Conceptual Art Essay Example | Topics and Well Written Essays - 750 words

Conceptual Art - Essay Example This essay stresses that as far as the work of Sol LeWitt is concerned, he has done a major contribution towards the growth and development of this form of artistic work. The artists will secondary to the process of conceptual art; he can not enforce his thoughts in such a way that the message of the work lost its meaning. Sol LeWitt is also of the opinion that artist can deceive the perception of the general public through conceptual art, therefore, viewers of the work should have a clear direction about how to evaluate and criticize the work of their favorite authors. Sol LeWitt also prescribed some measures by the help of which one can evaluate the work efficiently and effectively. Sol LeWitt also believed endorses the concept sharing method in order to nourish the baby concept of one artist. Since sharing among the people can also help them to produce better final art work. There is also a chance that artist might not understand his work. Therefore artist also needs guidance afte r the completion of their first draft. Conceptual art is a mechanical process, therefore, its steps should be followed properly in order to avoid any problems. As far as conceptual art is concerned one should also realize the fact that artist might be in the face of finding the truth while developing his product. Therefore it is necessary to make changes once the artist has come to know that he can improve things in a much better way. When the art is restricted to the words like sculpture and painting then the artist has a very little chance.... This essay analyzes that there is also a chance that artist might not understand his work. Therefore artist also needs guidance after the completion of their first draft. Conceptual art is a mechanical process, therefore, its steps should be followed properly in order to avoid any problems. As far as conceptual art is concerned one should also realize the fact that artist might be in the face of finding the truth while developing his product. Therefore it is necessary to make changes once the artist has come to know that he can improve things in a much better way (Peter). Sol LeWitt is also of the opinion that when the art is restricted to the words like sculpture and painting then the artist has a very little chance to produce high-quality work by using his imagination. Therefore it can be implied that artist should not be restricted by traditional boundaries since it can affect his overall productivity while developing the conceptual artwork. The work of an artist may be compared to the conductor in the sense that it may or may not reach to the audience in a way which was perceived by the artist. This paper makes a conclusion that however, this problem can be resolved if an artist tries to link his artistic work to the contemporary issues of the society. The artist should also try to follow the conventions of art so that he can develop quality work. Artists should always come up with new ideas in order to remain competitive in this industry. Sol LeWitt has produced a great work himself as well.

Amazon Case Study Example | Topics and Well Written Essays - 1000 words

Amazon - Case Study Example The operating cash flow for amazon increase to 31% to record $5.47 billion in the financial year ended December 2013 compared to $4.18 billion in the year 2012. For the net sales for the fourth quarter in year ended 2013, the company recorded a 20% increase that is equivalent to $25.59 billion compared to $21.27 billion in the financial year 2012 fourth quarter. The company in overall accrued a net sales of $74.45 billion in the financial year 2013 which was a 22% increase compared to $61.09 billion in the year 2012 ("Why Amazon Is A Lousy Business"). The company is divided into two segments that include; the North American segment and the international segment. The North American segment is involved in export sales from the www.amazon.ca and www.amazon.com, all of which are based in U.S. The international segment is involved in international websites that are involved in export sales to customers in Canada and U.S. from abroad (Amazon.com, Inc.). Jeff Bezos is an American entrepreneur and the founder of amazon.com and acts as the company’s chief executive. He was born in 1964 January 12th in Albuquerque in New Mexico. He studied in Princeton University where he was awarded a computer science and electrical engineering bachelor’s degree in 1986. He worked in several firms in Wall Street that included Bankers Trust, Fitel and lastly in D.E. Shaw investment firm where he was acknowledged as the youngest vice president in the year 1990. He later quit the job in 1994 to go to Seattle where he had seen an opportunity in internet market which by then was untapped market. Despite the successes that Bezos achieved after founding amazon.com, he was first faced with the challenge of marketing budget. He had to use word of mouth to make the business succeed and be recognized by many people. Another challenge he faced in his first years is the skepticism that people had about him. They saw him as a failure, someone who left a good job to start something

Microeconomics Research Proposal Example | Topics and Well Written Essays - 1750 words - 1

