Maxima and Minima are important topics of math’s Calculus. It is the approach for finding maximum or minimum value of any function or any event. It is practically very helpful as it helps in solving the complex problems of science and commerce. It can be with one variable of with more than one variable. These can be done with the help of simple geometry and math functions. Finding the maxima and minima, both absolute and relative, of various functions represents an important class of problems solvable by use of differential calculus. The theory behind finding maximum and minimum values of a function is based on the fact that the derivative of a function is equal to the slope of the tangent.

Analytical definition

A real-valued function f defined on a real line is said to have a local (or relative) maximum point at the point xâˆ-, if there exists some Îµ > 0 such that f(xâˆ-) â‰¥ f(x) when |x âˆ’ xâˆ-| < Îµ. The value of the function at this point is called maximum of the function. Similarly, a function has a local minimum point at xâˆ-, if f(xâˆ-) â‰¤ f(x) when |x âˆ’ xâˆ-| < Îµ. The value of the function at this point is called minimum of the function. A function has a global (or absolute) maximum point at xâˆ- if f(xâˆ-) â‰¥ f(x) for all x. Similarly, a function has a global (or absolute) minimum point at xâˆ- if f(xâˆ-) â‰¤ f(x) for all x. The global maximum and global minimum points are also known as the arg max and arg min: the argument (input) at which the maximum (respectively, minimum) occurs.

If you need assistance with writing your essay, our professional essay writing service is here to help!

Restricted domains: There may be maxima and minima for a function whose domain does not include all real numbers. A real-valued function, whose domain is any set, can have a global maximum and minimum. There may also be local maxima and local minima points, but only at points of the domain set where the concept of neighbourhood is defined. A neighbourhood plays the role of the set of x such that |x âˆ’ xâˆ-| < Îµ.

A continuous (real-valued) function on a compact set always takes maximum and minimum values on that set. An important example is a function whose domain is a closed (and bounded) interval of real numbers (see the graph above). The neighbourhood requirement precludes a local maximum or minimum at an endpoint of an interval. However, an endpoint may still be a global maximum or minimum. Thus it is not always true, for finite domains, that a global maximum (minimum) must also be a local maximum (minimum).

## Finding Functional Maxima And Minima

Finding global maxima and minima is the goal of optimization. If a function is continuous on a closed interval, then by the extreme value theorem global maxima and minima exist. Furthermore, a global maximum (or minimum) either must be a local maximum (or minimum) in the interior of the domain, or must lie on the boundary of the domain. So a method of finding a global maximum (or minimum) is to look at all the local maxima (or minima) in the interior, and also look at the maxima (or minima) of the points on the boundary; and take the biggest (or smallest) one.

Local extrema can be found by Fermat’s theorem, which states that they must occur at critical points. One can distinguish whether a critical point is a local maximum or local minimum by using the first derivative test or second derivative test.

For any function that is defined piecewise, one finds maxima (or minima) by finding the maximum (or minimum) of each piece separately; and then seeing which one is biggest (or smallest).

## Examples

The function x2 has a unique global minimum at x = 0.

The function x3 has no global minima or maxima. Although the first derivative (3×2) is 0 at x = 0, this is an inflection point.

The function x-x has a unique global maximum over the positive real numbers at x = 1/e.

The function x3/3 âˆ’ x has first derivative x2 âˆ’ 1 and second derivative 2x. Setting the first derivative to 0 and solving for x gives stationary points at âˆ’1 and +1. From the sign of the second derivative we can see that âˆ’1 is a local maximum and +1 is a local minimum. Note that this function has no global maximum or minimum.

The function |x| has a global minimum at x = 0 that cannot be found by taking derivatives, because the derivative does not exist at x = 0.

The function cos(x) has infinitely many global maxima at 0, ±2Ï€, ±4Ï€, â€¦, and infinitely many global minima at ±Ï€, ±3Ï€, â€¦.

The function 2 cos(x) âˆ’ x has infinitely many local maxima and minima, but no global maximum or minimum.

The function cos(3Ï€x)/x with 0.1 â‰¤ x â‰¤ 1.1 has a global maximum at x = 0.1 (a boundary), a global minimum near x = 0.3, a local maximum near x = 0.6, and a local minimum near x = 1.0. (See figure at top of page.)

The function x3 + 3×2 âˆ’ 2x + 1 defined over the closed interval (segment) [âˆ’4,2] has two extrema: one local maximum at x = âˆ’1âˆ’âˆš15â„3, one local minimum at x = âˆ’1+âˆš15â„3, a global maximum at x = 2 and a global minimum at x = âˆ’4.

