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:

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:

a. ? = 1.5 (increases by 1.5)

b. ? = -3 (decreases by 3)

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.

Solution: slope: y-intercept:

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 / 0 / 1 / 2 / 3 / 4
Annual Sales, y / 2.8 / 4.1 / 5.3 / 6.2 / 7.6

a.  Plot the annual sales (y) versus the year (x).

Solution:

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).

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.

- 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?

Solution: Þ y - 0 = -5200(x – 5) Þ y = -5200x + 26000

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:

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.

Solution: (3000, 20) (5200, 13)

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:

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?

Solution: (2500, 18) (1800, 15)

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 Þ x = 3 Þ 2(3) + 3y = 12 Þ 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.

Solution: Solve simultaneously: (2981, $20.07)