This term paper presents concise explanations of the subject's general principles and uses worked examples freely to expand the ideas about solving the problems by suitable methods. Each example shows the method of obtaining the solution and includes additional explanatory techniques. For some topics, where it would have been difficult to understand a solution given on a single problem, the solution has been drawn in step-by-step form. All the figures used have been taken from Google Book search.
The term paper covers the necessary definitions on MAXIMA AND MINIMA OF THE FUNCTIONS and some of its important applications. It covers the topic such as types of other method for solving the big problem in a shortcut method known . The aspects of how to develop some of the most commonly seen problems is also covered in this term paper. The motive of this term paper is make the reader familiar with the concepts of application of maxima and minima of the function and
where this is used. Focus has been more on taking the simpler problem so
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that the concept could be made clearer even to the beginners to engineering mathematics.
MAXIMA AND MINIMA
DEFINITION
In mathematics, a point x*is a local maximumof a function fif there exists some ε >0such that f(x*) ≥ f(x)for all xwith |x-x*| <ε. Stated less formally, a local maximum is a point where the function takes on its largest value among all points in the immediate vicinity. On a graph of a function, its local maxima will look like the tops of hills.
A local minimumis a point x*for which f(x*) ≤ f(x)for all xwith |x-x*| <ε. On a graph of a function, its local minima will look like the bottoms of valleys.
A global maximumis a point x*for which f(x*) ≥ f(x)for all x. Similarly, a global minimumis a point x*for which f(x*) ≤ f(x)for all x. Any global maximum (minimum) is also a local maximum (minimum); however, a local maximum or minimum need not also be a global maximum or minimum.
The concepts of maxima and minima are not restricted to functions whose domain is the real numbers. One can talk about global maxima and global minima for real-valued functions whose domain is any set. In order to be able to define local maxima and local minima, the function needs to take real values, and the concept of neighborhood must be defined on the domain of the function. A neighborhood then plays the role of the set of x such that |x - x*| <ε.
One refers to a local maximum/minimum as to a local extremum(or local optimum), and to a global maximum/minimum as to a global extremum(or global optimum).
LOCAL MAXIMA AND MINIMA
Functions can have "hills and valleys": places where they reach a minimum or maximum value.
It may not be the minimum or maximum for the whole function, but locally it is.
You can see where they are,
but how do we define them?
Local Maximum
First we need to choose an interval:
Then we can say that a local maximum is the point where:
The height of the function at "a" is greater than (or equal to) the height anywhere else in that interval.
Or, more briefly:
f(a) ≥ f(x) for all x in the interval
In other words, there is no height greater than f(a).
Note: f(a) should be inside the interval, not at one end or the other.
Local Minimum
Likewise, a local minimum is:
f(a) ≤ f(x) for all x in the interval
The plural of Maximum is Maxima
The plural of Minimum is Minima
Maxima and Minima are collectively called Extrema
Global (or Absolute) Maximum and Minimum
The maximum or minimum over the entire function is called an "Absolute" or "Global" maximum or minimum.
There is only one global maximum (and one global minimum) but there can be more than one local maximum or minimum.
Assumingthis function continues downwards to left and right:
The Global Maximum is about 3.7
The Global Minimum is -Infinity
Maxima and Minima of Functions of Two Variables
Locate relative maxima, minima and saddle points of functions of two variables. Several examples with detailed solutions are presented. 3-Dimensional graphs of functions are shown to confirm the existence of these points. More on Optimization Problems with Functions of Two Variables in this web site.
Theorem
Let f be a function with two variables with continuous second order partial derivativesfxx, fyyand fxyat a critical point (a,b). Let
D = fxx(a,b) fyy(a,b) - fxy2(a,b)
If D >0 and fxx(a,b) >0, then f has a relative minimum at (a,b).
If D >0 and fxx(a,b) <0, then f has a relative maximum at (a,b).
If D <0, then f has a saddle point at (a,b).
If D = 0, then no conclusion can be drawn.
We now present several examples with detailed solutions on how to locate relative minima, maxima and saddle points of functions of two variables. When too many critical points are found, the use of a table is very convenient.
Example 1:Determine the critical points and locate any relative minima, maxima and saddle points of function f defined by
f(x , y) = 2x2+ 2xy + 2y2- 6x
.
Solution to Example 1:
Find the first partial derivatives fxand fy.
fx(x,y) = 4x + 2y - 6
fy(x,y) = 2x + 4y
The critical points satisfy the equations fx(x,y) = 0 and fy(x,y) = 0 simultaneously. Hence.
4x + 2y - 6 = 0
2x + 4y = 0
The above system of equations has one solution at the point (2,-1).
