Math 1324 – Final Exam Sample.
1.
The revenue generated by selling x units of a product is given by the quadratic
function Rx   0.03x 2  63x , where the revenue R ( x ) is measured in dollars. How
many units to be sold to obtain maximum revenue, and what is this maximum
revenue?
Answer. This is a quadratic function concave downward. The maximum value is
b
63

 1,050 and
indicated by its vertex (h , k ) where h  
2a
2 0.03
k  R1,050  33,075 . R( x)  0.03x
There are 1,050 units to be sold to obtain the maximum revenue of $33,075.
2.
Let the revenue function be Rx   0.03x 2  63x and the cost function be
C x   12 x  1,500 , where x represent the number of units of a product being
produced and sold. Find the average cost function and the profit function.
Answer.
Total cost
12 x  1,500

number of units
x
2
Profit  Revenue - Cost   0.03x  63x  12 x  1,500  0.03x 2  51x  1,500
Average cost 


3.
Given f x   2 x  3 and g x   x 3  1 . Find f  g .
Answer.
 f  g x  f g x  f x 3  1  2x 3  1  3  2x 3  5
OR
 f  g x  f g x  2g x  3  2x 3  1  3  2x 3  5
4.
Given
a.
b.
c.
Px   2 x 2 x  2x  3 .
Find the x-intercepts and y-intercept.
Find the degree and the leading coefficient
Sketch its graph
Answer.
a. x – intercepts : ( 0 , 0 ) , ( 2 , 0 ) , and (– 3 , 0 ).
b. Degree is 4. Leading coefficient is – 2.
c.
5.
y-intercept: ( 0 , 0 )
Find the intercepts and the asymptotes, and then sketch a graph of
x2
x2
R x  
 2
x  3x  1 x  2 x  3
Answer.
2

x – intercepts : ( – 2 , 0 ) . y-intercept:  0, 
3

H.A. are x  3 and x  1
V.A. is y  0
6.
Sketch a graph of f x   2 x1  3 . State the basic function and the transformations
applied.
Answer. Basic function is y  2 x . Transformations: left by 1 units, then down by 3 units
7.
Sketch a graph of f x   log 2 x  3  2 . State the basic function and the
transformations applied.
Answer. Basic function is y  log 2 x . Transformations: right by 3 units, then up by 2
units
Find the domain of f x   log 2 5x  3  2 .
8.
Answer.
5x  3  0
5x  3
3
x
5
3 
D   ,
5 
Find the inverse of f x   2 x1  3 .
9.
Answer.
y  2 x 1  3
y  3  2 x 1
log 2  y  3  x  1
x  log 2  y  3  1
f
1
10.
x   log 2 x  3  1
Rewrite ln 2  2 ln x  ln 5  3 ln y into a single logarithmic expression.
Answer.
 2x 2 
2x 2
ln 2  2 ln x  ln 5  3 ln y  ln 2  ln x 2  ln 5  ln y 3  ln  3   ln 3
5y
 5y 
 
11.
Rewrite log
 
7a 4
into a sum/difference of simple logarithms containing only
3b 5
single variables.
Answer.
7a 4
log 5  log 7  log a 4  log 3  log b 5  log 7  4 log a  log 3  5 log b
3b
12.
The growth rate function of an investment is G(t )  5 log 3 (2t  7)  3 (%) where
t is the number of years. Find the growth rate after 8 years (using a calculator and
change of base formula)
Answer.
G10  5 log 3 2  8  7   3  5 log 3 23  3  5 
13.
ln 23
 3  17.27%
ln 3
$40,000 is invested at the annual rate of 1.2% compounded quarterly.
a. Find the value of the investment after 8 years.
b. How long will it take so that the value is $50,000?
 r
At   P1  
 n
nt
Answer. P  40,000
r  0.012
 0.012 
a. A8  40,0001 

4 

n4
4 8 
 44,024.04
b. Solve for t where At   50,000
 0.012 
50,000  40,0001 

4 

4t
50,000  0.012 
 1 

40,000 
4 
5
4t
 1.003
4
5
4t
ln  ln 1.003
4
ln 5  ln 4  4t ln 1.003
ln 5  ln 4
t
 18.62 years
4 ln 1.003
4t
14.
Solve the following equations
10
a.
3
x 3
5e  2
b. 4 ln x  1  5  13
Answer.
a.
10  3 5e x 3  2


10  15e x 3  6
16  15e x 3
16
 e x 3
15
16
ln  ln e x 3
15
ln 16  ln 15  x  3
x  ln 16  ln 15  3
b.
4 ln  x  1  5  13
4 ln  x  1  8
ln  x  1  2
x 1  e2
x  e2  1
15.
Use Gauss Elimination with Back-Substitution to solve the system below
x
 3z   2
3x
 y  2z 
2x  2 y
z 
5
4
Answer.
 1 0  3  2
 1 0  3  2
 1 0  3  2
3 1  2



3 R1 R 2
 2 R1 R 3
5  
0 1
7 11  
0 1
7 11

2 2
2 2
0 2
1
4
1
4
7
8
 1 0  3  2
 1 0  3  2


2 R 2 R 3
R 3 /( 7 )
 0 1
7
11  0 1
7 11
0 0  7  14
0 0
1
2
x
 3z   2
y  7z 
11
z 
2
y  7(2)  11
y  3
x  3(2)  2
x4
Solution is 4,3,2
16.
Set up a linear system for the following problem
A sum of $200,000 were invested in three accounts that paid simple interest at
the rates of 2%, 3% and 1.5%. After one year the total interest earned was
$3,256. The interest earned from the 2% account was $320 more than the
interest earned from the 1.5% account. How much was invested in each
account?
Answer. Let x , y , and z be the amount invested in the 2% , 3% , and 1.5% respectively.
Then
z 
200,000
0.02 x  0.03 y  0.015 z 
3,256
x
0.02 x
y
 320  0.015 z
17.
Find the optimal values of z  4 x  5 y with constraints
x  0
y 
0
x  3 y  15
4 x  y  16
Answer.
Vertices
(0,5)
(3,4)
(0,0)
(4,0)
z-values
25
32
0
16
Maximum value of z is 32. Minimum value of z is 0.
18.
Set up a linear programming problem for the following
For a TV factory that produces LCD TV and Plasma TV we have the
following data
Profit ($)
One LCD
One Plasma
100
160
Assembling (hours)
2
5
Testing (hours)
1
2
Packaging (hours)
1
1
Each month the factory can assign at most 3000 hours for assembling, 1400 hours for testing,
and 900 hours for packaging. Find the monthly optimal level of production (the number of TV
of each type should be produced)
Answer. Let P be the monthly profit, x and y be the number of LCD and Plasma TV produced
in that month. Then
Maximize: P = 100 x + 160 y (Profit)
Constraints:
x0
y0
2 x  5 y  3000 (Assembing)
x  2 y  1400 (Testing)
x  y  900 (Packaging )