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Applied Calculus
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 1 of 78
Course Contents

Functions

Derivatives

Applications of Derivative

Techniques of Differentiation

Logarithmic Functions and Applications

The Definite Integrals

The Trigonometric Functions
© 2010 Pearson Education Inc.
…
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 2 of 78
Chapter 0
Functions
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 3 of 78
Chapter Outline

Functions and Their Graphs

Some Important Functions

The Algebra of Functions

Zeros of Functions

The Quadratic Formula and Factoring

Exponents and Power Functions

Functions and Graphs in Applications
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 4 of 78
Rational & Irrational Numbers
Definition
Rational Number: A number
that may be written as a finite
or infinite repeating decimal,
in other words, a number that
can be written in the form m/n
such that m, n are integers
Irrational Number: A number
that has an infinite decimal
representation whose digits
form no repeating pattern
© 2010 Pearson Education Inc.
Example

2
 0.2857142857 14...
7
 0.285714
3  1.73205
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 5 of 78
The Number Line
The Number Line
A geometric representation of the real numbers is shown below.

-6
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-5
-4
-3
-2
-1
2
7
0
3
1
2
3
4
5
6
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 6 of 78
Open & Closed Intervals
Definition
Example
Infinite Interval: The set of
numbers that lie between a
given endpoint and the
infinity
Closed Interval: The set of
numbers that lie between
two given endpoints,
including the endpoints
themselves
Open Interval: The set of
numbers that lie between
two given endpoints, not
including the endpoints
themselves
© 2010 Pearson Education Inc.
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
2
3
4
5
6
3
4
5
6
4, 
x4
-6
-5
-4
-3
-2
-1
0
1
[−1, 4]
1  x  4
-6
-5
-4
-3
-2
-1
0
1
2
(−1, 4)
1  x  4
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 7 of 78
Functions
• A function f is a rule that assigns to each value of a real variable x exactly one
value of another real variable y.
• The variable x is called the independent variable and the variable y is called the
dependent variable.
• We usually write y = f (x) to express the fact that y is a function of x. Here f (x) is
the name of the function.
x
y = f (x)
y
EXAMPLES:
f x   x 2  4
1
g x  
x 3
h x   x 2  1
k t   e  2t  sin 2 t 
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 8 of 78
Functions in Application
EXAMPLE
When a solution of acetylcholine is introduced into the heart muscle of a frog, it
diminishes the force with which the muscle contracts. The data from experiments of
the biologist A. J. Clark are closely approximated by a function of the form
R x  
100 x
b x
where x is the concentration of acetylcholine (in appropriate units), b is a positive
constant that depends on the particular frog, and R(x) is the response of the muscle to
the acetylcholine, expressed as a percentage of the maximum possible effect of the
drug.
(a) Suppose that b = 20. Find the response of the muscle when x = 60.
(b) Determine the value of b if R(50) = 60 – that is, if a concentration of x = 50 units
produces a R = 60% response.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 9 of 78
Functions in Application
SOLUTION
R60  
100  60
 75
20  60
Replace b with 20 and x with 60.
Therefore, when b = 20 and x = 60, R (x) = 75%.
100 x
b x
100  50
R50  
b  50
100  50
60 
b  50
b  50   60  5000  b  50 
b  50
60b  3000  5000
(b)
R x  
This is the given function.
Replace x with 50.
Replace R(50) with 60.
Multiply both sides by b + 50.
Distribute on the left side.
60b  2000
Subtract 3000 from both sides.
b  33.3
Divide both sides by 60.
Therefore, b = 33.3 when R (50) = 60.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 10 of 78
Functions
EXAMPLE
If f (x) = x2 + 4x + 3, find f (a − 2).
SOLUTION
f x   x 2  4 x  3
This is the given function.
f a  2  a  2  4a  2  3
Replace each x with a – 2.
f a  2  a2  4a  4  4a  2  3
Evaluate (a – 2)2 = a2 – 4a + 4.
f a  2  a2  4a  4  4a  8  3
Remove parentheses and distribute.
2
f a  2  a 2  1
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Combine like terms.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 11 of 78
Domain of a Function
Definition
Domain of a Function:
The set of acceptable
values for the variable x.
Example
The domain of the function
f x   x  2
is
x20
x  2
[- 2 ,)
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 12 of 78
Domain of a Function
Definition
Example
Domain of a Function: The The domain of the function
2x
set of acceptable values for
f x   2
x 4
the variable x.
is
x2  4  0
x  2
x  R \  2
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 13 of 78
Domain of a Function
Definition
Example
The domain of the function
Domain of a Function:
The set of acceptable
x2
f x  
values for the variable x.
3 x
is
3 x  0
x 3
( ,3)
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 14 of 78
Graphs of Functions
Definition
Example
Graph of a Function: The set of
all points (x, f (x)) where x is
the domain of f (x). Generally,
this forms a curve in the xyplane.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 15 of 78
The Vertical Line Test
Definition
Example
Vertical Line Test: A curve
in the xy-plane is the graph
of a function if and only if
each vertical line cuts or
touches the curve at no
more than one point.
Although the red line intersects
the graph no more than once (not
at all in this case), there does
exist a line (the yellow line) that
intersects the graph more than
once. Therefore, this is not the
graph of a function.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 16 of 78
Graphs of Equations
EXAMPLE
1