We now need to find the second order partial derivatives fxx(x,y), fyy(x,y) and fxy(x,y).
fxx(x,y) = 4
fxx(x,y) = 4
fxy(x,y) = 2
We now need to find D defined above.
D = fxx(2,-1) fyy(2,-1) - fxy2(2,-1) = ( 4 )( 4 ) - 22= 12
Since D is positive and fxx(2,-1) is also positive, according to the above theorem function f has a local minimum at (2,-1).
The 3-Dimensional graph of function f given above shows that f has a local minimum at the point (2,-1,f(2,-1)) = (2,-1,-6).
Example 2:Determine the critical points and locate any relative minima, maxima and saddle points of function f defined by
f(x , y) = 2x2- 4xy + y4+ 2
.
Solution to Example 2:
Find the first partial derivatives fxand fy.
fx(x,y) = 4x - 4y
fy(x,y) = - 4x + 4y3
Determine the critical points by solving the equations fx(x,y) = 0 and fy(x,y) = 0 simultaneously. Hence.
4x - 4y = 0
- 4x + 4y3= 0
The first equation gives x = y. Substitute x by y in the equation - 4x + 4y3= 0 to obtain.
- 4y + 4y3= 0
Factor and solve for y.
4y(-1 + y2) = 0
y = 0 , y = 1 and y = -1
We now use the equation x = y to find the critical points.
(0 , 0) , (1 , 1) and (-1 , -1)
We now determine the second order partial derivatives.
fxx(x,y) = 4
fyy(x,y) = 12y2
fxy(x,y) = -4
We now use a table to study the signs of D and fxx(a,b) and use the above theorem to decide on whether a given critical point is a saddle point, relative maximum or minimum.
critical point (a,b)
(0,0)
(1,1)
(-1,1)
fxx(a,b)
4
4
4
fyy(a,b)
0
12
12
fxy(a,b)
-4
-4
-4
D
-16
32
32
saddle point
relative minimum
relative minimum
A 3-Dimensional graph of function f shows that f has two local minima at (-1,-1,1) and (1,1,1) and one saddle point at (0,0,2).
Example 3:Determine the critical points and locate any relative minima, maxima and saddle points of function f defined by
f(x , y) = - x4- y4+ 4xy
.
Solution to Example 3:
First partial derivatives fxand fyare given by.
fx(x,y) = - 4x3+ 4y
fy(x,y) = - 4y3+ 4x
We now solve the equations fy(x,y) = 0 and fx(x,y) = 0 to find the critical points..
- 4x3+ 4y = 0
- 4y3+ 4x = 0
The first equation gives y = x3. Combined with the second equation, we obtain.
- 4(x3)3+ 4x = 0
Which may be written as .
x(x4- 1)(x4+ 1) = 0
Which has the solutions.
x = 0 , -1 and 1.
We now use the equation y = x3to find the critical points.
(0 , 0) , (1 , 1) and (-1 , -1)
We now determine the second order partial derivatives.
fxx(x,y) = -12x2
The First Derivative: Maxima and Minima
Consider the function f(x)=3x4−4x3−12x2+3 on the interval [−23]. We cannot find regions of which f is increasing or decreasing, relative maxima or minima, or the absolute maximum or minimum value of f on [−23] by inspection. Graphing by hand is tedious and imprecise. Even the use of a graphing program will only give us an approximation for the locations and values of maxima and minima. We can use the first derivative of f, however, to find all these things quickly and easily.
Increasing or Decreasing?
Let f be continuous on an interval I and differentiable on the interior of I.
If f(x)0 for all xI, then f is increasing on I.
If f(x)0 for all xI, then f is decreasing on I.
Example
The function f(x)=3x4−4x3−12x2+3 has first derivative f(x) = = = 12x3−12x2−24x 12x(x2−x−2) 12x(x+1)(x−2) Thus, f(x) is increasing on (−10)(2) and decreasing on (−−1)(02).
Relative Maxima and Minima
Relative extrema of f occur at critical points of f, values x0 for which either f(x0)=0 or f(x0) is undefined.
First Derivative Test
Suppose f is continuous at a critical point x0.
If f(x)0 on an open interval extending left from x0 and f(x)0 on an open interval extending right from x0, then f has a relative maximum at x0.
If f(x)0 on an open interval extending left from x0 and f(x)0 on an open interval extending right from x0, then f has a relative minimum at x0.
If f(x) has the same sign on both an open interval extending left from x0 and an open interval extending right from x0, then f does not have a relative extremum at x0.
In summary, relative extrema occur where f(x) changes sign.
Example
Our function f(x)=3x4−4x3−12x2+3 is differentiable everywhere on [−23], with f(x)=0 for x=−102. These are the three critical points of f on [−23]. By the First Derivative Test, f has a relative maximum at x=0 and relative minima at x=−1 and x=2.
