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Maxima and minima - Applications of Differentiation - Local and Global(Absolute) Maxima and Minima | 11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation

Chapter: 11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation

Local and Global(Absolute) Maxima and Minima

Local Maximum and local Minimum, Absolute maximum and absolute minimum

Local and Global(Absolute) Maxima and Minima

Definition 6.1

Local Maximum and local Minimum

A function f has a local maximum (or relative maximum) at c if there is an open interval (a,b) containing c such that f(c)≥f(x) for every x (a,b)

Similarly, f has a local minimum at c if there is an open interval (a,b) containing c such that f(c) ≤ f(x) for every x ! (a,b).

 

Definition 6.2

Absolute maximum and absolute minimum

A function f has an absolute maximum at c if f(c)≥f(x) for all x in domain of f. The number f(c) is called maximum value of f in the domain. Similarly f has an absolute minimum at c if f(c)≤f(x) for all x in domain of f and the number f(c) is called the minimum value of f on the domain. The maximum and minimum value of f are called extreme values of f.

NOTE

Absolute maximum and absolute minimum values of a function f on an interval (a,b) are also called the global maximum and global minimum of f in (a,b).

Criteria for local maxima and local minima

Let f be a differentiable function on an open interval (a,b) containing c and suppose that f ‘’ (c) exists.

(i) If f (c) = 0 and f ‘’ (c) > 0, then f has a local minimum at c.

(ii) If f (c) = 0 and f ‘’ (c) < 0,then f has a local maximum at c.

NOTE

In Economics, if y = f(x) represent cost function or revenue function, then the point at which dy/dx = 0, the cost or revenue is maximum or minimum.

 

Example 6.25

Find the extremum values of the function f(x)=2x3+3x2–12x.

Solution :

Given

f(x) = 2x3+3x2–12x … (1)

f (x) = 6x2+6x–12

f''(x) = 12x + 6

f (x) = 0 6x2+6x–12 =0

6(x2+x–2)= 0

6(x+2)(x–1)= 0

x = –2 ; x = 1

When

x= –2

f ‘’ (–2) = 12(–2) + 6

= –18 < 0

 f(x) attains local maximum at x = – 2 and local maximum value is obtained from (1) by substituting the value x = – 2

f(–2) = 2 (–2)3+3(–2)2–12(–2)

 = –16+12+24

 = 20.

When

x = 1

f ‘’ (1) = 12(1) + 6

 = 18.

f(x) attains local minimum at x = 1 and the local minimum value is obtained by substituting x = 1 in (1).

f(1) = 2(1) + 3(1) – 12 (1)

= –7

Extremum values are – 7 and 20.

 

Example 6.26

Find the absolute (global) maximum and absolute minimum of the function f(x)=3x5–25x3+60x+1 in the interval [–2,2]

Solution :

f(x) = 3x5-25x3+60x+1 … (1)

f’ (x) = 15x4–75x2+60

= 15(x4–5x2+4)

f (x) = 0 15(x4–5x2+4)= 0

(x2–4)(x2–1)= 0

x = ±2 (or)  x = ±1

of these four points 2, ±1 [2, 1]  and 2   [2,1]

From (1)

f(–2) =  3 ( - 2 )5 – 25(- 2 )3 + 60 ( - 2 ) + 1

= –15

When x= 1

f(1) =  3 ( 1 )5 – 25(1 )3 + 60 ( 1 ) + 1

= 39

When x = – 1

f(-1) =  3 ( -1 )5 – 25(-1 )3 + 60 ( -1 ) + 1

= –37.

Absolute maximum is 39

Absolute minimum is -37

 

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11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation : Local and Global(Absolute) Maxima and Minima | Maxima and minima - Applications of Differentiation

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11th Business Mathematics and Statistics(EMS) : Chapter 6 : Applications of Differentiation


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