f
x

x


 x  2 ?
Is the point (3, 12) on the graph of the function
2

SOLUTION
1

f x    x   x  2
2

This is the given function.
1

f 3   3  3  2
2

Replace x with 3.
1

12   3  3  2
2

Replace f (3) with 12.
12  2.55
Simplify.
12  12.5 false
Multiply.
Since replacing x with 3 and f (x) with 12 did not yield a true statement in the original
function, we conclude that the point (3, 12) is not on the graph of the function.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 17 of 78
Linear Equations
Equation
Example
y = mx + b
(This is a linear function)
x=a
(This is not the graph of a
function)
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 18 of 78
Linear Equations
CONTINUED
Equation
Example
y=b
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 19 of 78
Piece-Wise Functions
EXAMPLE
1  x for x  3
Sketch the graph of the following function f x   
.
2
for
x

3

SOLUTION
We graph the function f (x) = 1 + x only for those values of x that are less than or equal
to 3.
6
4
2
0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
-2
-4
-6
Notice that for all values of x greater than 3, there is no line.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 20 of 78
Piece-Wise Functions
CONTINUED
Now we graph the function f (x) = 2 only for those values of x that are greater than 3.
6
5
4
3
2
1
0
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Notice that for all values of x less than or equal to 3, there is no line.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 21 of 78
Piece-Wise Functions
CONTINUED
Now we graph both functions on the same set of axes.
-6
© 2010 Pearson Education Inc.
-5
-4
-3
-2
6
5
4
3
2
1
0
-1 -1 0
-2
-3
-4
-5
-6
1
2
3
4
5
6
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 22 of 78
Quadratic Functions
Definition
Example
Quadratic Function:
A function of the form
f x   ax 2  bx  c
where a, b, and c are
constants and a  0.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 23 of 78
Polynomial Functions
Definition
Example
Polynomial Function: A
function of the form
f x   an x n  an 1 x n 1    a0
f x   17 x3  x 2  5
where n is a nonnegative
integer and a0, a1, ..., an are
given numbers.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 24 of 78
Rational Functions
Definition
Rational Function: A
function expressed as the
quotient of two
polynomials.
© 2010 Pearson Education Inc.
Example
3x  x 4
g x   2
5x  x  1
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 25 of 78
Power Functions
Definition
Power Function: A
function of the form
f x   x r .
© 2010 Pearson Education Inc.
Example
f x   x5.2
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 26 of 78
Absolute Value Function
Definition
Example
Absolute Value Function:
The function defined for all
f x   x
numbers x by
f x   x ,
f  1 2   1 2  1 2
such that |x| is understood to
be x if x is positive and –x if
x is negative
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 27 of 78
Adding Functions
EXAMPLE
Given f  x  
2
and
x3
g x  
1
, express f (x) + g(x) as a rational function.
x2
SOLUTION
f (x) + g(x) =
2
1


x3 x 2
x2 2
1 x3




x 2 x3 x 2 x3

2x  4
x 3

x  2x  3 x  2x  3
2x  4  x  3

x  2x  3

3x  1
x2  x  6
© 2010 Pearson Education Inc.