Absolute Maxima and Minima
If f has an extreme value on an open interval, then the extreme value occurs at a critical point of f.
If f has an extreme value on a closed interval, then the extreme value occurs either at a critical point or at an endpoint.
According to the Extreme Value Theorem, if a function is continuous on a closed interval, then it achieves both an absolute maximum and an absolute minimum on the interval.
Example
Since f(x)=3x4−4x3−12x2+3 is continuous on [−23], f must have an absolute maximum and an absolute minimum on [−23]. We simply need to check the value of f at the critical points x=−102 and at the endpoints x=−2 and x=3: f(−2) f(−1) f(0) f(2) f(3) = = = = = 35 −2 3 −29 30 Thus, on [−23], f(x) achieves a maximum value of 35 at x=−2 and a minimum value of -29 at x=2.
We have discovered a lot about the shape of f(x)=3x4−4x3−12x2+3 without ever graphing it! Now take a look at the graph and verify each of our conclusions.
APPLICATION
The terms maxima and minima refer to extreme values of a function, that is, the maximum and minimum values that the function attains. Maximum means upper bound or largest possible quantity. The absolute maximum of a function is the largest number contained in the range of the function. That is, if f(a) is greater than or equal to f(x), for all x in the domain of the function, then f(a) is the absolute maximum. For example, the function f(x) = -16x2 + 32x + 6 has a maximum value of 22 occurring at x = 1. Every value of x produces a value of the function that is less than or equal to 22, hence, 22 is an absolute maximum. In terms of its graph, the absolute maximum of a function is the value of the function that corresponds to the highest point on the graph. Conversely, minimum means lower bound or least possible quantity. The absolute minimum of a function is the smallest number in its range and corresponds to the value of the function at the lowest point of its graph. If f(a) is less than or equal to f(x), for all x in the domain of the function, then f(a) is an absolute minimum. As an example, f(x) = 32x2 - 32x - 6 has an absolute minimum of -22, because every value of x produces a value greater than or equal to -22.
In some cases, a function will have no absolute maximum or minimum. For instance the function f(x) = 1/x has no absolute maximum value, nor does f(x) = -1/x have an absolute minimum. In still other cases, functions may have relative (or local) maxima and minima. Relative means relative to local or nearby values of the function. The terms relative maxima and relative minima refer to the largest, or least, value that a function takes on over some small portion or interval of its domain. Thus, if f(b) is greater than or equal to f(b ± h) for small values of h, then f(b) is a local maximum; if f(b) is less than or equal to f(b ± h), then f(b) is a relative minimum. For example, the function f(x) = x4 -12x3 - 58x2 + 180x + 225 has two relative minima (points A and C), one of which is also the absolute minimum (point C) of the function. It also has a relative maximum (point B), but no absolute maximum.
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. When the values of a function increase as the value of the independent variable increases, the lines that are tangent to the graph of the function have positive slope, and the function is said to be increasing. Conversely, when the values of the function decrease with increasing values of the independent variable, the tangent lines have negative slope, and the function is said to be decreasing. Precisely at the point where the function changes from increasing to decreasing or from decreasing to increasing, the tangent line is horizontal (has slope 0), and the derivative is zero. (With reference to figure 1, the function is decreasing to the left of point A, as well as between points B and C, and increasing between points A and B and to the right of point C). In order to find maximum and minimum points, first find the values of the independent variable for which the derivative of the function is zero, then substitute them in the original function to obtain the corresponding maximum or minimum values of the function. Second, inspect the behavior of the derivative to the left and right of each point. If the derivative Figure 1. Illustration by Hans & Cassidy. Courtesy of Gale Group. is negative on the left and positive on the right, the point is a minimum. If the derivative is positive on the left and negative on the right, the point is a maximum. Equivalently, find the second derivative at each value of the independent variable that corresponds to a maximum or minimum; if the second derivative is positive, the point is a minimum, if the second derivative is negative the point is a maximum.
A wide variety of problems can be solved by finding maximum or minimum values of functions. For example, suppose it is desired to maximize the area of a rectangle inscribed in a semicircle. The area of the rectangle is given by A = 2xy. The semicircle is given by x2 + y2 = r2, for y ≥ 0, where r is the radius. To simplify the mathematics, note that A and A2 are both maximum for the same values of x and y, which occurs when the corner of the rectangle intersects the semicircle, that is, when y2 = r2 - x2. Thus, we must find a maximum value of the function A2 = 4x2(r2 -x2) = 4r2x2 - 4x4. The required condition is that the derivative be equal to zero, that is, d(A2)/dx = 8r2x - 16x3 = 0. This occurs when x = 0 or when x = 1â„2(r √ +2 ). Clearly the area is a maximum when x = 1â„2(r √ +2 ). Substitution of this value into the equation of the semicircle gives y = 1â„2(r √ +2 ), that is, y = x. Thus, the maximum area of a rectangle inscribed in a semicircle is A = 2xy = r2.
