Download 1 1500 39.625 2 1 1500 79.25 1500 79.25 79.25 1500 0 yyyyyyyyy + =

Precalc Notes
2.3: Applications of Equations
Day 1
Real-life situations are usually described verbally, but they must be interpreted and expressed as
equivalent mathematical statements. The following guideline may be helpful.
1. Read the problem carefully, and determine what is asked for.
2. Label the unknown quantities with variables.
3. Draw a picture of the situation, if appropriate.
4. Translate the verbal statements in the problem and the relationships between the known and
unknown quantities into mathematical language.
5. Consolidate the mathematical information into an equation in one variable that can be solved
or an equation in two variables that can be graphed.
6. Solve for at least one of the unknown quantities.
7. Find all remaining unknown quantities by using the relationships given in the problem.
8. Check and interpret all quantities found in the original problem.
Ex. 1: The average of two real numbers is 39.625, and one number is 1500 times the reciprocal
of the other. Find the two numbers.
Two numbers: x and y
x y
 2  39.625


 x  1500  1 

 y
We can substitute for x:
1
1500    y
 y
 39.625
2


1
1500    y  79.25  y
 y


2
1500  y  79.25 y
y 2  79.25 y  1500  0
y
79.25 
 79.25 
2 1
2
 4 11500 
79.25  280.5625
2
79.25  16.75

2
 31.25 or 48

The two numbers are 31.25 and 48.
Ex. 2: The width of a rectangle is three times its height. If it has an area of 60.75 square feet,
what are its dimensions?
3x
x
x  3x   60.75
3 x 2  60.75
x 2  20.25
x  4.5
The width is 4.5 feet, and the height is 13.5 feet.
Ex. 3: A storage locker in the shape of a rectangular prism with a square floor has a volume of
70,000 cubic inches. If the surface area of the four walls and the top is 10,000 square
inches, what are the dimensions of the locker?
x
x
Each wall has surface area: xh
The top has surface area: x2
2
 x h  70, 000

2
4 xh  x  10, 000
h
70, 000
x2
 70, 000  2
4x 
  x  10, 000
2
 x

280, 000
 x 2  10, 000
x
h
The equation we get is not quadratic,
so let’s try to solve by graphing.
280, 000
 x 2  10, 000  0
x
280, 000
y1 
 x 2  10, 000
x
Then find the x-intercept(s)
x  30.971 or 80.851
If x = 30.971 in., the dimensions are: 30.971 in. x 30.971 in. x 72.977 in.
If x = 80.851 in., the dimensions are: 80.851 in. x 80.851 in. x 10.708 in. (not a very good
locker shape)
Ex. 4: A real estate investment yields a return of 10% per year and a certificate of deposit (CD)
pays 6% interest per year. How much of $7500 should be put in the real estate investment
and how much should be put in the CD to obtain a return of 7% on the entire $7500?
Real estate investment: x
CD: 7500 – x
0.1 x   0.06  7500  x   0.07  7500 
0.1x  450  0.06 x  525
0.04 x  75
x  1875
$1875 should be invested in real estate, and $5625 should be put in the certificate of
deposit.
Ex. 5: A canoeist paddled a total of 11 hours on a 48-kilometer round trip to Bear Lake. The
upstream trip to Bear Lake followed a period of heavy rain, and the current he paddled
against was 3 kilometers per hour. For the downstream return trip, he paddled with the
current that had slowed to 2 kilometers per hour. At what speed could he paddle the
canoe in still water?
Rate  Time  Distance, or alternatively, Time 
Rate
r–3
r+2
Time
24
r 3
24
r2
Distance
Rate
Distance
24
24
We know the total time is 11 hours, so we can set up this equation in one variable:
24
24

 11
r 3 r  2
24
 24


 11  r  3 r  2 

 r 3 r  2

24  r  2   24  r  3  11 r  3 r  2 
48r  24  11r 2  11r  66
11r 2  59r  42  0
11r  7  r  6   0
r
7
or 6
11
The negative answer doesn’t make sense, so we take the positive one.
The canoeist could paddle 6 km/h in still water.
Day 2
Ex. 6: A walkway of uniform width is to be built around a square swimming pool with 18-foot
edges. How wide should the walkway be to make the area of the walkway’s surface 400
square feet?
Area of outer square – Area of pool = Area of walkway
18  2 x 18  2 x   18
2
 400
324  72 x  4 x 2  324  400
18 + 2x
4 x 2  72 x  400  0
x 2  18 x  100  0
x
18 
18
2
 4 1 100 
2 1
18 + 2x
18  724
2
 4.454 or -22.454

The negative answer doesn’t make sense, so we take the positive one. The walkway should be
about 4.454 feet wide.
Ex. 7: A container in the shape of a rectangular prism with no top that has a volume of 650 cubic
inches is to be constructed from a 28 x 20-inch piece of sheet metal by cutting squares of
equal size from each corner and folding up the flaps. What size square should be cut from
each corner?
x
V  lwh
 x  20  2 x  28  2 x   650
x
This equation is not quadratic, so let’s solve for zero, and
then look at the graph to find the x-intercepts.
20 – 2x
28 – 2x
 x  20  2 x  28  2 x   650  0
y1   x  20  2 x  28  2 x   650
x  1.542, 6.677, or 15.781
The largest x-intercept doesn’t make sense in the context of the problem because it would make
sides of the box negative lengths. The other two answers are both reasonable.
The square should either be 1.542 in. x 1.542 in. or 6.677 in. by 6.677 in.
Ex. 8: A chemist has 30mL of a 40% acid solution in a test tube. How much of the solution
should be poured off and replaced with pure acid so that the new mixture is 70% acid?
Amount of acid poured off and replaced: x
Amount of acid left after pouring off x mL + x mL acid added = Amount of acid in final mixture
0.4  30  x   x  0.7  30 
1.2  0.4 x  x  21
0.6 x  9
x  15
The chemist should pour off 15mL of the solution and replace it with pure acid.