© Scarborough
MATH 141 Week In Review 1 KEY
Spring 2013
1
1. The demand for a VCR is 8000 units when the price is $230. If the price drops down to $210 then 9000
are sold. The manufacturer will not market VCRs if the price drops to $100. For each $60 increase in
price, the manufacturer will supply an additional 1300 VCRs. How many VCR's are made at the
equilibrium point?
Demand
(x, p)
(8000, 230)
Supply
(x, p)
(0, 100)
(9000, 210)
(0+1300, 100+60) = (1300, 160) or m 
210  230
1

9000  8000 50
1
x  8000
p  230 
50
1
p  230 
x  160
50
1
p
x  390 price in $ for x VCRs demanded
50
m
160  100 3

1300  0 65
3
x  0
p  100 
65
3
p  100 
x
65
3
p
x  100 price in $ for x VCRs supplied
65
m
Demand = Supply at the equilibrium point
1
3
x  390 
x  100
50
65
 43
x  290
650
188500
x
 4384
43
Therefore 4384 VCRs are made at the equilibrium point.
2. A tractor is originally purchased for $48,000. After 8 years, the tractor is worth $19360.
a. Find a linear equation for the value, V, of the tractor as a function of time, t, in years.
(
):
(
),
(
)
(
( )
)
dollars in value of the tractor t years after purchase
b. What is the rate of depreciation of the tractor?
Since m = –3580, the depreciation rate is $3580 per year.
c. What is the tractor worth after 10 years?
V 10   35800  48000  $12, 200
60
3

1300 65
© Scarborough
MATH 141 Week In Review 1 KEY
Spring 2013
2
3. It cost a company $22,500 to make 50 gadgets and $26,000 to make 120 gadgets. This company sells the
gadgets for $80 each. What is the profit function? Compute the profit or loss when 2500 gadgets are
produced and sold.
C ( x)  cx  F is the linear cost function.
R( x)  sx is the linear revenue function.
P( x)  R( x)  C( x) is the linear profit function.
Cost
 x, C  (50, 22500) (120, 26000)
26000  22500
 50
120  50
C  22500  50( x  50) or C  26000  50  x  120 
m
C  x   50 x  20000 dollars in cost for x gadgets
Revenue
R( x)  80 x dollars in revenue for x gadgets
Profit
P( x)  R( x)  C ( x)  80 x  50 x  20000   80 x  50 x  20000
Therefore P( x)  30 x  20000 dollars in profit/loss for x gadgets
P  2500   30  2500  20000  55000
Therefore there is a profit of $55,000 when 2500 gadgets are produced and sold.
© Scarborough
MATH 141 Week In Review 1 KEY
Spring 2013
3
4. A GPS manufacturer has a fixed monthly production cost of $59,985. If 360 GPS’s are produced and
sold, there is a loss of $43,785. A GPS is sold for $210. What is the break-even quantity?
Cost function:
Revenue function:
Profit function:
C ( x)  cx  F
R( x)  sx
P( x)  R( x)  C( x)
C ( x)  cx  59985
R( x)  210 x
P( x)  210 x   cx  59985
P(360)  210  360   c  360  59985  43785
360c  59400
c  165
Note: At the break-even point R( x)  C ( x) and P( x)  45x  59985  0 .
R( x )  C ( x )
OR
210 x  165x  59985
45x  59985
x  1333
Therefore the break-even quantity is 1333 GPS’s.
P( x)  45x  59985  0
45x  59985
x  1333
© Scarborough
MATH 141 Week In Review 1 KEY
Spring 2013
4
5. Due to recycling by Aggies, trash has been decreasing in Maroon City. The table gives the amount of
trash in thousands of pound over the time period of 2000 to 2007. Let x = 0 represent the year 2000 and
y represent thousands of pounds of trash.
Year
x
Trash, in thousands of pounds
2000
0
400
2002
2
360
2004
4
320
2005
5
310
2006
6
290
L1
L2
a. Find the equation of the least-squares (best fit, regression) line, in slope-intercept form, for this data.
Give your coefficients to five decimal places. State any special features or programs you use on your
calculator.
LinReg L1, L2, Y1
y = –18.18966 x + 397.84483 thousands of pounds of trash for x years after 2000
b. Use the unrounded-coefficients of the linear regression line to estimate the pounds of trash, to the
nearest whole number, that will be produced in the year 2009.
Y1(9) = 234.137931 or Y1(2009 – 2000) = 234.137931 and (234.137931)(1000) = 234137.931
OR
1000 * Y1(9) = 234137.931 or 1000 * Y1(2009 – 2000) = 234137.931
234137.931  234138 pounds of trash
c. What is the value, to 5 decimal places, and interpretation of the correlation coefficient?
r  –0.99715, which indicates a strong negative relation between the data points and the regression
line.
NOTE: Clear/Reset your calculator: MEM (2nd +), Reset, ALL, Reset
To turn on the correlation coefficient: Catalog (2nd 0), DiagnosticOn, Enter, Enter
© Scarborough
MATH 141 Week In Review 1 KEY
Spring 2013
5
6. The table shows the thousands of bacteria in a culture based upon the temperature.
x, degrees C
y, thousands of bacteria
35
80
40
65
45
52
50
40
55
20
a. What is the best-fitting (linear regression) line that represents the data? Do not round your
coefficients.
LinReg L1, L2, Y1
y = –2.9x + 181.9 thousands of bacteria for x degrees C
b. Using the line of best fit, how many bacteria, to the nearest whole number, would you predict if the
temperature was 38 degrees C?
Y1(38) = 71.7
this is in thousands
Predict 71,700 bacteria at 38 degrees C
c. Using the line of best fit, at what temperature, to the nearest degree, would you expect to have
30,000 bacteria?
Y2 = 30
Intersect
(52.37931…., 30)
OR
30 = –2.9 x + 181.9
1519
x=
 52.37931034
29
At 52 degrees C, would predict to have 30,000 bacteria.
d. What is the value, to 3 decimal places, and interpretation of the correlation coefficient?
r  –0.996, which indicates a strong negative relation between the data points and the regression
line.
© Scarborough
MATH 141 Week In Review 1 KEY
Spring 2013
6
7. If a water recirculation system is priced at $690, the quantity demanded is 310. For each $210 drop in
price, an additional 310 are demanded. Suppliers of the water recirculation system will supply 930
systems if the price is $540 and will supply 1550 if the price is $660.
a. Find the linear demand equation.
(x, p): (310, 690)
210 21
21
or use the point (310 + 310, 690 – 210) = (620, 480) p  690 
m

