1. KAY174 MATHEMATICS II Prof. Dr. Doğan Nadi Leblebici

2. APPLICATIONS OF EQUATIONS AND INEQUALITIES PURPOSE: TO LEARN HOW TO TRANSLATE RELATIONSHIPS STATED IN THE PROBLEMS INTO MATHEMATICAL SYMBOLS WHICH IS CALLED “MODELING”.

3. SOME BUSINESS TERMS FIXED COST (OVERHEAD) is the sum of all costs that are independent of the level of production, such as rent, insurance, etc. VARIABLE COST is the sum of all costs that are dependent on the level of output, such as labor and material. TOTAL COST is the sum of variable cost and fixed cost. Total Cost = Variable Cost + Fixed Cost TOTAL REVENUE is the price per unit of output times the number of units sold. Total Revenue = (Price per unit)(Number of units sold) PROFIT is total revenue minus total cost. Profit = Total Revenue – Total Cost

4. EXAMPLE Problem: The Kurek Plastic Inc. prouces Product G for which the variable cost per unit is TL 6 and fix cost is TL 80.000. Each unit has a selling price of TL 10. Determine the number of units that must be sold for the company to earn a profit of TL 60.000. Solution: Profit = Total Revenue – Total Cost; n=units sold Total Revenue = (Price per unit)(Number of units sold) Total Cost = Variable Cost + Fixed Cost 60.000=10n-(6n+80.000) → 60.000=10n-6n-80.000 4n=140.000 n=35.000

5. EXAMPLE Problem: A company manufactures women sportswear and is planning to sell its new model of sportswear set to retail outlets. The cost to the retailer will be TL 33 per set. As convenience to the retailer, the manufacturer will attach a price tag to each set. What amount should be marked on the price tag so that the retailer may reduce this price by 20 percent during a sale and still make a profit of 15 percent on the cost? Solution: Selling Price = Cost Per Set + Profit Per Set; p=the tag price Selling Price = p -.2p P -.2p = 33 + (.15).(33).8p = 37.95 p = 47.4375 → 47.44

6. EXAMPLE Problem: A total of TL 10.000 was invested in two business ventures, A and B. At the end of the first year, A and B yielded returns of 6 percent and percent, respectively, on the original investments. How was the original amount allocated if the total amount earned was TL 588.75? Solution: Amount invested in A = x and amount invested in B = 10.000 – x (.06)x + (.0575)(10.000 – x) = 588.75.06x + 575 –.0575x = 588.75.0025x = 13.75 x = 5.500 Amount invested in A = 5.500 and amount invested in B = 4.500

7. LINEAR INEQUALITIES Suppose a and b are two points on the real number line. Then either a and b coincide, or a lies to the left of b, or a lies to the right of b. Another inequality symbol “≤” is read as “less than or equal” and thus symbol “≥” is read as “greater than or equal”. An inequality is a statement that one number is less than another number. b abaab

8. LINEAR INEQUALITIES RULES If a<b, then a(+/-)c<b(+/-)c If a 0, then a/c<b/c If a 0, then a(-c)>b(-c) and If a<b and a=c, then c<b If a>b>0 and n>0, then and

9. LINEAR INEQUALITIES A linear inequality in the variable x is an inequality which can be written in the form ax + b < 0 or ax + b ≤ 0 Examples: 2(x – 3)<4 → 2x – 6 <4 → 2x<10 → x<5 3 – 2x≤6 → -2x≤3 → x≥-3/2 2(x – 4) – 3 > 2x – 1 → 2x – 8 – 3>2x – 1 → -11>-1 → No solution Φ 5 -3/2

10. LINEAR INEQUALITIES If a≤x≤b, it is called a closed interval. For closed interval we use [-------] If a<x<b, it is called a open interval. For open interval we use (--------) (a,b] means a<x≤b [a,b) means a≤x<b [a,∞) means x≥a (a,∞) means x>a (-∞,a] means x≤a (-∞,a) means x<a (-∞,∞) means -∞<∞

