Intermediate Algebra (11th edition)

OBJECTIVE 4 Use quadratic functions to solve problems involving maxi-

mum or minimum value. The vertex of the graph of a quadratic function is either

the highest or the lowest point on the parabola. It provides the following information.

1. The y-value of the vertex gives the maximum or minimum value of y.

2. The x-value tells where the maximum or minimum occurs.

SECTION 9.6 More About Parabolas and Their Applications 545

In many applied problems we must find the greatest or least value of some

quantity. When we can express that quantity in terms of a quadratic function,

the value of kin the vertex 1 h, k 2 gives that optimum value.

PROBLEM-SOLVING HINT

Finding the Maximum Area of a Rectangular Region

A farmer has 120 ft of fencing to enclose a rectangular area next to a building. (See

FIGURE 15.) Find the maximum area he can enclose and the dimensions of the field

when the area is maximized.

EXAMPLE 6

x 120 – 2x

x

FIGURE 15

Let the width of the field.

Sum of the sides is 120 ft.

Combine like terms.

Subtract 2x.

The area is given by the product of the length and width.

Area length width

Distributive property

To determine the maximum area, use the vertex formula to find the vertex of the

parabola given by. Write the equation in standard form.

Then

and.

The graph is a parabola that opens down, and its vertex is Thus, the max-

imum area will be This area will occur if x, the width of the field, is 30 ft

and the length is

120 - 21302 =60 ft. NOW TRY

1800 ft^2.

130 , 18002.

a 1302 =- 213022 + 1201302 =- 219002 + 3600 = 1800

x=

- b

2 a

=

- 120

21 - 22

=

- 120

- 4

= 30 ,

a 1 x 2 =- 2 x^2 + 120 x a=-2, b=120, c= 0

a 1 x 2 = 120 x- 2 x^2

a 1 x 2 = 120 x- 2 x^2

a 1 x 2 = 1120 - 2 x 2 x = #

a 1 x 2

length= 120 - 2 x

2 x+ length= 120

x+x+ length= 120

x=

NOW TRY
EXERCISE 6
Solve the problem in
Example 6if the farmer has
only 80 ft of fencing.

NOW TRY ANSWER

6. The field should be 20 ft
   by 40 ft with maximum
   area 800 f t^2.