Maxima and Minima | Engineering Mathematics - Civil Engineering (CE) PDF Download

Definition of Maxima and Minima

1. The function f is said to have a maximum value in I, if there exists a point c in such that f(c) > f(x), for all x I. The value f(c) is called the maximum value of f(x) in I and the point c is called a point of maximum value of f(x) in I.
2. The function f is said to have a minimum value in I, if there exists a point c in I such that f(c) < f(x), for all x I. The value f(c), in this case, is called the minimum value of f(x) in I and the point c, in this case is called a point of minimum value of
   f(x) in I.
3. The function f is said to have an extreme value in I if there exists a point c in I such that f(c) is either maximum value or a minimum value of f(x) in I.
4. The number f(c) in this case, is called an extreme value of f(x) in I and the point c is called an extreme point.Here Point A, C are Local Minima and B, D are Local Maxima.

Notes 

1. Concave Downwards indicates Maxima of the function i.e.
2. Concave Upwards indicates Minima of the function i.e.
3. If f ‘(x) does not change sign as x increases through c, then c is neither a point of Local maxima nor a point of local minima. In fact, such a point is called point of inflection. So the condition for point of inflection is f ‘’(x) = 0 
4. Similarly the necessary condition for the existing of either Maxima or Minima is f ‘(x) = 0. 

Example 1: Find all the points of local maxima and local minima of the function f given by f(x) = 2x³ - 6x² + 6x + 5
Solution: Given that f(x) = 2x³ - 6x² + 6x + 5
f ‘(x) = 6x² - 12x + 6 = 6(x - 1)²
f ‘(x) = 0 at x = 1
Observe that f ‘(x) > 0 for all x R and in particular f ‘(x) > 0, for values close to 1 and to the left and right of 1.
Hence x = 1 is a point of inflection.

Definition of Local Maxima and Minima 

Let f be a function defined on an interval I and c I. Let f be twice differentiable at c.
Then

1. x = c is a point of local maxima if f ‘(c) = 0 and f ‘'(c) < 0. Then the value f(c) is local maximum value of f(x).
2. x = c is a point of local minima if f ‘(c) = 0 and f ‘'(c) > 0. In this case, f(c) is local minimum value of f(x).

Example 2: Find local maximum and local minimum values of the function f given by
f(x) = 3x⁴ + 4x³ - 12x² + 12
Solution: f(x) = 3x⁴ + 4x³ - 12x² + 12  
f ‘(x) = 12x³ + 12x² – 24x = 12x(x – 1)(x + 2)  
f ‘(x) = 0 at
x = 0, x = 1 and x = –2
f ‘'(x) = 36x² + 24x – 24
f ‘'(0) = –24 < 0
f ‘'(1) = 36 > 0
f ‘'(–2) = 144 – 48 – 24 = 72 > 0
x = 0 is a point of local maxima and local maximum value of f(x) at x = 0 is f(0) = 12 while x = 1 and x = –2 are local minimum points.
Local minimum values are f(1) = 7 and f(–2) = 20

Absolute Maxima and Absolute Minima

1. The graph gives a continuous function defined on a closed interval [a, d]. Observe that the function f has a local minima at x = b and local minimum vales is f(b), the function also has a local maxima at x = c and local maximum values is f(c).  
2. Also form the graph, it is evident that f has absolute maximum value f(a) and absolute minimum value f(d). Further note that absolute maximum (minimum) value of f(x) is different from local maximum (minimum) value of f(x). 
3. This absolute maximum value is nothing but global maxima and absolute minimum value is nothing but global minima.

Method to Find Global Maxima and Minima

Step 2: Take the end points of the interval.  

Step 3: At all these points (listed in step 1 and step 2) calculate the values of f(x).  

Step 4: Identity the maximum and minimum values of f(x) out of the values will be the absolute maximum (greatest) value of f(x) and the minimum (least) value of f(x).

Example 3: Find the absolute maximum and minimum values of a function f(x) given by   f(x) = 2x³ – 15x² + 36x + 1  in the interval [1, 5]

Solution: f '(x) = 6x² – 30x + 36 = 6(x – 3)(x – 2)
f '(x) = 0 gives x = 2, x = 3
We shall evaluate the f(x) at x = 2 and x = 3 and at the end points.
f (1) = 24
f(2) = 29
f(3) = 28
f(5) = 56
So absolute maximum value is 56 at x = 5
Absolute minimum value is 24 at x = 1

PYQs: Competitive Exams

Q. Find the maximum value of the following function:

$f\left(x\right)=-3x2+6x+5f(x)\; =\; -3x^2\; +\; 6x\; +\; 5$f(x)=−3x2+6x+5

Solution:

We need to find the maximum value of the quadratic function $f\left(x\right)=-3x2+6x+5f(x)\; =\; -3x^2\; +\; 6x\; +\; 5$f(x)=−3x2+6x+5.

Step 1: Differentiate the function to find the critical points.

The first derivative of $f\left(x\right)f(x)$f(x) is:

$f\prime \left(x\right)=-6x+6f\text{'}(x)\; =\; -6x\; +\; 6$f′(x)=−6x+6

Step 2: Set the derivative equal to zero to find the critical points.

$-6x+6=0-6x\; +\; 6\; =\; 0$−6x+6=0

Solving for $xx$x, we get:

$x=1x\; =\; 1$x=1

Step 3: Check if this critical point is a maximum or minimum by using the second derivative.

The second derivative of $f\left(x\right)f(x)$f(x) is:

$f\prime \prime \left(x\right)=-6f\text{'}\text{'}(x)\; =\; -6$f′′(x)=−6

Since $f\prime \prime \left(x\right)=-6f\text{'}\text{'}(x)\; =\; -6$f′′(x)=−6, which is negative, this indicates that the function has a maximum at $x=1x\; =\; 1$x=1.

Step 4: Calculate the maximum value by substituting $x=1x\; =\; 1$x=1 into the original function.

$f\left(1\right)=-3\left(1\right)2+6\left(1\right)+5=-3+6+5=8f(1)\; =\; -3(1)^2\; +\; 6(1)\; +\; 5\; =\; -3\; +\; 6\; +\; 5\; =\; 8$f(1)=−3(1)2+6(1)+5=−3+6+5=8

Final Answer:

The maximum value of the function is $8\backslash mathbf\{8\}$8.