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MATH103 Mathematics for Business and Economics - I Chapter 1 Applications of Equation and inequality Procedure for Solving Word Problems 1. 2. 3. 4. 5. Read the problem carefully and introduce a variable to represent an unknown quantity in the problem. Identify other quantities in the problem (known or unknown) and express unknown quantities in terms of the variable you introduced in the first step. Write a verbal statement using the conditions stated in the problem and then write an equivalent mathematical statement (equation or inequality.) Solve the equation or inequality and answer the questions posed in the problem. Check that the solution solves the original problem. Section Some Business Terms: EXAMPLE 1 PROFIT Anderson Company produces a product for which the variable cost per unit is $6 and fixed cost is $80,000. Each unit has a selling price of $10. How many units must be sold for the company to earn a profit of $60,000? EXAMPLE 2 Analyzing Investments EXAMPLE 3 Analyzing Investments Tyrick invests $15,000, some in stocks and the rest in bonds. If he invests twice as much in stocks as he does in bonds, how much does he invest in each? Solution Step 2 Let x = the amount invested in stocks. The rest of the $15,000 investment ($15,000 – x) is invested in bonds. We have one more important piece of information to use: Twice the amount Amount invested = invested in bonds, in stocks, x 15,000 – x EXAMPLE 3 Step 3 Analyzing Investments x 2 15, 000 x Replace the verbal description with algebraic expressions. Step 4 x 30, 000 2x 3x 30, 000 x 10, 000 Distributive property Add 2x to both sides. Divide both sides by 3. Step 5 Tyrick invests $10,000 in stocks and $15,000 – $10,000 = $5000 in bonds. Step 6 Tyrick’s total investment is $10,000 + $5000 = $15,000, and $10,000 (stocks) is twice $5,000 (bonds). EXAMPLE 4 Solving Problems Involving Simple Interest Ms. Sharpy invests a total of $10,000 in blue-chip and technology stocks. At the end of a year, the blue-chips returned 12% and the technology stocks returned 8% on the original investments. How much was invested in each type of stock if the total interest earned was $1060? Solution Step 1 Step 2 We are asked to find two amounts: that invested in blue-chip stocks and that invested in technology stocks. If we know how much was invested in blue-chip stocks, then we know that the rest of the $10,000 was invested in technology stocks. Let x = amount invested in blue-chip stocks. Then 10,000 – x = amount invested in technology stocks. Invest P r t I = Prt Blue x 0.12 1 0.12x Tech 10,000 – x 0.08 1 0.08(10,000 – x) Interest from Blue-chip Step 3 Step 4 Interest from + technology = Total Interest 0.12x 0.08 10, 000 x 1060 12x 8 10, 000 x 106, 000 12x 80, 000 8x 106, 000 4x 26, 000 $ in blue-chip stocks x 6500 10, 000 x 10, 000 6500 $ in technology stocks 3500 Step 5 Ms. Sharpy invests $3500 in technology stocks and $6500 in blue- chip stocks. Step 6 $3500 $6500 $10,000 12% of $6500 $780 8% of $3500 $280 Total interest earned = $1060. EXAMPLES] A fence is to enclose a rectangular area of 800 square feet. The length of the plot is twice the width. How much fencing must be used? Solution: Let w = width of the plot then 2w = length of the plot So the width of the plot is 20 ft and the length of the plot is 40ft. So 20+20+40+40 = 120 ft of fencing must be used. EXAMPLES] A builder makes concrete by mixing 1 part clay, 3 parts sand and 5 parts crushed stone. If 765 cubic feet of concrete are needed, how much of each ingredient is needed? Solution: Let x = amount of clay, 3x = amount of sand and 5x = amount of crushed stone. So 85 cubic feet of clay, 255 cubic feet of sand and 425 cubic feet of stone. EXAMPLES] A rectangular plot 4 meters by 8 meters is to be used for a garden. The owner wants a border so that 12 square meters of the plot is left for flowers. How wide should the border be? Solution: Let X = the width of the border then 4-2x = the width of the plot and 8-2x = the length of the plot So the border should be 1m wide