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Mathematics 116 test two solutions
Thursday, November 4, 2004
1. Find a formula for a linear function f (x) which has function values f (2) = 70 and
f (4) = 90.
x
2
4
y
70
90
y − 70 = 10(x − 2)
90 − 70
m=
= 10
4−2
y = 10x + 50
f (x) = 10x + 50
2. Calculate the minimum value of g(x) = 2x2 − 80x + 860, and the value of x at which the
minimum value occurs.
occurs at x = −
b
−80
=−
= 20
2a
2(2)
maximum value = g(20)
= 2(20)2 − 80(20) + 860 = 60
3. Sketch a graph of the quadratic function
y = −.04x2 + 2x + 5
Include in your sketch the exact location of the vertex and x & y intercepts.
vertex:
x=−
b
2
=−
= 25
2a
2(−.04)
y
y = y(25) = −.04(25)2 + 2(25) + 5 = 30
30
y-intercept: (0, 5)
x-intercepts:
x=
−b ±
√
b2 − 4ac
2a
5
x
-2.39
x=
−2 ±
q
4 − 4(−.04)(5)
2(−.04)
x ≈ −2.39 or 52.39
25
52.39
page two
4.
p
(500, 5)
!bà
!!
5
4
3
2
1
!
!!
!
!!supply
!!
HH
H
HH
H !!
!H
!H
!
H
!
HH
!
!
!
H demand
HH
H
H
100
200
300
400
(a) Find an equation for the supply line
m=
(b) Find an equation for the demand line
1
5−1
=
500 − 0
125
equation:
q
500
m=
0−4
1
=−
400 − 0
100
equation:
p=
1
q+1
125
p=−
1
q+4
100
(c) Find the equilibrium point (the price & quantity at which the two lines intersect)
1
1
q+1=−
q+4
125
100
multiply by 500
4q + 500 = −5q + 2000 =⇒ q =
500
≈ 166.67
3
find p:
p=−
1
1 500
7
q+4=−
+ 4 = ≈ 2.33
100
100 3
3
page three
5. Calculate the solution set of each system of equations:
(a)



x − 2y + 3z = −3
3x + y + 5z = 2

 −x + 3y + 2z = −1





(
(b)
x − y + 2z = 4
x+y−z = 0
)
Add the equations together to eliminate
y:
First eliminate the x:
add equations 1 and 3:
2x + z = 4
y + 5z = −4 (equation 4)
add equation 2 and three times
equation 3
let x = r = any real number
then z = 4 − 2x = 4 − 2r
and from equation 2:
y = −x + z = −r + (4 − 2r) = −3r + 4
10y + 11z = −1 (equation 5)
solution: (x, y, z) = (r, 4 − 3r, 4 − 2r)
Now solve for z by multiplying
equation 4 by −10 and adding
to equation 5:
(
−10y − 50z = 40
10y + 11z = −1
)
−39z = 39 ⇒ z = −1
From equation (4):
y = −5z − 4 = 1
From equation (1):
x = 2y − 3z − 3 = 2
solution: (x, y, z) = (2, 1, −1)
page four
6. Suppose that the price of a bus ride is linearly related to the number of bus riders. At a
price of one dollar per bus ride, it is found that the bus company transports 400, 000 riders
each day. When the price of a ride is $1.50, the bus company transports only 150, 000 riders.
What price will maximize the bus company’s revenue? (where R = pq)
summary of data (q in millions of riders and p in dollars):
q
p
0.4 1.00
0.15 1.50
Find the equation of the linear relation between q and p:
m=
1.50 − 1.00
= −2
.15 − .40
(p − 1.00) = −2(q − 0.4)
p = −2q + 1.8
Now an equation for the revenue can be found (R in millions of dollars):
R = pq = (−2q + 1.8)q = −2q 2 + 1.8q
The maximum value of the revenue occurs at (use vertex formula)
q=−
1.8
= 0.45 millions = 450, 000
2(−2)
p = −2q + 1.8 = −2(0.45) + 1.8 = $0.90