## Functions of more than one variable

For functions of more than one variable, similar conditions apply. For example, in the (enlargeable) figure at the right, the necessary conditions for a local maximum are similar to those of a function with only one variable. The first partial derivatives as to z (the variable to be maximized) are zero at the maximum (the glowing dot on top in the figure). The second partial derivatives are negative. These are only necessary, not sufficient, conditions for a local maximum because of the possibility of a saddle point. For use of these conditions to solve for a maximum, the function z must also be differentiable throughout. The second partial derivative test can help classify the point as a relative maximum or relative minimum.

In contrast, there are substantial differences between functions of one variable and functions of more than one variable in the identification of global extrema. For example, if a differentiable function f defined on the real line has a single critical point, which is a local minimum, then it is also a global minimum (use the intermediate value theorem and Rolle’s Theorem to prove this by reduction ad absurdum). In two and more dimensions, this argument fails, as the function

shows. Its only critical point is at (0,0), which is a local minimum with Æ’(0,0) = 0. However, it cannot be a global one, because Æ’(4,1) = âˆ’11.

The global maximum is the point at the top

## In relation to sets

Maxima and minima are more generally defined for sets. In general, if an ordered set S has a greatest element m, m is a maximal element. Furthermore, if S is a subset of an ordered set T and m is the greatest element of S with respect to order induced by T, m is a least upper bound of S in T. The similar result holds for least element, minimal element and greatest lower bound.

In the case of a general partial order, the least element (smaller than all other) should not be confused with a minimal element (nothing is smaller). Likewise, a greatest element of a partially ordered set (poset) is an upper bound of the set which is contained within the set, whereas a maximal element m of a poset A is an element of A such that if m â‰¤ b (for any b in A) then m = b. Any least element or greatest element of a poset is unique, but a poset can have several minimal or maximal elements. If a poset has more than one maximal element, then these elements will not be mutually comparable.

In a totally ordered set, or chain, all elements are mutually comparable, so such a set can have at most one minimal element and at most one maximal element. Then, due to mutual comparability, the minimal element will also be the least element and the maximal element will also be the greatest element. Thus in a totally ordered set we can simply use the terms minimum and maximum. If a chain is finite then it will always have a maximum and a minimum. If a chain is infinite then it need not have a maximum or a minimum. For example, the set of natural numbers has no maximum, though it has a minimum. If an infinite chain S is bounded, then the closure Cl(S) of the set occasionally has a minimum and a maximum, in such case they are called the greatest lower bound and the least upper bound of the set S, respectively.

The diagram below shows part of a function y = f(x).

The Point A is a local maximum and the Point B is a local minimum. At each of these points the tangent to the curve is parallel to the x-axis so the derivative of the function is zero. Both of these points are therefore stationary points of the function. The term local is used since these points are the maximum and minimum in this particular region. There may be others outside this region.

function f(x) is said to have a local maximum at x = a, if $ is a neighbourhood I of ‘a’, such that f(a) f(x) for all x I. The number f(a) is called the local maximum of f(x). The point a is called the point of maxima.

Note that when ‘a’ is the point of local maxima, f(x) is increasing for all values of x < a and f (x) is decreasing for all values of x > a in the given interval.

At x = a, the function ceases to increase.

A function f(x) is said to have a local minimum at x = a, if $ is a neighbourhood I of ‘a’, such that

f(a) f(x) for all x I

Here, f(a) is called the local minimum of f(x). The point a is called the point of minima.

Note that, when a is a point of local minimum f (x) is decreasing for all x < a and f (x) is increasing for all x > a in the given interval. At x = a, the function ceases to decrease.

If f(a) is either a maximum value or a minimum value of f in an interval I, then f is said to have an extreme value in I and the point a is called the extreme point.

## Monotonic Function maxima and minima

A function is said to be monotonic if it is either increasing or decreasing but not both in a given interval.

Consider the function

The given function is increasing function on R. Therefore it is a monotonic function in [0,1]. It has its minimum value at x = 0 which is equal to f (0) =1, has a maximum value at x = 1, which is equal to f (1) = 4.

Here we state a more general result that, ‘Every monotonic function assumes its maximum or minimum values at the end points of its domain of definition.’

Note that ‘every continuous function on a closed interval has a maximum and a minimum value.’