EXAMPLES] A company wants to know the total sales units required to earn a profit of $100,000. The unit selling price is $20, the variable cost per unit is $15, and the total fixed costs are $600,000. Find the sales units required. Solution:Let q = the number of units then the cost = 15q+600000 and revenue = 20q So 140,000 units would need to be sold. EXAMPLES] Some workers went on strike for 46 days. Before this they earned $7.50 per hour and worked 260 eight-hour days a year. What percent increase is needed in yearly income to make up for the lost time in 1 year? Solution: Let x=the percent increase then x/100 is the decimal form of the percentage So about 17.7% increase will be needed. EXAMPLES]$20000 is invested in two enterprises so that the total income per year will be $1440.One enterprise pays 6% annually the other 7%. How much was invested into each enterprise? Solution: Let x = amount invested in 6% enterprise and 20000-x = amount in 7% enterprise So $4000 was invested at 6% and $16000 was invested at 7% EXAMPLES]The price of a product is p dollars each. Suppose a manufacturer will supply 3p2-4p units of product to the market and consumers will demand 24-p2 units. Find the value of p for which the supply will equal the demand. Solution: So the price is $3 per product unit. EXAMPLES] A circular ventilation duct with diameter 140 mm is attached to a square duct system. To ensure smooth air flow the areas of the circle and square sections must be equal. Find the length of the square side of the section to the nearest millimeter. Solution: Let x = the length of the side of the square then As= x2 and As=Ac So the length of the side of the square section is 124mm. Section 1.2 An inequality is a statement that one algebraic expression is less than, or is less than or equal to, another algebraic expression. If the equality symbol = in a linear equation is replaced by an inequality symbol (<, >, ≤, or ≥), the resulting expression is called a first-degree, or linear, inequality. For example x 5 1 3x 2 2 is a linear inequality. A linear inequality in one variable is an inequality that is equivalent to one of the forms ax + b < 0 or ax + b ≤ 0, where a and b represent real numbers and a ≠ 0. Linear Inequalities in One Variable Interval Notation is used to write solution sets of inequalities. Note: A parenthesis is used to indicate an endpoint in not included. A square bracket indicates the endpoint is included. Interval Notation Type of Interval Set Graph Interval Notation {x| a < x } (a, ) ( a {x | a < x < b} (a, b) ( a {x | x < b} (-, b) {x | x is a real number} (-, ) Open Interval ) b ) b Interval Notation Type of Interval Set Interval Notation Graph {x| a x} [a, ) [ a {x | a < x b} (a, b] ( a ] b {x | a x b} [a, b) [ a ) b {x | x b} (-, b] Half-open Interval ] b Interval Notation Type of Interval Closed Interval Set {x| a x b} Interval Notation [a, b] Graph [ a ] b • Example 1: Solve k – 5 > 1 k – 5 +5 > 1 + 5 k>6 Solution set: (6,) • Example 2: Solve 5x + 3 4x – 1 and graph the solution set. 5x – 4x -1 –3 x -4 [ Solution set: [-4,) -4 • Example 3: Solve -2x < 10 x > -5 Solution set: (-5,) • Example 4: Solve 2x < -10 x < -5 Solution set: (-,-5) • Example 5: Solve –9m < -81 and graph the solution set m>9 ( Solution set: (9, ) 9 Solving and Graphing Linear Inequalities EXAMPLE 6 Solve the inequality 3(x-1) < 5(x + 2) - 5 Solution: 3(x-1) < 5(x + 2) - 5 3x - 3 < 5x + 10 - 5 Distribute the 3 and the 5 3x - 3 < 5x + 5 Combine like terms. -2x < 8 Subtract 5x from both sides, and add 3 to both sides x > -4 Notice that the sense of the inequality reverses when we divide both sides by -2. x > -4 is equivalent to (-4, ∞) ( -4 x EXAMPLE 7 Solving and Graphing Linear Inequalities Solve the inequality 2(x-3) < 4 Line Notation Solution Set = { x: x < 5 } Set Notation (-∞, 5) Interval Notation EXAMPLE 7 Solving and Graphing Linear Inequalities Solve the inequality 3-2x 6 Set Notation Solution Set = { x: x >= 5 } Line Notation EXAMPLE 8 Solving and Graphing Linear Inequalities Solve the inequality 2(x-4)-3 > 2x-1 Example 9: Solve 6(x-1) + 