11. APPLICATIONS OF INEQUALITIES Problem: For a manufacturer of thermostats, the combined cost for labor and material is TL 4 per thermostat. Fixed cost are TL 60.000. If the selling price of a thermostat is TL 7, how many must be sold for the company to earn profit? Solution: Number of thermostat that must be sold = n. Total cost of thermostat that must be sold = 4n + 60.000 Total revenue of thermostat that must be sold = 7n Profit = Total Revenue - Total Cost → 7n – (4n + 60.000)>0 3n>60.000 n>20.000 Number of thermostat that must be sold is 20.001

12. APPLICATIONS OF INEQUALITIES Problem: A publishing company finds that the cost of publishing each copy of a certain magazine is TL 0.38. The revenue from dealers is TL 0,35 per copy. The advertising revenue is 10 percent of the revenue received from dealers for all copies sold beyond 10.000. What is the least number of copies which must be sold so as to have a profit for the company? Solution: Number of copies that must be sold = x. The revenue from dealers =.35x + (.10)[(.35)(x-10.000)] Profit = Total Revenue - Total Cost →.35x + (.10)[(.35)(x-10.000)] -.38x>0.35x +.035x-350 -.38x>0.005x – 350>0 x>70.000 Number of copies that must be sold is 70.001

13. ABSOLUTE VALUE Sometimes it is useful to consider, on the real number line, the distance between a number x and 0. We call this distance the absolute value of x and denote it by IxI. For example, I5I=5 and I- 5I=5 because both 5 and -5 are five units from 0. Similarly I0I=0. Example Ix-3I=2 x-3=2 or x-3=-2 → x=5 or x=1 I7-3xI=5 7-3x=5 or 7-3x=-5 → x=4 or x=2/3

14. FUNCTIONS AND GRAPHS PURPOSE: TO UNDERSTAND RELATIONSHIP BETWEEN DEPENDENT AND INDEPENDENT VARIABLES IN A FORMULA.

15. FUNCTIONS In 1964, Leibniz, one of the developers of calculus, introduced the word function into the mathematical vocabulary. A function is a rule that assigns to each input number exactly one output number. The set of all input numbers to which the rule applies is called the domain of the function. The set of all output numbers is called the range. y = x + 2 For each value of x, there is a value for y. x is independent variable and y is dependent variable.

16. FUNCTIONS Examples y 2 =x Let’s say x=9 → y 2 =9 → y=±3 Thus y has two values, -3 and 3. Hence y is not a function of x. x=y 2 Let’s say y=3 → x=9 → Hence x is a function of y. Suppose that the equation p=100/q describes the relationship between the price per unit, p, of a certain product and the number of units, q, of that product that consumers will buy (that is, demand) per week. This equation is called demand equation for the product. If q is an input number, then to each value of q there is assigned exactly one output number p.

17. FUNCTIONS Examples q→100/q=p 20→100/20=5 Thus price is a function of quantity demanded. Since q can not be 0 or negative, domain is all values of q such that q>0. f(q)=100/q → f(q)=p f(5)=100/5 → f(5)=20 → p=20

18. FUNCTIONS f(x)=2 is a constant function. y=f(x)=-3x 2 + x -5 is a quadratic function. y=f(x)=-c n x n + c n-1 x n-1 + c 1 x + c0 is a polynomial function. c n is the leading coefficient. n is called the degree of function. f(x)=IxI is an absolute value function.

19. COMBINATIONS OF FUNCTIONS If f and g are functions, we can combine them to create new functions. For example, f(x)=x 2 and g(x)=x + 1 Adding f(x) and g(x) in the obvious way gives f(x) + g(x) = x 2 + (x + 1) This sum defines a new function – Let’s call it H H(x)=f(x) + g(x) = x 2 + (x + 1)

20. GRAPHS IN RECTANGULAR COORDINATES A rectangular (or Cartesian) coordinate system allows us to specify and locate points in a plane. It also provides a geometric way to represent equations in two variables as well as functions.