## Theorem on First Derivative Test

(First Derivative Test)

Let f (x) be a real valued differentiable function. Let a be a point on an interval I such that f ‘(a) = 0.

(a) a is a local maxima of the function f (x) if

i) f (a) = 0

ii) f(x) changes sign from positive to negative as x increases through a.

That is, f (x) > 0 for x < a and

f (x) < 0 for x > a

(b) a is a point of local minima of the function f (x) if

i) f (a) = 0

ii) f(x) changes sign from negative to positive as x increases through a.

That is, f (x) < 0 for x < a

f (x) > 0 for x > a

## Working Rule for Finding Extremum Values Using First Derivative Test

Let f (x) be the real valued differentiable function.

Step 1:

Find f ‘(x)

Step 2:

Solve f ‘(x) = 0 to get the critical values for f (x). Let these values be a, b, c. These are the points of maxima or minima.

Arrange these values in ascending order.

Step 3:

Check the sign of f'(x) in the immediate neighbourhood of each critical value.

Step 4:

Let us take the critical value x= a. Find the sign of f ‘(x) for values of x slightly less than a and for values slightly

greater than a.

(i) If the sign of f ‘(x) changes from positive to negative as x increases through a, then f (a) is a local maximum value.

(ii) If the sign of f ‘(x) changes from negative to positive as x increases through a, then f (a) is local minimum value.

(iii) If the sign of f (x) does not change as x increases through a, then f (a) is neither a local maximum value not a minimum value. In this case x = a is called a point of inflection.

## Maxima and Minima Example

Find the local maxima or local minima, if any, for the following function using first derivative test

f (x) = x3 – 6×2 + 9x + 15

## Solution to Maxima and Minima Example

f (x) = x3 – 6×2 + 9x + 15

f ‘ (x) = 3×2 -12x + 9

= 3(x2- 4x + 3)

= 3 (x – 1) (x – 3)

Thus x = 1 and x = 3 are the only points which could be the points of local maxima or local minima.

Let us examine for x=1

When x<1 (slightly less than 1)

f ‘(x) = 3 (x – 1) (x – 3)

= (+ ve) (- ve) (- ve)

= + ve

When x >1 (slightly greater than 1)

f ‘(x) = 3 (x -1) (x – 3)

= (+ ve) (+ ve) (- ve)

= – ve

The sign of f ‘(x) changes from +ve to -ve as x increases through 1.

x = 1 is a point of local maxima and

f (1) = 13 – 6 (1)2 + 9 (1) +15

= 1- 6 + 9 + 15 =19 is local maximum value.

Similarly, it can be examined that f ‘(x) changes its sign from negative to positive as x increases through the point x = 3.

x = 3 is a point of minima and the minimum value is

f (3) = (3)3- 6 (3)2+ 9(3) + 15

= 15

## Theorem on Second Derivative Test

Let f be a differentiable function on an interval I and let a I. Let f “(a) be continuous at a. Then

i) ‘a’ is a point of local maxima if f ‘(a) = 0 and f “(a) < 0

ii) ‘a’ is a point of local minima if f ‘(a) = 0 and f “(a) > 0

iii) The test fails if f ‘(a) = 0 and f “(a) = 0. In this case we have to go back to the first derivative test to find whether ‘a’ is a point of maxima, minima or a point of inflexion.

## Working Rule to Determine the Local Extremum Using Second Derivative Test

Step 1

For a differentiable function f (x), find f ‘(x). Equate it to zero. Solve the equation f ‘(x) = 0 to get the Critical values of f (x).

Step 2

For a particular Critical value x = a, find f “‘(a)

(i) If f ”(a) < 0 then f (x) has a local maxima at x = a and f (a) is the maximum value.

(ii) If f ”(a) > 0 then f (x) has a local minima at x = a and f (a) is the minimum value.

(iii) If f ”(a) = 0 or , the test fails and the first derivative test has to be applied to study the nature of f(a).

## Example on Local Maxima and Minima

Find the local maxima and local minima of the function f (x) = 2×3 – 21×2 +36x – 20. Find also the local maximum and local minimum values.

Solution:

f ‘(x) = 6×2 – 42x + 36

f ‘(x) = 0

x = 1 and x = 6 are the critical values

f ”(x) =12x – 42

If x =1, f ”(1) =12 – 42 = – 30 < 0

x =1 is a point of local maxima of f (x).

