Geometry Problems To solve these applications, we continue to use a six-step problem-solving strategy.
■ Geometry Problems
■ The Pythagorean Theorem
■ Height of a Projected Object
■ Modeling with Quadratic Equations
5
5
5
5 5
5
5
5 3x – 10
x – 10 x
3x
Figure 5
Lial: College Algebra Fig: 7821_01_FG006 First Pass: 2011-04-29
5 x – 10 3x – 10
Figure 6
EXAMPLE 1 Solving a Problem Involving Volume
A piece of machinery produces rectangular sheets of metal such that the length is three times the width. Equal-sized squares measuring 5 in. on a side can be cut from the corners so that the resulting piece of metal can be shaped into an open box by folding up the flaps. If specifications call for the volume of the box to be 1435 in.3, find the dimensions of the original piece of metal.
SOLUTION
Step 1 Read the problem. We must find the dimensions of the original piece of metal.
Step 2 Assign a variable. We know that the length is three times the width.
Let x =the width (in inches) and thus, 3x = the length.
The box is formed by cutting 5 +5 = 10 in. from both the length and the width. See Figure 5. The width of the bottom of the box is x- 10, the length of the bottom of the box is 3x- 10, and the height is 5 in. (the length of the side of each cut-out square). See Figure 6.
Step 3 Write an equation. The formula for volume of a box is V= lwh.
Volume = length *width *height
1435 = 13x -1021x - 102152
(Note that the dimensions of the box must be positive numbers, so 3x- 10 and x - 10 must be greater than 0, which implies x7 103 and x 7 10. These are both satisfied when x 7 10.)
Step 4 Solve the equation from Step 3.
1435= 15x2 - 200x + 500 Multiply.
0 = 15x2 - 200x - 935 Subtract 1435 from each side.
0 = 3x2- 40x - 187 Divide each side by 5.
0 = 13x+ 1121x- 172 Factor.
3x + 11= 0 or x- 17= 0 Zero-factor property x = -11
3 or x = 17 Solve each equation.
Step 5 State the answer. Only 17 satisfies the restriction x710. Thus, the dimensions of the original piece should be 17 in. by 31172= 51 in.
Step 6 Check. The length and width of the bottom of the box are 51 - 2152 = 41 in. Length and 17 - 2152 = 7 in. Width
The height is 5 in. (the amount cut on each corner), so the volume is V =lwh =41 *7 * 5= 1435 in.3, as required.
■✔ Now Try Exercise 27.
The width cannot be negative.
151
1.5 Applications and Modeling with Quadratic Equations
The Pythagorean Theorem Example 2 requires the use of the Pythag- orean theorem for right triangles. Recall that the legs of a right triangle form the right angle, and the hypotenuse is the side opposite the right angle.
Pythagorean Theorem
In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
a2 + b2 = c2 Leg b
Hypotenuse Leg a c
PROBLEM-SOLVING HINT As seen in Example 1, discard any solu- tion that does not satisfy the physical constraints of a problem.
2x + 30 x
2x + 20 x is in meters.
Figure 7
EXAMPLE 2 Applying the Pythagorean Theorem
A piece of property has the shape of a right triangle. The longer leg is 20 m longer than twice the length of the shorter leg. The hypotenuse is 10 m longer than the length of the longer leg. Find the lengths of the sides of the triangular lot.
SOLUTION
Step 1 Read the problem. We must find the lengths of the three sides.
Step 2 Assign a variable.
Let x = the length of the shorter leg (in meters).
Then 2x+ 20= the length of the longer leg, and
12x+ 202 + 10, or 2x+ 30= the length of the hypotenuse.
See Figure 7. Step 3 Write an equation.
a2 + b2 = c2
x2 + 12x +2022 = 12x+ 3022 Substitute into the Pythagorean theorem.
The hypotenuse is c.
Step 4 Solve the equation.
x2+ 14x2 +80x+ 4002= 4x2 +120x +900 Square the binomials.
Remember the middle terms.
x2 - 40x - 500= 0 Standard form
1x -5021x +102 = 0 Factor.
x- 50= 0 or x + 10= 0 Zero-factor property x= 50 or x= -10 Solve each equation.
Step 5 State the answer. Because x represents a length, -10 is not reasonable.
The lengths of the sides of the triangular lot are
50 m, 21502+ 20= 120 m, and 21502+ 30 =130 m.
Step 6 Check. The lengths 50, 120, and 130 satisfy the words of the problem and also satisfy the Pythagorean theorem.
■✔ Now Try Exercise 35.
Height of a Projected Object If air resistance is neglected, the height s (in feet) of an object projected directly upward from an initial height of s0 feet, with initial velocity v0 feet per second, is given by the following equation.
s = −16 t2 +v0 t + s0
Here t represents the number of seconds after the object is projected. The coef- ficient of t2, -16, is a constant based on the gravitational force of Earth. This constant varies on other surfaces, such as the moon and other planets.
EXAMPLE 3 Solving a Problem Involving Projectile Height
If a projectile is launched vertically upward from the ground with an initial velocity of 100 ft per sec, neglecting air resistance, its height s (in feet) above the ground t seconds after projection is given by
s = -16t2 + 100t.
(a) After how many seconds will it be 50 ft above the ground?
(b) How long will it take for the projectile to return to the ground?
SOLUTION
(a) We must find value(s) of t so that height s is 50 ft.
s = -16t2 + 100t
50= -16t2 + 100t Let s=50.