3x -x –3(x + 2) and graph the solution set Step 1: 6x-6 + 3x -x –3x – 6 9x - 6 –4x – 6 Step 2: 13x 0 [ Step 3: x 0 Solution set: [0, ) 0 Example 10: Solve 1 3 (m 3) 2 ( m 8) 4 4 1 3 3 m 2 m6 4 4 4 1 11 3 24 m m 4 4 4 4 1 3 24 11 m m 4 4 4 4 2 13 m 4 4 13 m 2 Solution set: [ 13 , ) 2 [ -13/2 and graph the solution set EXAMPLE 8 Solving and Graphing Linear Inequalities Solve the inequality Line Notation 3 s 2 1 2 s 4 2 Section 1.3 Applications of Inequalities EXAMPLES] A company manufactures a product that has a unit selling price of $20 and a unit cost of $15. If fixed costs are $600,000, determine the least number of units that must be sold for the company to have a profit. Solution: Let the x be the number of units then 20x = total revenue and 15x+600000 = total cost So we need at least 120001 units to make a profit EXAMPLES] A company invests $30,000 of surplus funds at two annual rates: 5% and 6%. The company wishes to have an annual yield of no less than 6%.What is the least amount of money the company can invest at 6%? Solution: Let x = amount at 6% then 30000-x= amount at 5% So need to invest at least $25714.29 EXAMPLES]The cost of publication of each copy of a magazine is $0.65. It is sold to dealers for $0.60 each and the amount received for advertising is 10% of the amount received for all magazines issued beyond 10000. What is the least number of magazines that can be published without loss? Solution:Let x = the number of units then 0.65x = total cost and 0.6x+0.1(0.6)(x-10000) = Total revenue So at least 60,000 magazines must be published. EXAMPLES]At present a manufacturer has 2500 units of a product in stock. This month, the product sells for $4 a unit. Next month the unit price will increase by $0.50. The manufacturer wants the total revenue received from the sale of 2500 units to be no less than $10,750. What is the maximum number of units that can be sold this month? Solution:Let x = the number of units then the total revenue is 4x+4.5(2500-x) So need to sell a max of 1000 units this month. Given Find the roots by equating to zero, =0 Solve by using either Factoring or Quadratic formula Suppose and Suppose = are the roots are the roots Solving Quadratic Inequalities: 1st Method Ex: Solve Solution : Solution set Solving Quadratic Inequalities: 2nd Method 2 x Example : Consider the inequality x 6 0 Solution: We can find the values where the quadratic equals zero by solving the equation, x2 x 6 0 x 3x 2 0 x 3 0 or x 2 0 x 3 or x 2 2 x For the quadratic inequality, x 6 0 we found zeros 3 and –2 by solving the equation x 2 x 6 0 . Put these values on a number line and we can see three intervals that we will test in the inequality. We will test one value from each interval. -2 3 Interval ,2 2, 3 3, Test Point x 3 Evaluate in the inequality True/False x2 x 6 0 3 3 6 9 3 6 6 0 True 2 x2 x 6 0 x0 02 0 6 0 0 6 6 < 0 False x2 x 6 0 x4 42 4 6 16 4 6 6 0 True Thus the intervals ,2 or 3, make up the solution set for the quadratic inequality, x 2 x 6 0 . In summary, one way to solve quadratic inequalities is to find the zeros and test a value from each of the intervals surrounding the zeros to determine which intervals make the inequality true. Example:Solve 2 x 2 3x 1 0 Solution: First find the zeros by solving the equation, 2 x 2 3x 1 0 2 x 2 3x 1 0 2x 1x 1 0 2 x 1 0 or x 1 0 1 x or x 1 2 Now consider the intervals around the zeros and test a value from each interval in the inequality. The intervals can be seen by putting the zeros on a number line. 1/2 1 Interval Test Point Evaluate in Inequality True/False 2 x 2 3x 1 < 0 1 , 2 x0 20 30 1 0 0 1 1 0 2 False 2 x 2 3x 1 < 0 1 ,1 2 9 9 1 3 3 2 3 1 1 < 0 8 4 8 4 4 2 x 3 4 True 2 x 2 3x 1 < 0 1, x2 Thus the interval the inequality 22 32 1 8 6 1 3 0 2 1 ,1 2 False makes up the solution set for 2 x 2 3x 1 0 . Ex: Solve Solution : There are no real roots, thus Solution set is empty set, {} Ex: Solve Solution : No real roots Solution set = Practice Problems x 2 5 x 24 0 12 x x 2 0 3x 2 5 x 2 < 0 5 x 2 13x 6 < 0 9 x2 0 2 x 2 5x 1 < 0 16 x 2 1 0 x 2 5 x < 4 3x 2 2 x 1 0 x2 2x 4