0$: $-0.2x + 2 = 0$, so $x = 10$.
3. 📈 Evaluate at critical points and endpoints:
   • $P(0) = -0.1(0)^2 + 2(0) + 5 = 5$
   • $P(10) = -0.1(10)^2 + 2(10) + 5 = 15$
   • $P(15) = -0.1(15)^2 + 2(15) + 5 = 12.5$
4. 🏆 Conclusion: The maximum profit is $15,000, which occurs when 10 units are sold.

Example 2: Minimizing Cost

The cost $C(x)$ (in dollars) of producing $x$ items is given by $C(x) = x^3 - 6x^2 + 5$ on the interval $[-1, 4]$. Find the minimum cost.

1. 🔍 Find the derivative: $C'(x) = 3x^2 - 12x$.
2. 📍 Find critical points: Set $C'(x) = 0$: $3x^2 - 12x = 0$, so $3x(x - 4) = 0$, giving $x = 0$ and $x = 4$.
3. 📈 Evaluate at critical points and endpoints:
   • $C(-1) = (-1)^3 - 6(-1)^2 + 5 = -2$
   • $C(0) = (0)^3 - 6(0)^2 + 5 = 5$
   • $C(4) = (4)^3 - 6(4)^2 + 5 = -27$
4. 🏆 Conclusion: The minimum cost is $-27, which occurs when $x = 4$.

Example 3: Temperature Variation

The temperature $T(t)$ in a room over time $t$ (in hours) is modeled by $T(t) = -t^3 + 8t$ on the interval $[0, 3]$. Find the maximum and minimum temperatures.

1. 🔍 Find the derivative: $T'(t) = -3t^2 + 8$.
2. 📍 Find critical points: Set $T'(t) = 0$: $-3t^2 + 8 = 0$, so $t^2 = \frac{8}{3}$, giving $t = \pm \sqrt{\frac{8}{3}} \approx \pm 1.63$. Since we are on the interval $[0, 3]$, we consider only $t = \sqrt{\frac{8}{3}}$.
3. 📈 Evaluate at critical points and endpoints:
   • $T(0) = -(0)^3 + 8(0) = 0$
   • $T(\sqrt{\frac{8}{3}}) = -(\sqrt{\frac{8}{3}})^3 + 8(\sqrt{\frac{8}{3}}) \approx 8.71$
   • $T(3) = -(3)^3 + 8(3) = -3$
4. 🏆 Conclusion: The maximum temperature is approximately 8.71, which occurs at $t = \sqrt{\frac{8}{3}}$, and the minimum temperature is -3, which occurs at $t = 3$.

📝 Conclusion

The Extreme Value Theorem is a powerful tool for finding absolute maximum and minimum values of continuous functions on closed intervals. By understanding its principles and applying it systematically, you can solve a wide range of optimization problems in various fields.