1)
where x = number of booxs produced
Fixed cost = C = 50,000
Variable cost = 2.8x
C(x)= 2.8x+50,000
Revenue function is R(x)=10x
We have to find x where C(x)=R(x)
2.8x+50,000=10x
7.2x=50,000
x=50,000/7.2
x = 6944.4445
Answer: 6944
2) 1580
We have to find x where 158/89= x/890 then
x = (158/89)890 = 1580
Answer: 1580
3)
a) Let V=mt+n , we have to find m and n
slope = m = (82,000-155,000)/10=-7,300
Then V=-7,300t+n
At t=0 V=155,000 then n=155,000
Answer: V = -7300t+155000
b) if t=6 then V=-7,300(6)+155,000=111,200
Answer: 111200
c)
We have to find t where V=80,000
-7,300t+155,0000=80,000
7,300t=75,000
t = 75,000/7,300 = 10.27 (in the 11th year)
Answer: 11
d) Slope is < 0 so the line must go down and V=155,000 at t=0
Answer: A
4)
A)
We must to find p=mx+b
And we know:
x
p
7,100
2.24
7,500
2.33
Slope = m = (2.33-2.24)/(7,500-7,100)=0.000225
p = 0.000225x+b
When x = 7,100 p=2.24 then:
2.24 = 0.000225(7,100)+b
b = 2.24-0.000225(7,100)=0.6425
Answer: p = 0.000225x+0.6425
B)
We must to find p=mx+b
And we know:
x
p
7,700
2.24
7,600
2.33
Slope = m = (2.24-2.33)/(7,600-7,700)=-0.0009
p = -0.0009x+b
When x = 7,700 p=2.24 then:
2.24 = -0.0009(7,700)+b
b = 2.24 + 0.0009(7,700)=9.17
Answer: p = -0.0009x+9.17
C)
We have to find (x,p) where supply=demand
0.000225x+0.6425=-0.0009x+9.17
0.001125x=8.5275
x = 8.5275/0.001125=7,580
using supply equation: p = 0.000225(7,580)+0.6425=2.348
Answer: (7580,2.348)
D) Supply has a positive slope and when x=7,000
p= 0.000225(7,000)+0.6425=2.2175 so supply line must be between 2 and 3 when x=7,000
Answer: A
5)
A)
Answer: w=2.1h+53
B)
If we have y=mx+n
Rate of change = m units of y per units of x
W = 2.1h+53
Slope=rate of change = 2.1 kg per inch
Answer: B
C) height = 5´5” then h=5 then w=2.1(5)+53=63.5
Answer: 63.5
D)
We have to find h where w=62 then 2.1h+53=62
2.1h=9 then h=9/2.1=4.2857
Answer: h=4.3
6)
A) revenue values are increasing and if you see those values the graph is B
Answer: B
B) R=26.1t+203, period ending 1/31/10 is in t=8
Then predicted value is 26.1(8)+203=411.8
Answer: 412
7) p(x)=4,000-50x
If x = 12 then p=4,000-50(12)=3,400
If x = 20 then p = 4,000- 50(20)= 3,000
Answer: 3,400 and 3,000
8)
A) R(x)=xp(x)=x(90-3x)=-3x2+90x
Answer: R(x)=-3x2+90x, domain: D
B)
R(4)=4(90-3(4))=312
R(8)=8(90-3(8))=528
R(12)=12(90-3(12))=648
R(16)=16(90-3(16))=672
R(20)=20(90-3(20))=600
Answer: 312,528,648,672,600
C) (Sorry) Correct answer is A, you did it right
9)
A=Pert, r=0.0315, t =11, P=25,000
A=25,000e0.0315(11) =35,352.73735
Answer: 35,352.74
10)
a) r=0.0405, P=25,000,t=1, n=12
A= P(1+r/n)nt = 25,000(1+0.0405/12)12 =26,031.51
b) r=0.0405, P=25,000,t=1, n=365
A= P(1+r/n)nt = 25,000(1+0.0405/365)365 =26,033.22
c) r=0.0405, P=25,000,t=1, n=4
A= P(1+r/n)nt = 25,000(1+0.0405/4)4 =26,027.98
11)
We have to find t where 2P=P(1.+0.2103)t
(1.2103)t = 2
tLn(1.2103)=Ln(2)
t = Ln(2)/Ln(1.2103)=3.6315
Answer: 4
12)
a) We must to find t where 3,700=2,000(1+0.09/4)4t
(1.0225)4t =1.85
4tLn(1.0225)=Ln(1.85)
t = Ln(1.85)/4Ln(1.0225) = 6.91
Answer: 6.91
b) We must to find t where 3,700=2,000e0.09t
e0.09t = 1.85
Ln(e0.09t)=Ln(1.85)
0.09t = Ln(1.85)
t = Ln(1.85)/0.09=6.8354
Answer: 6.84