Math 141 Week in Review
Sections 1.1-1.4
9/5/05
Section 1.2.
1. Find the slope of the line shown in the figure.
Solution: Find two points on the line:
(0, 6) and (3, 0)
Use the slope formula:
y y 2  y1 6  0
m


 2
x x2  x1 0  3
2. Given the equation 3x - 2y = 7, answer the following questions:
a. If x increases by 1 unit, what is the corresponding change in y?
b. If x decreases by 2 units, what is the corresponding change in y?
Solution: The relationship of the change in y to the change in x is about slope:
3 ?
a. m  
? = 1.5 (increases by 1.5)
2 1
3
?
b. m  
? = -3 (decreases by 3)
2 2
3. Find an equation of the vertical line that passes through (0, 8).
Find an equation of the vertical line that passes through (2, 7).
Solution: A vertical line has all x-values the same, so the equation is x=0 for the
first one, and x=2 for the second one.
4. Write the equation in slope-intercept form, and find the slope and y-intercept of
the corresponding line: y + 6 = 0.
Solution: Put the equation in slope-intercept form: y = 0x – 6. The slope is 0,
and the y-intercept is (0, -6).
5. Write the equation in slope-intercept form, and find the slope and y-intercept of
the corresponding line: 8x + 5y – 24 = 0.
8
24
8
 24 
Solution: y   x 
slope: 
y-intercept:  0, 
5
5
5
 5 
6. Find an equation of the line passing through the point (c, d) with undefined slope.
Solution: A line with undefined slope is vertical. A vertical line has all x-values
the same. So the equation is x = c.
7. Find an equation of the line passing through the point (c, d) with slope 0.
Solution: A line with slope 0 is horizontal. The equation is y = d.
8. Sketch the straight line by finding the x- and y-intercepts: 3x – 5y = 20
Solution:
9. A mathematical model for a pharmaceutical company’s sales, in billions of
dollars, is given by S = 5.74 + 0.97x where x = 0 corresponds to 1988.
a. What is the slope of the line? What does it represent?
Solution: 0.97; On average, the increase in sales each year is $.97 billion.
b. What is the S-intercept of the line? What does it represent?
Solution: (0, 5.74); In 1988, the sales were $5.74 billion.
10. The sales (in millions of dollars) of a company’s equipment sales from 2000
through 2004 is given below (x = 0 corresponds to 2000).
Year x
Annual Sales, y
0
2.8
1
4.1
2
5.3
a. Plot the annual sales (y) versus the year (x).
Solution:
3
6.2
4
7.6
b. Draw a straight line L through the points corresponding to 2000 and 2004.
c. Derive an equation of the line L.
Solution: See graph for a) and b).
7.6  2.8
m
 1.2
40
y – 2.8 = 1.2(x – 0)
y = 1.2x + 2.8
d. Use the equation found in part (c) to estimate the annual sales of equipment
in 2002.
Solution: y = 1.2(2) + 2.8
y = $5.2 million
Section 1.3
1. Determine whether the equation defines y as a linear function of x. If so, write it
in the form y = mx + b.
3x = 2y - 7
Solution: yes; y = (3/2)x + 7/2
2. Determine whether the equation defines y as a linear function of x.
x - 5y = 2
Solution: no
3. An automobile purchased for use by the manager of a firm at a price of $26,000
is to be depreciated using the straight-line method over 5 yr. What will be the
book value of the automobile at the end of 2 yr?
0  26000
 5200  y - 0 = -5200(x – 5)  y = -5200x + 26000
Solution: m 
50
After 2 years: y = -5200(2) + 26000 = $15,600
4. A camera manufacturer has a monthly fixed cost of $26,000 and a production
cost of $12 for each camera manufactured. The cameras sell for $18 each.
a. What is the cost function?
Solution: C(x) = 26000 + 12x
b. What is the revenue function?
Solution: R(x) = 18x
c. What is the profit function?
Solution: P(x) = 18x – (26000 + 12x) = 6x - 26000
d. Compute the profit (loss) corresponding to production levels of 2000,
6000, and 10,000 cameras, respectively.
Solution: P(2000) = 6(2000) – 26000 = -14000
P(6000) = 6(6000) – 26000 = 10000
P(100000) = 6(10000) – 26000 = 34000
5. Sketch the equation of the demand curve 4p + 5x – 60 = 0, where x represents
the quantity demanded in units of 1000 and p is the unit price in dollars.
Determine the quantity
demanded corresponding to the
unit price $12.
Solution:
4 p  5 x  60
5
p   x  15
4
5
12   x  15
4
x  2.4
2400 units
6. The quantity demanded for a certain computer chip is 3000 units when the unit
price is set at $20. The quantity demanded is 5200 units when the unit price is
$13. Find the demand equation if it is known to be linear.
20  13
7