0 = -16t2 + 100t -50 Standard form
0 = 8t2 - 50t+ 25 Divide by -2.
t = -b{ 2b2 -4ac
2a Quadratic formula
t = -1-502 { 21-5022 -41821252 2182
Substitute a=8, b= -50, and c=25.
t = 50{ 21700
16 Simplify.
t ≈ 0.55 or t≈ 5.70 Use a calculator.
Both solutions are acceptable. The projectile reaches 50 ft twice—once on its way up (after 0.55 sec) and once on its way down (after 5.70 sec).
Substitute carefully.
(b) When the projectile returns to the ground, the height s will be 0 ft.
s = -16t2 + 100t
0 = -16t2 + 100t Let s=0.
0 = -4t14t - 252 Factor.
-4t =0 or 4t -25 =0 Zero-factor property t =0 or t =6.25 Solve each equation.
The first solution, 0, represents the time at which the projectile was on the ground prior to being launched, so it does not answer the question. The pro- jectile will return to the ground 6.25 sec after it is launched.
■✔ Now Try Exercise 47.
Galileo Galilei (1564–1642) According to legend, Galileo dropped objects of different weights from the Leaning Tower of Pisa to disprove the Aristotelian view that heavier objects fall faster than lighter objects. He developed the formula d=16t2 for freely falling objects, where d is the distance in feet that an object falls (neglecting air resistance) in t seconds, regardless of weight.
153
1.5 Applications and Modeling with Quadratic Equations
LOOKING AHEAD TO CALCULUS In calculus, you will need to be able to write an algebraic expression from the description in a problem like those in this section. Using calculus techniques, you will be asked to find the value of the variable that produces an optimum (a maximum or minimum) value of the expression.
Modeling with Quadratic Equations
EXAMPLE 4 Analyzing Trolley Ridership The I-Ride Trolley service car-
ries passengers along the Inter- national Drive resort area of Orlando, Florida. The bar graph in Figure 8 shows I-Ride Trolley ridership data in millions. The quadratic equation
y = -0.00525x2 +0.0913x + 1.64 models ridership from 2000 to 2013, where y represents rider- ship in millions, and x= 0 rep- resents 2000, x= 1 represents 2001, and so on.
(a) Use the model to determine ridership in 2011. Compare the result to the actual ridership figure of 2.1 million.
(b) According to the model, in what year did ridership reach 1.8 million?
SOLUTION
(a) Because x= 0 represents the year 2000, use x= 11 to represent 2011.
y= -0.00525x2 + 0.0913x + 1.64 Given model y= -0.0052511122 + 0.09131112 +1.64 Let x=11.
y≈2.0 million Use a calculator.
The prediction is about 0.1 million (that is, 100,000) less than the actual figure of 2.1 million.
(b) y= -0.00525x2+ 0.0913x + 1.64 Given model 1.8= -0.00525x2+ 0.0913x + 1.64 Let y=1.8.
0= -0.00525x2+ 0.0913x - 0.16 Standard form x= -0.0913{210.091322 - 41-0.0052521-0.162
21-0.005252 Quadratic formula
x≈2.0 or x≈15.4 Use a calculator.
0 0.5 1.0 1.5 2.0
Millions
’00 ’01 ’02 ’03 ’04 ’05 ’06 ’07’08’09 ’10 Year
Source: I-Ride Trolley, International Drive Master Transit, www.itrolley.com
Ridership
’12 ’13
’11
Figure 8
Solve this equation
for x.
The year 2002 corresponds to x= 2.0. Thus, according to the model, rider- ship reached 1.8 million in the year 2002. This outcome closely matches the bar graph and seems reasonable.
The year 2015 corresponds to x= 15.4. Round down to the year 2015 because 15.4 yr from 2000 occurs during 2015. There is no value on the bar graph to compare this to, because the last data value is for the year 2013.
Always view results that are beyond the data in a model with skepticism, and realistically consider whether the model will continue as given. The model predicts that ridership will be 1.8 million again in the year 2015.
■✔ Now Try Exercise 49.
CONCEPT PREVIEW Answer each question.
1. Area of a Parking Lot For the rectangular parking area of the shopping center shown, with x in yards, which one of the following equations says that the area is 40,000 yd2?
2x + 200
x
A. x12x+ 2002= 40,000 B. 2x +212x +2002 =40,000 C. x+12x+2002=40,000 D. x2+12x+20022=40,0002 2. Diagonal of a Rectangle If a rectangle is r feet long and
s feet wide, which expression represents the length of its diagonal in terms of r and s?
A. 2rs B. r+ s C. 2r2+s2 D. r2+ s2 3. Sides of a Right Triangle To solve for the lengths of the
right triangle sides, which equation is correct?
A. x2=12x- 222+ 1x+ 422 B. x2+1x +422= 12x- 222 C. x2=12x- 222- 1x+ 422 D. x2+ 12x- 222=1x+422 4. Area of a Picture The mat and frame
around the picture shown measure x inches across. Which equation says that the area of the picture itself is 600 in.2?
A. 2134-2x2 +2121-2x2 =600 B. 134-2x2121- 2x2= 600 C. 134 -x2121- x2= 600 D. x13421212= 600
5. Volume of a Box A rectangular piece of metal is 5 in. longer than it is wide. Squares with sides 2 in. long are cut from the four corners, and the flaps are folded upward to form an open box. Which equation indicates that the volume of the box is 64 in.3?
2
2
2
2 2
2
2
2 x
x + 5
2 x – 4 x + 1
A. 1x +121x- 42122= 64 B. x1x +52122= 64 C. 1x +121x- 42= 64 D. x1x +52 =64