There are numerous practical applications in which it is desired to find the maximum or minimum value of a particular quantity. Such applications exist in economics, business, and engineering. Many can be solved using the methods of differential calculus described above. For example, in any manufacturing business it is usually possible to express profit as a function of the number of units sold. Finding a maximum for this function represents a straightforward way of maximizing profits. In other cases, the shape of a container may be determined by minimizing the amount of material required to manufacture it. The design of piping systems is often based on minimizing pressure drop which in turn minimizes required pump sizes and reduces cost. The shapes of steel beams are based on maximizing strength.
Finding maxima or minima also has important applications in linear algebra and game theory. For example, linear programming consists of maximizing (or minimizing) a particular quantity while requiring that certain constraints be imposed on other quantities. The quantity to be maximized (or minimized), as well as each of the constraints, is represented by an equation or inequality. The resulting system of equations or inequalities, usually linear, often contains hundreds or thousands of variables. The idea is to find the maximum value of a particular variable that represents a solution to the whole system. A practical example might be minimizing the cost of producing an automobile given certain known constraints on the cost of each part, and the time spent by each laborer, all of which may be interdependent. Regardless of the application, though, the key step in any maxima or minima problem is expressing the problem in mathematical terms.
FINDING THE MAXIMA AND MINIMA OF THE FUNCTION WITH CONSTRAINED CONDITIOIN
Lagrange's Method of Multipiers. Let F(x, y, z) and Φ(x, y, z) be functions defined over some region R of space. Find the points at which the function F(x, y, z) has maximums and minimums subject to the side condition Φ(x, y, z) = 0. Lagrange's method for solving this problem consists of forming a third function G(x, y, z) given by
17) G(x, y, z) = F(x, y, z) + λΦ(x, y, z) ,
where λ is a constant (i.e. a parameter) to which we will later assign a value, and then finding the maxima and minima of the function G(x, y, z). A reader might quickly ask, "Of what interest are the maxima and minima of the function G(x, y, z)? How does this help us solve the problem of finding the maxima and minima of F(x, y, z)?" The answer is that examination of 17) shows that for those points corresponding to the solution set of Φ(x, y, z) = 0 the function G(x, y, z) is equal to the function F(x, y, z) since at those points equation 17) becomes
G(x, y, z) = F(x, y, z) + λ·0 .
Thus, for the points on the surface Φ(x, y, z) = 0, functions F and G are equal so the maxima and minima of G are also the maxima and minima of F. The procedure for finding the maxima and minima of G(x, y, z) is as follows: We regard G(x, y, z) as a function of three independent variables and write down the necessary conditions for a stationary point using 1) above:
18) F1 + λΦ1 = 0 F2 + λΦ2 = 0 F3 + λΦ3 = 0
We then solve these three equations along with the equation of constraint Φ(x, y, z) = 0 to find the values of the four quantities x, y, z, λ. More than one point can be found in this way and this will give us the locations of the stationary points. The maxima and minima will be among the stationary points thus found.
Let us now observe something. If equations 18) are to hold simultaneously, then it follows from the third of them that λ must have the value
If we substitute this value of λ into the first two equations of 18) we obtain
F1Φ3 - F3Φ1 = 0 F2Φ3 - F3Φ2 = 0
or
We note that the two equations of 19) are identically the same conditions as 8) above for the previous method. Thus using equations 19) along with the equation of constraint Φ(x, y, z) = 0 is exactly the same procedure as the previous method in which we used equations 8) and the same constraint.
One of the great advantages of Lagrange's method over the method of implicit functions or the method of direct elimination is that it enables us to avoid making a choice of independent variables. This is sometimes very important; it permits the retention of symmetry in a problem where the variables enter symmetrically at the outset.
Lagrange's method can be used with functions of any number of variables and any number of constraints (smaller than the number of variables). In general, given a function F(x1, x2, ... , xn) of n variables and h side conditions Φ1 = 0, Φ2 = 0, .... , Φh = 0, for which this function may have a maximum or minimum, equate to zero the partial derivatives of the auxiliary function F + λ1Φ1 + λ2Φ2 + ...... + λhΦh with respect to x1, x2, ... , xn , regarding λ1, λ2, ..... ,λh as constants, and solve these n equations simultaneously with the given h side conditions, treating the λ's as unknowns to be eliminated.
The parameter λ in Lagrange's method is called Lagrange's multiplier.
