M a t h 152B
Chapter 10.4: Application Problems
Objectives:
• Solve systems of nonlinear equations
C h a p t e r 10.4
Steps for Solving Application Problems:
1. Read, throw out nonsense numbers
2. Assign a variable (What is it asking for?)
3. Write an equation
4. Solve the equation
5. Check, does it make sense?
Solve Systems of Nonlinear Equations
Ex: A rectangle has a perimeter of 70 meters and an area of 300 square meters. Find the dimensions of the
rectangle.
Let
Lei
X
(.^
Answer:
=
=
4
l / y
gctoi;
t«W>Jf
v=20 X=l5
15 m = width of rectangle
20 m = length of rectangle
^—•
You try:
1. A rectangle has a perimeter of 28 feet and an area of 48 square feet. Find the dimensions of the
rectangle.
Let ><
Let U
Equations:
Answer:
6 ft = width of rectangle
8 ft = length of rectangle
Ex: A rectangle has an area of 117 square centimeters with a diagonal of ^250 centimeters. Find the
length and width of the rectangle.
Let
K
=
l<.1c\\\4 \^J^\ai4^^
Let„_U^ =
Equations:
_
^
J
^
K'^k^XW
15
cr
Answer:
9 cm = width of rectangle
13 cm = length of rectangle
'~pI^\L£{
iK
^>^'
vJC^CDA
^
\
You try:
^^
1. A rectangle has an area of 3000 square feet. If the diagonal across the rectangle is 85 feet, then find the
dimensions of the rectangle.
Let
Let
X
= [<J\(M\
- I
| e : t X n j \ > ^ \^( tr\
Equations:
= ^ 5
Answer:
40 ft - width of rectangle ^
75 ft = length of rectangle
Ex: A rectangle that is made up of two right triangles has a diagonal that measures 26 inches, where the
perimeter of each of the triangles is 60 inches. Find the dimensions of the rectangle
Let
x; = (Jcl^A d
gdrav-'i:.^
(5 X^\)^^^"^ --^^-'^
Equations:
Answer:
10 in = width of rectangle
24 in = length of rectangle
XJ^^^^
' ^
You try:
1. A right triangle has a perimeter of 36 meters and a hypotenuse of 15 meters. Find the length of the legs
of the triangle.
Let
Let
Equations:
-
Sim
j
<^
^
\^f^
^cj ^
^
K^ll^
2-
Answer:
12 m = side of triangle
Ex: A ball is thrown upward from the ground. Its height above the ground at any time, /, is given by the
formula/?(?) = -16r^ +12& . At the same time, another ball is thrown upward from the top of a 60-foot-tall
building. Its height above the ground at any time, t, is given by the formula/z(0 = -16/^ +6& + 60. Find the
time at which the two balls v^U be the same height above the ground.
Let t = tk t -iaUs^r
-^r ^pt.b9fm
Equations:
•vbO
Answer:
1 sec. = time it takes for both to be the same height.
You try:
1. A ball is thrown downward from a height of 950 feet. The height of the ball above the ground at any
time, t, is found by the formula/7(0 = -16/^ -10/ +950. At the instant, another ball is thrown upward
from the top of a 750-foot-tall building. The height above the ground at any time, /, is found by the
formula/?(/) = -16/^^ + 80? + 750. Find the time at which the two balls will be the same height above the
ground.
Let 1. -
tt-tate-fer
b#yki bt.
'xw
Equations:
Answer:
about 2.22 sec. = time it takes for both to be the same height
Ex: Simple interest is calculated using the simple interest formula, interest = principle*rate*time or / = prt.
If you invest a certain principal at a specific interest rate for 1 year, the interest you would obtain is $10.50.
If you increase the principal by $25 and decrease the interest rate by 1%, the interest remains the same.
Find the principal and the interest rate.
Let
r
= Ic^.
Let _ p _ _ _ = ^innripdl
Equations:
pr(0=IO.^
— _
jaso(.ci)
-l^S
7%=rate
$150 - principal
Answer:
- 2560\
t—
-.105'^.^Sr-.AS:.
F
= 0
^
You try:
1. Simple interest is calculated using the simple interest formula, interest = principle*rate*time or / =prt.
If you invest principal at a specific interest rate for 1 year, the interest you would obtain is $72, If you
increase the principal by $120 and decrease the interest rate by 2%, the simple interest remains the
same. Find the principal and the interest rate.
Let
a?
t
Let^p
=
Equations:
i:
Answer:
TV
foOL
\ri -)
12% ="rate
$600 = principal
Ex: The cost, C , of manufacturing and selling x units of a product
revenue, R, is 7? = I2JC - 0.2x^. Find the break-even value of x.
Let X
Equations:
Answer:
isC = 0.6^:^+9, and the corresponding
= imlj^foj Luik^ il hi^i^%m\
^
about 1 unit and 14 units = number of units to break-even
You try:
1.
The cost, C , of manufacturing and selling J: units of a product isC = 80x+900, and the corresponding
revenue, R,\s,R^ \2Qx- Q.2x^. Find the break-even value of x.
Let
Equations:
= rmhf f)f ii/iik h jjiki^-emx
^ ^ | ^
,M
Answer:
about 26 units and 174 units = number of units to break-even
.'W
^
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