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Mathematical Applications

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 Sub Topics Applied mathematics is the branch of mathematics which deals with applications of mathematics to the real world problems. It is a journey into the mathematical ideas and computational methods for solving deterministic and stochastic optimization problems. We can see applied mathematics applications in fourier transform, laplace transform, linear algebra, vector calculus, probability, statistics, ordinary differential equations, partial differential equations, matrices and determinants, geometry etc.,

Mathematics Application And Concepts

Given below are some of the important concepts used in mathematical applications.
Calculus
Calculus provides tools to solve many problems and is derived from latin word which means counting. In modern mathematics calculus plays a very important role. Change (increase or decrease) in quantity such as motion. There are different applications of derivatives used in different situations where calculus plays an significant role in finding the derivative.

Statistics

Statistics is a mathematical science pertaining to the collection, analysis, interpretation or explanation, and presentation of data. The word statistics can either be singular or plural. In plural form it refers to a quantity. Statistical method help us to summarize a collection of data and is a method of analyzing statistical data. Probability theory contains more of mathematics. However, there are topics in statistics which are independent of probability.

Probability
Probability of an outcome in a sample space is that chance or relative frequency for the outcome to occur. A probability model lists the possible outcomes of the experiment and the associated probability of each outcome is displayed in a table by a graph.
A probability model should satisfy the following:
1) If E is an event then the probability that E has occurrred is the sum of the probabilities of the simple sample points that make the event.

2) Probability of an impossible event is zero and if an event is a certainity then its probability should be 1.

3) Any probabilities assigned must be a non negative number.

4) Impossible event has a probability of 0.

5) For any event E, $0 \leq$ P(E) \leq 1$. 6) Probability of sample space should be equal to 1. Math Application Problems Back to Top Example 1: The cost function of a firm is given by$c(x)$= 300$x$- 10$x$$^{2} + \frac{1}{3} x$$^{3}$where$C$is the total cost for$x$units.Calculate the output at which the average cost is minimum. Solution: Total cost = 300$x$- 10$x$$^{2} +\frac{1}{3} x$$^{3}$Average cost (AC) =$\frac{Total\ cost}{x}$= 300 - 10$x$+$\frac{x^{2}}{3}\frac{d}{dx}(AC)$= -10 +$\frac{2x}{3}$Put$\frac{d}{dx}(AC)$= 0 to find the average cost. =>$\frac{2x}{3}$= 10$x$= 15 Now,$\frac{d^{2}}{dx^2}(AC)$=$\frac{2}{3}\geq$0 Therefore, average cost is minimum at$x$= 15. Example 2:The demand function of a company is given by$P(q)$= 800 - 0.4$q$and the total cost function is given by$c(q)$= 80$q$+ 8000, where$q$is the level of output and P is the price per unit. Find the level of output and the price charged which maximises the profit. Solution: Demand function$P$= 800 - 0.4$q$Total Cost$C$= 80$q$+ 8000 Marginal Cost$MC$= 80 Demand function = 800 - 0.4$q$Total revenue = Qty$P$=$q$(800 - 0.4$q$) = 800q - 0.4$q^{2}$Marginal revenue = 800 - 0.8$q$Profit maximum$MR$=$MC$800 - 0.8$q$= 80$q$=$\frac{720}{0.8}q$= 900 units Total profit = Total Revenue - Total Cost = 800$q$- 0.4$q^{2}$- 80$q$- 8000 = 720q - 0.4$q^{2}$- 8000 Plug in$q$= 900 = 720 (900) - 0.4 (900)$^{2}$- 8000 = 3,16,000 Example 3: A spherical snowball is forming so that its volume is increasing at the rate of 8 cc/sec. Find the rate at which the radius of the snowball is increasing when the ball is 6 cm in diameter. Solution:$\frac{dv}{dt}$= 8$r$= 3$V$=$\frac{4}{3}$$\pir^{3} \frac{dv}{dt} = \frac{4}{3}$$\pi\times$3$\times$r$^{2}\frac{dr}{dt}$8 = 4$\pi\times$9$\times\frac{dr}{dt}\frac{dr}{dt}$=$\frac{2}{9\pi}\$ cm/sec, is the rate at which the radius of the snowball is increasing when the ball is 6 cm in diameter.

Example 4: How many 4 digit numbers can be formed by using the digits 2, 3, 7, 8, 9 which are divisible by 3?

Solution : If a number is divisible by 3, then the sum of all digits should be divisible by 3.

Sum of 2, 3, 7, 8, 9 is 29.

29 - 9 = 20 which is not divisible by 3.

29 - 3 = 26 which is not divisible by 3.

Therefore, we have to exclude the digits 3 and 9.

If we exclude 3, the remaining four digits can be performed in 4! ways = 24.

If we exclude 9, the remaining four digits can be performed in 4! ways = 24.

Therefore, the number of 4 digit numbers that can be performed are 24 + 24 = 48.