Transcript 6.7
§ 6.7
Formulas and Applications of Rational Equations
Applications of Rational Equations
EXAMPLE Solve the formula for
R
:
I
E R
r
.
SOLUTION
I
E R
r
R
r
I
R
r
E R
r
This is the original equation.
Multiply both sides by the LCD,
R
+
r
.
I
R
r
E I
E R
r
Simplify.
Divide both sides by
R
+
r
.
Blitzer,
Intermediate Algebra
, 4e – Slide #104
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.
(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?
Blitzer,
Intermediate Algebra
, 4e – Slide #105
Applications of Rational Equations
CONTINUED SOLUTION (a) The cost function,
C
, is the sum of the fixed cost and the variable costs.
C
20 , 000 20
x
Fixed cost is $20,000.
Variable cost: $20 for each canoe produced.
costs
divided by the number of canoes produced
.
C
20 , 000 20
x x
Blitzer,
Intermediate Algebra
, 4e – Slide #106
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,
x
.
40 20 , 000 20
x x
Substitute 40 for
C
. 40
x
20 , 000 20
x
20
x
20 , 000
x
1 , 000 Multiply both sides by the LCD,
x
.
Subtract 20
x
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
Applications of Rational Equations
Time in Motion
t
d r
Time traveled Distance traveled Rate of travel Blitzer,
Intermediate Algebra
, 4e – Slide #108
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
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.
Train 1 Train 2 Distance
140 200
Rate
x x
+ 5
Time
140
x x
200 5 We are now ready to write an equation that describes the problems’ conditions.
Blitzer,
Intermediate Algebra
, 4e – Slide #110
Applications of Rational Equations
CONTINUED Time of the first train plus time of the second train equals 140
x x
200 5 9 9 hours.
4) Solve the equation and answer the question.
140
x x
200 5 9 This is the equation for the problems’ conditions.
x
x
5 140
x x
200 5
x
x
5 9
x
x
5 140
x
x
x
5
x
200 5
x
x
5 9 Multiply both sides by the LCD,
x
(
x
+ 5).
Use the distributive property on both sides.
Blitzer,
Intermediate Algebra
, 4e – Slide #111
Applications of Rational Equations
CONTINUED 140
x
5 200
x
9
x
x
5 140
x
700 200
x
9
x
2 45
x
340
x
700 9
x
2 45
x
0 9
x
2 295
x
700 0
x
35 9
x
20 0
x
35 35
x
0 9
x
20 20
x
9 Simplify.
Use the distributive property.
Combine like terms.
Subtract 340
x
+ 700 from both sides.
Factor the right side.
Set each variable factor equal to zero.
Solve for
x
.
Blitzer,
Intermediate Algebra
, 4e – Slide #112
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.
Time of the first engine Distance Rate 140 35 4 hours Time of the second engine Distance 200 5 hours Rate 40 The total time is 4 + 5 = 9 hours. This checks correctly.
Blitzer,
Intermediate Algebra
, 4e – Slide #113
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).
Blitzer,
Intermediate Algebra
, 4e – Slide #114
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.
unknown quantities.
There are no other Blitzer,
Intermediate Algebra
, 4e – Slide #115
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.
First Crew Second Crew Third Crew Fractional part of job completed in 1 hour
1/10 1/15 1/20
Time working together
x x x
Fractional part of job completed in x hours
x
/10
x
/15
x
/20 Blitzer,
Intermediate Algebra
, 4e – Slide #116
Applications of Rational Equations
CONTINUED Because all three teams working together can complete the job in
x
hours,
x
10
x
15
x
20 1 .
4) Solve the equation and answer the question.
x
10
x
15
x
20 1 This is the equation for the problem’s conditions.
60
x
10
x
15
x
20 60 1 Multiply both sides by 60, the LCD.
6 60
x
10 4 60
x
15 3 60
x
20 60 Use the distributive property on each side.
Blitzer,
Intermediate Algebra
, 4e – Slide #117
Applications of Rational Equations
CONTINUED 6
x
4
x
3
x
60 13
x
60
x
4 .
6 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
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