Microeconomics - Research Proposal Example Microeconomics covers a rather broad area: demand and supply, indifference curve analysis, elasticity of supply and demand, production and cost, marginal analysis, market structure, pricing, and so on. This study will attempt to deal with the area of market structure which is very interesting because of its pervasiveness in our lives: Perfect competition, monopoly, monopolistic competition, and oligopoly. An understanding of market structure is essential not only for economists but also for individual businessmen and corporations, for it will help them greatly in dealing with numerous problems encountered in the competitive arena of business. The four major market structures often discussed by economists are perfect competition, which is production by numerous firms with identical/homogeneous products as well as the presence of conditions of free entry and exit; monopoly, which is production by a single firm; monopolistic competition, which is production by many firms with somewhat different or differentiated products; and oligopoly, which is production by several firms. Perfect competition is the ideal market condition envisioned by the great economist Adam Smith, and is often studied first because it is the easiest to understand and it can serve as starting point as well as a gauge by which to measure the performance of the other market structures. The demand curve of a perfectly competitive firm is horizontal because its output is but a small part of total production and it cannot affect the price. The conditions for perfect competition are: a) a large number of small firms and customers, b) homogeneity of product, c) freedom of entry and exit, and d) perfect information about available products and their prices. Many farming and fishing industries closely approximate perfect competition. At the other end of the spectrum is the

Home depot in chile Case Study Example | Topics and Well Written Essays - 500 words

Home depot in chile - Case Study Example Home Depot adopted the international strategy, when it entered the Chilean market in 1998. As part of that strategy, it tried to implement the same practices and marketing strategies, which they carried out in its United States operations. In the case of international strategy, the firm transfers its core competencies and operational strategies to the foreign subsidiary. (Aswathappa 355). One of the key operational strategies of Home Depot in U.S. is sizably selling Do-It-Yourself (DIY) products, and they transferred the same to Chile. Instead of multidomestic strategy (which focuses on customizing the product according to the local market) and global strategy (which focuses on selling a product based on a globally formulated strategy), Home Depot adopted international strategy. As it was its first market expansion out of North America, Home Depot did not have any global strategy. Also, they avoided the multidomestic strategy and did not customize its products according to the Chilea n people. So, in line with the international strategy, Home Depot put more of DIY products on its shelves. However, this strategy of transferring the same practices to Chile did not work first because Chilean people were less willing than in the United States to repair their houses by themselves. b.

Discuss how job redesign can improve organisational performance Essay

Discuss how job redesign can improve organisational performance - Essay Example This paper sheds light on how job-redesign improves the overall organizational performance, by enlisting a myriad of advantages that are linked with job redesign. 2. Job Redesign and Organizational Performance 2.1. Employee Motivation The basic objective behind job redesign is to motivate the employee so that he is willing to perform better and produce greater results than before (Zhang & Bartol 2010). Employee motivation brings about increased worker productivity because it guarantees high quality worker job experience. Smith (1994) discusses the reason why employee motivation is at all necessary, and comes up with the answer that organizations need to implement motivational strategies through job redesign to ensure their survival in the market, because when employees are motivated, they perform better. Lindner (1998), in his research about what motivates employees, found that job redesign includes motivating factors such as interesting work, good wages, appreciation, encouragement, job security, healthy workplace environment, promotions, participation in decision-making, and sympathy shown from the supervisors on personal problems. 2.2. Employee Empowerment Job redesign empowers employees, which leads to a boost in employees’ morale, job persistence, productivity, and organizational performance; and, the absence of these factors can destroy the whole organization (Kuo et al. 2010). Gitman and McDaniel (2007:323) have called this â€Å"investing in people† which includes four trends called employee â€Å"education and training, employee ownership, work-life benefits, and nurturing knowledge workers†. Bilton (2007:71) suggests that job redesign must empower the first-line entrepreneurs instead of controlling them, and that it should give â€Å"greater autonomy and flexibility† to the employee, as there are commercial reasons for it like achievement of business goals. 2.3. Job Satisfaction Tella, Ayeni and Popoola (2007, par.15) defin e job satisfaction as a comfortable feeling that results from â€Å"employees’ perception of how well their job provides those things that are viewed as important†. The most precious asset that a company may cherish is its employees, and it needs to be seriously concerned with all issues that may dissatisfy, frustrate or depress them. Without happy employees, no strategy is going to work, and no progress will be seen in the long run. After job redesigning, employees work harder because they have enough motivation to learn and excel (Tims & Bakker 2010). They help their employers to increase productivity and achieve customer satisfaction. Tietjen and Myers (1998) state that organizational support through job redesign guarantees improved turnover behavior among employees, because they want to stick to their jobs when they find that their employers are there to value their strengths and eliminate their weaknesses through proper training. This creates a trustworthy relatio nship between employees and employers, which also results in reduced workplace conflicts and enhanced organizational performance. Today, employers are spending thousands of compensation dollars to devise and implement benefit plans,

