§ 6.7

1. §6.7 Formulas and Applications of Rational Equations

2. Applications of Rational Equations EXAMPLE Solve the formula for R: SOLUTION This is the original equation. Multiply both sides by the LCD, R + r. Simplify. Divide both sides by R + r. Blitzer, Intermediate Algebra, 4e – Slide #104

3. Applications of Rational Equations EXAMPLE • A company is planning to manufacture small canoes. Fixed monthly cost will be $20,000 and it will cost $20 to produce each canoe. • Write the cost function, C, of producing x canoes. • Write the average cost function, , of producing x canoes. • How many canoes must be produced each month for the company to have an average cost of $40 per canoe? Blitzer, Intermediate Algebra, 4e – Slide #105

4. Applications of Rational Equations CONTINUED SOLUTION (a) The cost function, C, is the sum of the fixed cost and the variable costs. Fixed cost is $20,000. Variable cost: $20 for each canoe produced. (b) The average cost function, , is the sum of fixed and variable costs divided by the number of canoes produced. Blitzer, Intermediate Algebra, 4e – Slide #106

5. Applications of Rational Equations CONTINUED (c) We are interested in the company’s production level that results in an average cost of $40 per canoe. Substitute 40, the average cost, for and solve the resulting rational equation for x. Substitute 40 for Multiply both sides by the LCD, x. Subtract 20x from both sides. Divide both sides by 20. The company must produce 1,000 canoes each month for an average cost of $40 per canoe. Blitzer, Intermediate Algebra, 4e – Slide #107

6. Applications of Rational Equations Blitzer, Intermediate Algebra, 4e – Slide #108

7. Applications of Rational Equations EXAMPLE An engine pulls a train 140 miles. Then a second engine, whose average rate is 5 miles per hour faster than the first engine, takes over and pulls the train 200 miles. The total time required for both engines is 9 hours. Find the average rate of the first engine. SOLUTION 1) Let x represent one of the quantities. Let x = the rate of the first engine. 2) Represent other quantities in terms of x. Because the average rate of the second engine is 5 miles per hour faster than the average rate of the first engine, let x + 5 = the rate of the second engine. Blitzer, Intermediate Algebra, 4e – Slide #109

8. Applications of Rational Equations CONTINUED 3) Write an equation that describes the conditions. By reading the problem again, we discover that the crucial idea is that the time for both engines’ trips is 9 hours. Thus, the time of the first engine plus the time of the second engine is 9 hours. The sum of the two times is 9 hours. The sum of the two times is 9 hours. The sum of the two times is 9 hours. We are now ready to write an equation that describes the problems’ conditions. Blitzer, Intermediate Algebra, 4e – Slide #110

9. Applications of Rational Equations CONTINUED time of the second train plus Time of the first train equals 9 hours. 4) Solve the equation and answer the question. This is the equation for the problems’ conditions. Multiply both sides by the LCD, x(x + 5). Use the distributive property on both sides. Blitzer, Intermediate Algebra, 4e – Slide #111

10. Applications of Rational Equations CONTINUED Simplify. Use the distributive property. Combine like terms. Subtract 340x + 700 from both sides. Factor the right side. Set each variable factor equal to zero. Solve for x. Blitzer, Intermediate Algebra, 4e – Slide #112

11. Applications of Rational Equations CONTINUED Because x represents the average rate of the first engine, we reject the negative value, -20/9. The rate of the first engine is 35 miles per hour. 5) Check the proposed solution in the original wording of the problem. Do the two engines’ trips take a combined 9 hours? Because the rate of the second engine is 5 miles per hour faster than the rate of the first engine, the rate of the second engine is 35 + 5 = 40 miles per hour. The total time is 4 + 5 = 9 hours. This checks correctly. Blitzer, Intermediate Algebra, 4e – Slide #113

12. Applications of Rational Equations Fractional part of the job done by the first person fractional part of the job done by the second person 1 (one whole job completed). + = Blitzer, Intermediate Algebra, 4e – Slide #114

13. Applications of Rational Equations EXAMPLE A hurricane strikes and a rural area is without food or water. Three crews arrive. One can dispense needed supplies in 10 hours, a second in 15 hours, and a third in 20 hours. How long will it take all three crews working together to dispense food and water? SOLUTION 1) Let x represent one of the quantities. Let x = the time, in hours, for all three crews to do the job working together. 2) Represent other quantities in terms of x. There are no other unknown quantities. Blitzer, Intermediate Algebra, 4e – Slide #115

14. Applications of Rational Equations CONTINUED 3) Write an equation that describes the conditions. Working together, the three crews can dispense the supplies in x hours. We construct a table to find the fractional part of the task completed by the three crews in x hours. Blitzer, Intermediate Algebra, 4e – Slide #116

15. Applications of Rational Equations CONTINUED Because all three teams working together can complete the job in x hours, 4) Solve the equation and answer the question. This is the equation for the problem’s conditions. Multiply both sides by 60, the LCD. 6 4 3 Use the distributive property on each side. Blitzer, Intermediate Algebra, 4e – Slide #117

16. Applications of Rational Equations CONTINUED Simplify. Combine like terms. Divide both sides by 13. Because x represents the time that it would take all three crews to get the job done working together, the three crews can get the job done in about 4.6 hours. 5) Check the proposed solution in the original wording of the problem. Will the three crews complete the job in 4.6 hours? Because the first crew can complete the job in 10 hours, in 4.6 hours, they can complete 4.6/10, or 0.46, of the job. Blitzer, Intermediate Algebra, 4e – Slide #118

17. Applications of Rational Equations CONTINUED Because the second crew can complete the job in 15 hours, in 4.6 hours, they can complete 4.6/15, or 0.31, of the job. Because the third crew can complete the job in 20 hours, in 4.6 hours, they can complete 4.6/20, or 0.23, of the job. Notice that 0.46 + 0.31 + 0.23 = 1, which represents the completion of the entire job, or one whole job. Blitzer, Intermediate Algebra, 4e – Slide #119