Maximum value = 2(1)3 – 21(1)2 + 36(1) – 20 = -3

If x = 6, f ”(6) = 72 – 42 = 30 > 0

x = 6 is a point of local minima of f (x)

Minimum value = 2(6)3 – 21 (6)2 + 36 (6)- 20

= -128

## Absolute Maximum and Absolute Minimum Value of a Function

Let f (x) be a real valued function with its domain D.

(i) f(x) is said to have absolute maximum value at x = a if f(a) ³ f(x) for all x Î D.

(ii) f(x) is said to have absolute minimum value at x = a if f(a) £ f(x) for all x Î D.

The following points are to be noted carefully with the help of the diagram.

Let y = f (x) be the function defined on (a, b) in the graph.

(i) f (x) has local maximum values at

x = a1, a3, a5, a7

(ii) f (x) has local minimum values at

x = a2, a4, a6, a8

(iii) Note that, between two local maximum values, there is a local minimum value and vice versa.

(iv) The absolute maximum value of the function is f(a7)and absolute minimum value is f(a).

(v) A local minimum value may be greater than a local maximum value.

Clearly local minimum at a6 is greater than the local maximum at a1.

## Theorem on Absolute Maximum and Minimum Value

Let f be a continuous function on an interval I = [a, b]. Then, f has the absolute maximum value and f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I.

## Theorem on Interior point in Maxima and Minima

Let f be a differentiable function on I and let x0 be any interior point of I. Then

(a) If f attains its absolute maximum value at x0, then f ‘ (x0)= 0

(b) If f attains its absolute minimum value at x0, then f ‘(x0) = 0.

In view of the above theorems, we state the following rule for finding the absolute maximum or absolute minimum values of a function in a given interval.

Step 1:

Find all the points where f ‘ takes the value zero.

Step 2:

Take the end points of the interval.

Step 3:

At all the points calculate the values of f.

Step 4:

Take the maximum and minimum values of f out of the values calculated in step 3. These will be the absolute maximum or absolute minimum values.

## Real life Problem Solving With Maxima And Minima

For a belt drive the power transmitted is a function of the speed of the belt, the law being

P(v) = Tv – av3

where T is the tension in the belt and a some constant. Find the maximum power if T = 600, a = 2 and v 12. Is the answer different if the maximum speed is 8?

Solution First find the critical points.

P = 600v – 2v3

And so

= 600 – 6v2

This is zero when v = ±10.

Commonsense tells us that v 0, and so we can forget about the critical point at -10.

So we have just the one relevant critical point to worry about, the one at x = 10. The two endpoints are v = 0 and v = 12.

We don’t hold out a lot of hope for v = 0, since this would indicate that the machine was switched off, but we calculate it anyway.

Next calculate P for each of these values and see which is the largest.

P(0) = 0 , P(10) = 6000 – 2000 = 4000 , P(12) = 7200 – 3456 = 3744

So the maximum occurs at the critical point and is 4000.

When the range is reduced so that the maximum value of v is down to 8, neither of the critical points is in range. That being the case, we just have the endpoints to worry about. The maximum this time is P(8) = 4800 – 1024 = 3776.

Our academic experts are ready and waiting to assist with any writing project you may have. From simple essay plans, through to full dissertations, you can guarantee we have a service perfectly matched to your needs.

A box of maximum volume is to be made from a sheet of card measuring 16 inches by 10. It is an open box and the method of construction is to cut a square from each corner and then fold.

Solution

Let x be the side of the square which is cut from each corner. Then AB = 16 – 2x, CD = 10 – 2x and the volume, V, is given by

V

= (16 – 2x)(10 – 2x)x

= 4x(8 – x)(5 – x)

And so

= 4(x3-13×2+40x)

=4(3×2-26x+40)

The critical points occur when

3×2 – 26x + 40 = 0

i.e. when

x =

## =

The commonsense restrictions are 5 x 0.So the only critical point in range is x = 2.

Now calculate V for the critical point and the two endpoints.

V(0) = 0 , V(2) = 144 , V(5) = 0

So the maximum value is 144, occurring when x = 2.

## Uses of Maxima and Minima in War.

Concepts of maxima and minima can be used in war to predict most probably result of any event. It can be very helpful to all soldiers as it help to save time. Maximum damage with minimum armor can be predicted via these functions. It can be helpful in preventive actions for military. It is use dto calculate ammunition numbers, food requests, fuel consumption, parts ordering, and other logical operations. It is also helpful in finding daily expenditure on war.