Solution: (3000, 20) (5200, 13) m 
3000  5200
2200
7
y  20  
( x  3000)
2200
7
325
y
x
2200
11
7. Sketch the equation of the supply curve ½x – ¾p + 8 = 0, where x represents the
quantity supplied in units of 1000 and p is the unit price in dollars. Determine the
number of units of the commodity the supplier will make available in the market
at the unit price $20.
Solution:
1
3
x p8  0
2
4
2
32
p  x
3
3
2
32
20  x 
3
3
x  14
14000 units
8. The manufacturer will make 2500 of the computer chips in problem #6 available
when the price is $18. At a unit price of $15, 1800 chips will be marketed. Find
the supply equation if the equation is known to be linear. How many chips will be
marketed when the unit price is $22?
18  15
3
m

Solution: (2500, 18) (1800, 15)
2500  1800 700
3
y  18 
( x  2500)
700
3
51
y
x
700
7
3
52
22 
x
700
7
x  3433 units
Section 1.4
1. Find the point of intersection of the pair of straight lines:
2x + 3y = 12
5x – 2y = 11
Solution: 2(2x + 3y = 12)
3(5x – 2y = 11)
4x + 6y = 24
15x – 6y = 33
19x = 57  x = 3  2(3) + 3y = 12  y = 2
(3, 2)
2. Find the break-even point for the firm whose cost function C and revenue
function R were found in Section 1.3, #4 above.
Solution: 6x – 26000 = 0
x  4333
R(x) = 18x
= 18(26000/6)
= $78,000
(4333, $78000)
3. A company manufactures microwave ovens. Each oven sells for $60. The
monthly fixed costs total $24,000, and the variable cost of producing each oven
is $8. Find the break-even point for the company.
Solution: C(x) = 24000 + 8x
R(x) = 60x
24000 + 8x = 60x
x  462
60(462) = $27,692.31
(462, $27,692.31)
4. The sales for Maddie’s Beauty Supply are expected to be given by S = 3.2 + .04t
thousand dollars t years from now. The annual sales of Jean’s Beauty Supply
are expected to be given by S = 1.4 + .05t thousand dollars t years from now.
When will Jean’s annual sales first surpass Maddie’s annual sales?
Solution: 1.4 + .05t > 3.2 + .04t
t > 180 years
5. Find the equilibrium quantity and price for the supply-and-demand equations,
where x represents the quantity demanded in units of 1000 and p is the unit price
in dollars: 4x + 5p – 50 = 0
and
6x – 3p + 15 = 0
Solution: 3(4x + 5p = 50)
5(6x – 3p = -15)
12x + 15p = 150
30x – 15p = -75
42x = 75
x = 25/14
4(25/14) + 5p = 50
p  $8.57
(1786, $8.57)
6. Find the equilibrium quantity and price for the computer chip company described
in Section 1.3, #6 and #8 above.
7
325
y
x
demand
2200
11
Solution:
Solve simultaneously: (2981, $20.07)
3
51
y
x
sup ply
700
7