1. § 6.7 Formulas and Applications of Rational Equations

2. Blitzer, Intermediate Algebra, 4e – Slide #104 Applications of Rational EquationsEXAMPLE 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.

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

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

5. Blitzer, Intermediate Algebra, 4e – Slide #107 Applications of Rational Equations (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. CONTINUED 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.

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

7. Blitzer, Intermediate Algebra, 4e – Slide #109 Applications of Rational EquationsEXAMPLE 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.

8. Blitzer, Intermediate Algebra, 4e – Slide #110 Applications of Rational Equations 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. CONTINUED DistanceRateTime Train 1140x Train 2200x + 5 The sum of the two times is 9 hours. We are now ready to write an equation that describes the problems’ conditions.

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

10. Blitzer, Intermediate Algebra, 4e – Slide #112 Applications of Rational EquationsCONTINUED 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.

11. Blitzer, Intermediate Algebra, 4e – Slide #113 Applications of Rational EquationsCONTINUED 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.

12. Blitzer, Intermediate Algebra, 4e – Slide #114 Applications of Rational Equations Work Problems In work problems, the number 1 represents one whole job completed. Equations in work problems are often based on the following condition: 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). +=

13. Blitzer, Intermediate Algebra, 4e – Slide #115 Applications of Rational EquationsEXAMPLE 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.

14. Blitzer, Intermediate Algebra, 4e – Slide #116 Applications of Rational Equations 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. CONTINUED Fractional part of job completed in 1 hour Time working together Fractional part of job completed in x hours First Crew1/10xx/10 Second Crew1/15xx/15 Third Crew1/20xx/20

15. Blitzer, Intermediate Algebra, 4e – Slide #117 Applications of Rational Equations Because all three teams working together can complete the job in x hours, CONTINUED 4) Solve the equation and answer the question. This is the equation for the problem’s conditions. Multiply both sides by 60, the LCD. Use the distributive property on each side. 643

16. Blitzer, Intermediate Algebra, 4e – Slide #118 Applications of Rational EquationsCONTINUED 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.

17. Blitzer, Intermediate Algebra, 4e – Slide #119 Applications of Rational EquationsCONTINUED 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.