 x  310
310
31
31
21
Therefore p 
x  900 price in dollars for x systems demanded
31
METHOD II (Calculator): Put points (310, 690) and (620, 480) in L1 and L2.
Then do LinReg L1, L2, Y1
b. Find the linear supply equation.
(x, p): (930, 540), (1550, 660)
660  540 120 6
m


1550  930 620 31
6
p  540   x  930 
31
6
Therefore p  x  360 price in dollars for x systems supplied
31
METHOD II (Calculator): Put points (930, 540) and (1550,660) in L3 and L4.
Then do LinReg L3, L4, Y2
c. Above what price will there be no demand?
21
p  x 
x  900
31
21
p  0 
 0   900  900
31
When priced at $900 or more, there will be no demand for the system.
METHOD II (Calculator): Y1(0) = 900
© Scarborough
MATH 141 Week In Review 1 KEY
Spring 2013
7
d. What quantity would be demanded if the system was free?
21
p  x 
x  900
31
21
0
x  900
31
21
x  900
31
9300
x
 1328.57
7
Therefore 1329 systems would be demanded if the system was free.
METHOD II (Calculator): Find the x-value of the zero/root [calc, zero] of Y1; a good viewing
window is [–200, 1600] by [–200, 1000]
e. Above what price will the system be marketed?
6
p  x   x  360
31
6
p  0    0   360  360
31
Therefore, when priced at $360 or more, the system will be marketed.
METHOD II (Calculator): Y2(0) = 360
f. If the system price is $840, how many systems will be marketed?
6
p  x   x  360
31
6
840  x  360
31
6
x  480
31
x  2480
Therefore, when the system price is $840, 2480 systems will be marketed.
METHOD II (Calculator): Let Y3 = 840. Find the x-value of the intersection point of Y2 and
Y3 [calc, intersection]; a good viewing window is [–200, 3000] by [–200, 1000]
© Scarborough
MATH 141 Week In Review 1 KEY
Spring 2013
8
g. Find and interpret the equilibrium point.
Solve the system of equations.
21
p
x  900
31
6
p  x  360 multiply equation by –1
31
21
x  900
31
6
p 
x  360 add the two equations
31
______________________
p
0
27
x  540
31
27
x  540
31
6
21
 620  360  480 or p  620   620  900  480
31
31
Therefore, the equilibrium point is (620, 480).
x  620
p  620  
When 620 water recirculation systems are produced and sold at a price of $480, both
consumers and producers are satisfied.
METHOD II (Calculator): Find the intersection point of Y1 and Y2 [calc, intersection]; a good
viewing window is [–200, 1600] by [–200, 1000]