Maxima and Minima

Maxima and Minima: Finding the Peaks and Valleys

Hey there, JEE aspirants! Welcome to the world of Maxima and Minima – a crucial concept in Application of Derivatives. In JEE Main, you'll often encounter problems that require you to find the maximum or minimum value of a function, be it to optimize a design or determine the range of a function. Mastering this topic gives you a powerful tool to tackle such problems efficiently. Let's dive in!

Understanding Local (Relative) Maxima and Minima

Imagine a roller coaster track. The peaks are the points of local maxima, and the valleys are the points of local minima. Local maxima is the highest value of the function within a small interval around a point, but not necessarily the highest value of the entire function. Similarly, a local minima is the lowest value of the function within a small interval, but not necessarily the lowest value of the entire function.

Think of a function $f(x)$. If there's an interval around a point $c$ such that $f(c) \geq f(x)$ for all $x$ in that interval, then $f(c)$ is a local maximum. If $f(c) \leq f(x)$ for all $x$ in that interval, then $f(c)$ is a local minimum.

Global (Absolute) Maxima and Minima

Now, let's talk about the highest and lowest points on the entire roller coaster track. These are the global maxima and global minima. The global maximum is the highest value the function attains over its entire domain, while the global minimum is the lowest value over the entire domain.

Formally, for a function $f(x)$, if $f(c) \geq f(x)$ for all $x$ in the domain of $f$, then $f(c)$ is the global maximum. If $f(c) \leq f(x)$ for all $x$ in the domain of $f$, then $f(c)$ is the global minimum.

First Derivative Test for Extrema

The first derivative test helps us find the local maxima and minima. A critical point $c$ of a function $f(x)$ is a point where $f'(c) = 0$ or $f'(c)$ is undefined. These points are potential locations for local extrema.

How does it work?

• If $f'(x)$ changes from positive to negative at $x = c$, then $f(c)$ is a local maximum.
• If $f'(x)$ changes from negative to positive at $x = c$, then $f(c)$ is a local minimum.
• If $f'(x)$ does not change sign at $x = c$, then $f(c)$ is neither a local maximum nor a local minimum (it's a point of inflection).

For example, consider the function $f(x) = x^3$. Its derivative is $f'(x) = 3x^2$. Here, $f'(0) = 0$, so $x = 0$ is a critical point. However, $f'(x)$ is always non-negative, so $x = 0$ is a point of inflection, not an extremum.

Second Derivative Test

The second derivative test provides a more direct way to determine if a critical point is a local maximum or minimum.

Intuition:

• If $f''(c) > 0$, the function is concave up at $x = c$, resembling a valley (local minimum).
• If $f''(c) < 0$, the function is concave down at $x = c$, resembling a peak (local maximum).

Example: Let $f(x) = x^2$. Then $f'(x) = 2x$ and $f''(x) = 2$. The critical point is $x = 0$. Since $f''(0) = 2 > 0$, $f(0) = 0$ is a local minimum.

What if $f''(c) = 0$? The test is inconclusive. You'll need to use the first derivative test or higher-order derivative tests.

Optimization Problems

Optimization problems involve finding the best possible solution (maximum or minimum) given certain constraints. These problems often appear in JEE Main.

Steps to solve:

1. Identify the objective function: This is the function you want to maximize or minimize.
2. Identify the constraints: These are the conditions that must be satisfied.
3. Express the objective function in terms of one variable: Use the constraints to eliminate other variables.
4. Find the critical points: Differentiate the objective function and set it to zero.
5. Apply the first or second derivative test: Determine if the critical points are maxima or minima.
6. Check endpoints (if applicable): For a closed interval, check the function's values at the endpoints.

Example: Find two numbers whose sum is 10 and whose product is maximum. Let the numbers be $x$ and $y$. We want to maximize $P = xy$, subject to $x + y = 10$. Then $y = 10 - x$, so $P = x(10 - x) = 10x - x^2$. $P'(x) = 10 - 2x$. Setting $P'(x) = 0$, we get $x = 5$. $P''(x) = -2 < 0$, so $x = 5$ gives a maximum. Then $y = 10 - 5 = 5$. The numbers are 5 and 5.

JEE-Specific Tricks

Symmetry: Exploit symmetry in functions to simplify optimization problems. If the function is symmetric about a line, the extremum often lies on that line.

Standard Inequalities: Remember AM-GM inequality and Cauchy-Schwarz inequality. They can be extremely useful in solving optimization problems quickly. For example, if $a, b > 0$, then $\frac{a+b}{2} \geq \sqrt{ab}$ with equality when $a=b$. In the previous example, we could have directly used AM-GM: $\frac{x+y}{2} \geq \sqrt{xy}$, so $\frac{10}{2} \geq \sqrt{xy}$, giving $25 \geq xy$. Thus, the maximum product is 25 when $x=y=5$.

That's it for Maxima and Minima! Practice these concepts with various problems, and you'll be well-prepared to ace those JEE questions. All the best!