The elderly man Essay Example for Free

The elderly man Essay My response to situation number four is not to charge the elderly man. First and foremost, a prosecutor’s duty entails determining what cases should be prosecuted. In effect, this means, acting as a â€Å"strainer. † This means that decisions are based on several factors such as â€Å"limited resources, difficulty in enforcement, and not to mention political and public pressures† In deciding against prosecution, the following factors were taken into consideration: age, public interest, and rehabilitation from imprisonment. From the nature of the case and the age of the accused, there would be little reason to imprison him. In addition, looking at the factual circumstances surrounding the commission of the crime, there is little reason to do so. The fact that he immediately turned himself in shows his understanding of right and wrong, in the same breath making a choice to commit such wrong to free his wife from her disease. Moreover, I took media attention in consideration. Being published on the front page of a newspaper indicates a high public in the case, which is the very basis for enacting laws. The article will elicit two kinds of reaction from the public. The first reaction is negative in that people would disagree to the act of ending his wife’s life, but the other reaction could be an outpouring of sympathy for an elderly man who only wanted his wife’s suffering to end. To quote the assigned text, â€Å"in some situations, prosecutors do not charge because of an outpouring of public sympathy or support for the accused, perhaps because of the type of crime or identity of the victim. † Furthermore, based on a study that looks at the prosecutor as â€Å"operating in an exchange system†, whether between the prosecutor and police officers or the courts, in which case considerations may include jail overcrowding and docket backlog, a prosecutor would think twice whether or not charges should be pressed. On the basis of the above reasons, taxpayers’ money would be better spent prosecuting individuals whose malicious or negligent actions have caused pain and suffering to the victims and their families.

Maxima And Minima Of Functions Mathematics Essay

Maxima And Minima Of Functions Mathematics Essay Maxima and Minima are important topics of maths 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 xà ¢Ã‹â€ -, if there exists some ÃŽÂ µ > 0 such that f(xà ¢Ã‹â€ -) à ¢Ã¢â‚¬ °Ã‚ ¥ f(x) when |x à ¢Ã‹â€ Ã¢â‚¬â„¢ xà ¢Ã‹â€ -| 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 à ¢Ã‹â€ Ã¢â‚¬â„¢ xà ¢Ã‹â€ -| 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 Fermats 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 (32) 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 à ¢Ã‹â€ Ã¢â‚¬â„¢ x has first derivative x2 à ¢Ã‹â€ Ã¢â‚¬â„¢ 1 and second derivative 2x. Setting the first derivative to 0 and solving for x gives stationary points at à ¢Ã‹â€ Ã¢â‚¬â„¢1 and +1. From the sign of the second derivative we can see that à ¢Ã‹â€ Ã¢â‚¬â„¢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,  ±2à Ã¢â€š ¬,  ±4à Ã¢â€š ¬, à ¢Ã¢â€š ¬Ã‚ ¦, and infinitely many global minima at  ±Ãƒ Ã¢â€š ¬,  ±3à Ã¢â€š ¬, à ¢Ã¢â€š ¬Ã‚ ¦. The function 2 cos(x) à ¢Ã‹â€ Ã¢â‚¬â„¢ x has infinitely many local maxima and minima, but no global maximum or minimum. The function cos(3à Ã¢â€š ¬x)/x with 0.1  Ãƒ ¢Ã¢â‚¬ °Ã‚ ¤Ã‚  x  Ãƒ ¢Ã¢â‚¬ °Ã‚ ¤Ã‚  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 + 32 à ¢Ã‹â€ Ã¢â‚¬â„¢ 2x + 1 defined over the closed interval (segment) [à ¢Ã‹â€ Ã¢â‚¬â„¢4,2] has two extrema: one local maximum at x = à ¢Ã‹â€ Ã¢â‚¬â„¢1à ¢Ã‹â€ Ã¢â‚¬â„¢Ãƒ ¢Ã‹â€ Ã… ¡15à ¢Ã‚ Ã¢â‚¬Å¾3, one local minimum at x = à ¢Ã‹â€ Ã¢â‚¬â„¢1+à ¢Ã‹â€ Ã… ¡15à ¢Ã‚ Ã¢â‚¬Å¾3, a global maximum at x = 2 and a global minimum at x = à ¢Ã‹â€ Ã¢â‚¬â„¢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 Rolles 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)  =  Ãƒ ¢Ã‹â€ Ã¢â‚¬â„¢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 à ¢Ã¢â‚¬ °Ã‚ ¤ 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 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 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 f (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) 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 62 + 9x + 15 Solution to Maxima and Minima Example f (x) = x3 62 + 9x + 15 f (x) = 32 -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 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) 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) (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) = 23 212 +36x 20. Find also the local maximum and local minimum values. Solution: f (x) = 62 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 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 ÃŽ D. (ii) f(x) is said to have absolute minimum value at x = a if f(a)  £ f(x) for all x ÃŽ 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 dont 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  ,  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  P(10) = 6000 2000 = 4000  ,  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  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. 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(x3-132+40x) =4(32-26x+40) The critical points occur when 32 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  ,  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  V(2) = 144  ,  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  Ã‚  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.