21. GRAPHS IN RECTANGULAR COORDINATES Using a rectangular coordinate system, we can geometrically represent equations in two variables. For example; y = x 2 + 2x -3

22. GRAPHS IN RECTANGULAR COORDINATES Example y = 2x + 3

23. GRAPHS IN RECTANGULAR COORDINATES Example s = 100/t

24. GRAPHS IN RECTANGULAR COORDINATES Example x = 2y 2

25. LINES, PARABOLAS, AND SYSTEMS PURPOSE: TO LEARN THE STEEPNESS TO INTERPRET RELATIONSHIP IN ECONOMICS.

26. LINES To measure the steepness of a line we introduce the notion of slope (m). Example Suppose that the line below shows the relationship between the price p of a widget (in TL) and the quantity q of widgets (in thousands) that consumers will buy at that price. Find and interpret the slope.

27. LINES Example We can use slope formula: The slope is negative, -3/4. This means that for each increase in quantity of 1 (thousand widgets), there corresponds a decrease in price of ¾ (TL per widget). Due to this decrease the line falls from left to right.

28. LINES y – y 1 = m(x – x 1 ) is the point-slope form of an equation of line through (x 1,y 1 ) and having slope m. Y = mx – b is the slope-intercept form of an equation of line with slope m and y-intercept (0,b). A function f is a linear function if and only if f(x) can be written in the form f(x) = ax + b, where a and b are constants and a ≠ 0.

29. PARABOLAS The graph of the quadratic function y = f(x) = ax2 + bx + c is called a parabola and has a shape such as the curves below. Vertex occurs at (-b/2a, f(-b/2a)).

30. PARABOLAS Example Suppose the demand function for a manufacturer’s product is p=f(q)=1000-2q, where p is the price (in TL) per unit when q units are demanded (per week) by consumers. Find the level of production that will maximize the manufacturer’s total revenue, and determine this revenue. Solution Total Revenue=(Price)(Quantity) r=(1000-2q)q → r=1000q -2q 2 → r is a quadratic formula a=-2, b=1000, c=0 → q=-b/2a maximum (vertex) q=-1000/2(-2)=250 → r=125.000

31. PARABOLAS Solution Total Revenue=(Price)(Quantity) r=(1000-2q)q → r=1000q -2q 2 → r is a quadratic formula a=-2, b=1000, c=0 → q=-b/2a maximum (vertex) q=-1000/2(-2)=250 → r=125.000

32. SYSTEMS OF LINEAR EQUATIONS We sometimes have to solve two equations that all equations in the set are satisfied simultaneously. For example; suppose that a company pays its salespeople on the basis of a certain percentage of the first TL 100.000 in sales, plus a certain percentage of any sales beyond TL 100.000. If a salesperson earned TL 17.000 on sales of TL 300.000 and another earned TL 12.500 on sales of TL 225.000, what are the two rates? x = rate for first 100.000 and y = rate for sales beyond 100.000 100.000x + 200.000y = 17.000 100.000x + 125.000y = 12.500

33. SYSTEMS OF LINEAR/NONLINEAR EQUATIONS The set of linear equations İs called a system of two linear equations in the variables (or unknowns) x and y. Its solution consists of values of x and y which satisfy both equations simultaneously.

34. SYSTEMS OF LINEAR/NON LINEAR EQUATIONS The left and right sides of equations can be substracted from the corresponding side of equations. 100.000x + 200.000y – (100.000x + 125.000y) = 17.000 – 12.500 75.000y = 4500 y =.06 and x =.05 Formulas in system can be nonlinear also. Solution of such systems should also satisfy both equations simultaneously.

35. SYSTEMS OF LINEAR/NON LINEAR EQUATIONS Example

36. SYSTEMS OF LINEAR/NON LINEAR EQUATIONS Example