Intermediate Algebra (11th edition)

OBJECTIVE 3 Solve applied problems using area formulas.

Solving an Area Problem

A rectangular reflecting pool in a park is

20 ft wide and 30 ft long. The gardener

wants to plant a strip of grass of uniform

width around the edge of the pool. She has

enough seed to cover 336 How wide

will the strip be?

Step 1 Readthe problem carefully.

Step 2 Assign a variable.The pool is shown in FIGURE 4.

Let the unknown width of the grass strip.

Then the width of the large rectangle (the width of the pool

plus two grass strips),

and the length of the large rectangle.

Step 3 Write an equation.Refer to FIGURE 4.

Area of large rectangle (length width)

or 600 Area of pool (in square feet)

The area of the large rectangle minus the area of the pool should equal

the area of the grass strip.

Area Area Area

of large of of

rectangle pool grass

Step 4 Solve. Multiply.

Standard form

Divide by 4.

Factor.

or Zero-factor property

or Solve for x.

Step 5 State the answer.The width cannot be ft, so the grass strip should be

3 ft wide.

Step 6 Check.If we can find the area of the large rectangle (which includes

the grass strip).

Area of pool and strip

The area of the pool is So, the area of the grass strip is

as required. The answer is correct. NOW TRY

OBJECTIVE 4 Solve applied problems using quadratic functions as models.

Some applied problems can be modeled by quadratic functions,which for real num-

bers a, b, and c, can be written in the form

ƒ 1 x 2 ax^2 bxc, with a0.

936 - 600 = 336 ft^2 ,

30 # 20 = 600 ft^2.

130 + 2 # 32120 + 2 # 32 = 36 # 26 =936 ft^2

x= 3 ,

- 28

x=- 28 x= 3

x+ 28 = 0 x - 3 = 0

1 x+ 2821 x- 32 = 0

x^2 + 25 x- 84 = 0

4 x^2 + 100 x- 336 = 0

600 + 100 x+ 4 x^2 - 600 = 336

130 + 2 x 2120 + 2 x 2 - 600 = 336

• =

336 ft^2 ,

30 #20,

130 + 2 x 2120 + 2 x 2 #

30 + 2 x=

20 + 2 x=

x=

ft^2.

EXAMPLE 4

SECTION 9.4 Formulas and Further Applications 525

Grass

x

x

x

x x

x

x

x 20 + 2x

30 + 2x

Pool

20

30

FIGURE 4

NOW TRY
EXERCISE 4
A football practice field is
30 yd wide and 40 yd long.
A strip of grass sod of uniform
width is to be placed around
the perimeter of the practice
field. There is enough money
budgeted for 296 sq yd of sod.
How wide will the strip be?

NOW TRY ANSWER

4. 2yd