Maxima and Minima

Imagine you're on a hike. As you climb, you reach the highest point of the hill—that's a maximum. When you descend into a valley, you hit the lowest point—that's a minimum. In calculus, we use these ideas to find the highest and lowest points on a graph. These points, called extrema, are where a function reaches its largest or smallest values. The highest value is maxima, and the lowest value is minima. Identifying maxima and minima helps solve real-life problems like optimizing business profits, minimizing costs, designing efficient structures, and finding the best routes. By analyzing the function's derivatives, we can pinpoint these critical points and make informed decisions based on them.

In this article, we will explore absolute maxima and minima, how to find local maxima and minima, and delve into examples and formulas used in these calculations.

1.0Maxima and Minima Definition

• Maxima are the highest points on a graph where the function reaches its peak.
• Minima are the lowest points where the function hits its lowest value.

These points can be further categorized into absolute (global) and local (relative) extrema.

2.0Absolute Maxima and Minima

• Absolute Maximum: The highest value a function achieves on its entire domain.
• Absolute Minimum: The lowest value a function reaches on its domain.

3.0Maxima and Minima of a Function

Let f be a function defined on an interval I:

1. f has a maximum value in I if there is a point c in I such that f (c) > f(x) for all x in I. Here, f(c) is the maximum value, and c is the point of maximum.
2. f has a minimum value in I if there is a point c in I such that f(c) < f(x) for all x in I. Here, f(c) is the minimum value, and c is the point of minimum.
3. f has an extreme value in I if there is a point c in I such that f(c) is either a maximum or a minimum value of f. Here, f(c) is an extreme value, and c is an extreme point.

4.0Local Maximum and Minimum 

Let f be a real-valued function and let c be an interior point in the domain of f:

1. c is a point of local maximum if there exists h > 0 such that f(c) ≥ f(x) for all x in (c – h, c + h), x ≠ c. The value f(c) is called the local maximum.
2. c is a point of local minimum if there exists h > 0 such that f (c) ≤ f (x) for all x in (c – h, c + h). The value f(c) is called the local minimum.

5.0Critical Points

A critical point of a function f(x) is a point x = c where either the derivative f′(c) = 0 or the derivative does not exist. Local Maxima or Minima are critical Point but not vice versa.

Finding Critical Points

To find the critical points of f(x):

Compute the first derivative f′(x).

Solve the equation f′(x)=0 to find the values of x.

Identify points where f′(x) does not exist, as these could also be critical points.

6.0First Derivative Test For Maxima and Minima

Let f be a function defined on an open interval I and continuous at a critical point c in I:

1. If f'(x) changes from positive to negative as x passes through c (i.e., f ′(x) > 0 just left of c and f'(x) < 0 just right of c), then c is a point of local maxima.
2. If f'(x) changes from negative to positive as x passes through c (i.e., f'(x) < 0 just left of c and f'(x) > 0 just right of c), then c is a point of local minima.
3. If f'(x) does not change sign as x passes through c, then c is neither a point of local maxima nor minima, but a point of inflection.

7.0Second Derivative Test For Maxima and Minima

Let f be a function defined on an interval I and let c ∈ I. Assume f is twice differentiable at c:

1. x = c is a point of local maxima if f ′(c) = 0 and f ″(c) < 0. In this case, f(c) is the local maximum value.
2. x = c is a point of local minima if f (c) ′ = 0 and f ″(c) > 0. In this case, f (c) is the local minimum value.
3. The test fails if f ′(c) = 0 and f ″(c) = 0. In this scenario, revert to the first derivative test to determine if c is a point of local maxima, local minima, or an inflection point.

8.0Global Extrema

Definition

Global maxima and minima are the highest and lowest points of the function on its entire domain.

Finding Global Extrema

To determine global extrema on a closed interval [a, b]:

1. Evaluate f(x) at the critical points within [a, b].
2. Evaluate f(x) at the endpoints a and b.
3. Compare these values to identify the global maximum and minimum.

9.0Real World Application

Finding the Best Price: Imagine you run a bakery. You want to find the price that will give you the maximum profit. By using calculus, you can determine the price at which your revenue minus your costs is at its highest. This involves finding the maximum point of your profit function.

10.0Solved Examples on Maxima and Minima

Example 1: Let $f(x)=x+\frac{1}{x}$; x ≠ 0. Discuss the local maximum and local minimum values of f(x).

