Some Application of Calculus

1446 WordsApr 16, 20116 Pages
Calculus: Calculus (Latin, calculus, a small stone used for counting) is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. This subject constitutes a major part of modern mathematics education. It has two major branches, Differential Calculus and Integral Calculus, which are related by the fundamental theorem of calculus. Calculus is the study of change, in the same way that geometry is the study of shape and algebra is the study of operations and their application to solving equations. A course in calculus is a gateway to other, more advanced courses in mathematics devoted to the study of functions and limits, broadly called mathematical analysis. Calculus has widespread…show more content…
Although when we are dealing with a curve it is a different story. Calculus allows us to find out how steeply a curve will tilt at any given time. This can be very useful in any area of study. 6) Calculating the Area of Any Shape: Although we do have standard methods to calculate the area of some shapes, calculus allows us to do much more. Trying to find the area on a shape like this would be very difficult if it wasn’t for calculus. Some Practical Examples: Example1: A manufacturer sells 500 units per week at 31 dollars per unit. If the price is reduced by one dollar, 20 more units will be sold. To maximize the revenue, find: (a) The selling price Let x be the number of one dollar reduction Price in dollars per unit: 31 − x (b) The number of units sold Number of units sold: 500 + 20x (c) The maximum revenue Revenue = (price per unit). (Number of units) R(x) = (31 − x). (500 + 20x) = −20x2 + 120x + 15500 Once the equation of the revenue function is found, use the first derivative to find C.N., Test the C.N. using the first or second derivative and then answer the following questions: (1) Find the selling price to maximize the revenue; (2) find the number of units sold to maximize the revenue; (3) Find the maximum revenue. R′(x) = −40x + 120 =) R′(x) = 0 =) −40x + 120 = 0 =) x = 3 Test the critical number x = 3 with the second derivative: R′′(x) = −40 < 0, relative maximum (1) The
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