Solution:  

Here, f'(x) = 1 – $\frac{1}{x^2}=\frac{x^2-1}{x^2}=\frac{(x-1)(x+1)}{x^2}$

Using number line rule, f(x) will have local maximum at x = –1 and local minimum at x = 1

∴ local maximum value of f(x) = –2 at x = –1

and local minimum value of f(x) = 2 at x = 1

Example 2: If f(x) = 2x³ – 3x² – 36x + 6 has local maximum and minimum at x = a and x = b respectively, then ordered pair (a, b) is -

(A) (3, –2) (B) (2, –3) (C) (–2, 3) (D) (–3, 2)

Ans. (C)

Solution: 

f(x) = 2x³ – 3x² – 36x + 6

f'(x) = 6x² – 6x – 36 & f''(x) = 12x – 6

Now f'(x) = 0 ⇒ 6(x² – x – 6) = 0

⇒ (x – 3) (x + 2) = 0 ⇒ x = –2, 3

f'' (– 2) = –30

∴ x = –2 is a point of local maximum

f''(3) = 30

∴ x = 3 is a point of local minimum

Hence, (–2, 3) is the required ordered pair. 

Example 3: Find point of local maxima and minima of f(x) = x⁵ – 5x⁴ + 5x³ – 1

Solution:  

f(x) = x⁵ – 5x⁴ + 5x³ – 1

f' (x) = 5x⁴ – 20x³ + 15x²

= 5x² (x² – 4x + 3)

= 5x² (x –1) (x – 3)

f'(x) = 0 ⇒ x = 0, 1, 3

f''(x) = 10x (2x² – 6x + 3)

But at x = 0, derivative sign is positive in its neighbourhood.

Now f''(1) < 0 ⇒ Maxima at x = 1

 f''(3) > 0 ⇒ Minima at x = 3

⇒ Neither maxima nor minima at x = 0.

Example 4: Let a cuboid having square base has area 6. Then find its maximum volume.

Solution: Total area = 2a² + 4ah = 6

$h=\frac{\left(6-2a^2\right)}{4a}\Rightarrow V=a^{2}h=\frac{a^2\left(6-2a^2\right)}{4a}\frac{dv}{da}=6-6a^2=0$

$a=1\&\frac{d^{2}v}{d{a}^2}=-ve$

Maximum V = 1

Example 5: The point of inflection for the curve $y=x^{\frac{5}{3}}$ is -

(A) (1, 1) (B) (0, 0) (C) (1, 0) (D) (0, 1)

Ans. (B)

Solution: 

Here $\frac{d^{2}y}{d{x}^2}=\frac{10}{9x^{1/3}}$

From the given points we find that (0, 0) is the point of the curve where-$\frac{d^{2}y}{d{x}^2}$ does not exist but sign of

$\frac{d^{2}y}{d{x}^2}$ changes about this point.

∴ (0, 0) is the required point.

Example 6: Find Local Maximum and Local Minimum of f(x) = x³ − 3x² + 4.

Solution: Consider f(x) = x³ − 3x² + 4.

f′(x) = 3x² − 6x.

f′(x) = 0 ⇒ 3x² – 6x =0

⇒ x = 0 or x = 2.

Compute f′′(x) = 6x − 6.

• At x=0, f′′(0) = −6 (local maximum).
• At x=2, f′′(2) = 6 (local minimum).

Example 7: Find Global Maximum and Global Minimum of g(x) = sin(x) on [0,2π].

Solution: Consider g(x) = sin(x) on [0,2π].

Find g′(x) = cos(x).

g′(x) = 0 ⇒ cos(x) = 0

we get $\mathrm{x}=\frac{\pi }{2},\frac{3\pi }{2}$

Evaluate g(x) at critical points and endpoints:

g(0) = 0

$g\left(\frac{\pi }{2}\right)=1$ (global maximum)

$g(\pi )=0g\left(\frac{3\pi }{2}\right)=-1$ (global minimum)

g(2π) = 0

11.0Practice Question Based on Maxima and Minima

1. The minimum value of the function defined by f(x) = max (x, x+1, 2–x) is

(A) 0 (B) 1/2 (C) 1 (D) 3/2

2. The point for the curve y = xe^x

(A) x = –1 is minimum (B) x = 0 is minimum

(C) x = –1 is maximum (D) x = 0 is maximum

3. For what value of x, x² ln(1/x) is maximum-

(A) e^–1/2 (B) e^1/2 (C) e (D) e^–1

4. Let f(x) = (x² – 1)ⁿ (x² + x + 1). f(x) has local extremum at x=1 if

(A) n = 2 (B) n = 3 (C) n = 4 (D) n = 6

5. If for a function f(x), f'(a) = 0, f''' (a) ≠ 0, then at x = a, f(x) is 

(A) Minimum (B) Maximum

(C) Not an extreme point (D) Extreme point