3x  1
x  2x  3
Replace f (x) and g(x) with the given
functions.
Multiply to get common
denominators.
Evaluate.
Add and simplify the numerator.
Evaluate the denominator.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 28 of 78
Subtracting Functions
EXAMPLE
f x  
Given
2
and
x3
g x  
1
, express f (x) − g(x) as a rational function.
x2
SOLUTION
f (x) − g(x) =
2
1


x3 x 2
x2 2
1 x3




x 2 x3 x 2 x3

2x  4
x 3

x  2x  3 x  2x  3

2 x  4  x  3
x  2x  3

x7
x7
 2
x  2x  3 x  x  6
© 2010 Pearson Education Inc.
Replace f (x) and g(x) with the given
functions.
Multiply to get common
denominators.
Evaluate.
Subtract.
Simplify the numerator and
denominator.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 29 of 78
Multiplying Functions
EXAMPLE
Given f  x  
2
1
and g  x  
, express f (x)g(x) as a rational function.
x3
x2
SOLUTION
f (x)g(x) =

2
1

x3 x 2
Replace f (x) and g(x) with the given
functions.

2 1
x  3x  2
Multiply the numerators and
denominators.

2
x2  x  6
© 2010 Pearson Education Inc.
Evaluate.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 30 of 78
Dividing Functions
EXAMPLE
Given f  x  
2
1
and g  x  
, express
x3
x2
f x 
as a rational function.
g x 
SOLUTION
f x 

g x 
2
 x3
1
x2

2 x2

x3 1
2 x  2
x  31
2x  4

x3

© 2010 Pearson Education Inc.
Replace f (x) and g(x) with the given
functions.
Rewrite as a product (multiply by
reciprocal of denominator).
Multiply the numerators and
denominators.
Evaluate.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 31 of 78
Composition of Functions
EXAMPLE
Table 1 shows a conversion table for men’s hat sizes for three countries. The function
1
g  x   8 x  1 converts from British sizes to French sizes, and the function f  x   x
8
converts from French sizes to U.S. sizes. Determine the function h (x) = f (g (x)) and
give its interpretation.
SOLUTION
h (x) = f (g (x))
This is what we will determine.
1
  g x 
8
In the function f, replace each
occurrence of x with g (x).
1
  8 x  1
8
Replace g (x) with 8x + 1.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 32 of 78
Composition of Functions
CONTINUED
1
1
  8x  1
8
8
1
 x
8
Distribute.
Multiply.
Therefore, h (x) = f (g (x)) = x + 1/8. Now to determine what this function h (x) means,
we must recognize that if we plug a number into the function, we may first evaluate
that number plugged into the function g (x). Upon evaluating this, we move on and
evaluate that result in the function f (x). This is illustrated as follows.
g (x)
British French
f (x)
French
U.S.
h (x)
Therefore, the function h (x) converts a men’s British hat size to a men’s U.S. hat size.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 33 of 78
Composition of Functions
EXAMPLE
Given f  x  
2
1


g
x

and
, find f (g (x)).
x3
x2
SOLUTION
f (g (x)) =

2
g x   3

2
1
3
x2
2
x2


1
3 x 2
x2
2x  4

 3x  5
© 2010 Pearson Education Inc.
Replace x by g(x) in the function
f (x)
Substitute.
Multiply the numerators and
denominators by x + 2.
Simplify.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 34 of 78
Zeros of Functions
Definition
Zero of a Function: For a
function f (x), all values of x
such that f (x) = 0.
© 2010 Pearson Education Inc.
Example
f x   x 2  1
0  x2 1
x  1
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 35 of 78
Quadratic Formula
Definition
Quadratic Formula: A
formula for solving any
quadratic equation of the
form ax 2  bx  c  0 .
The solution is:
 b  b  4ac
x
.
2a
2
There is no solution if
b 2  4ac  0.
© 2010 Pearson Education Inc.
Example
x 2  3x  2  0
a  1; b  3; c  2
x
x
 3 
32  41 2
21
 3  17
2
These are the
solutions/zeros of the
quadratic function
f x   x 2  3x  2.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 36 of 78
Graphs of Intersecting Functions
EXAMPLE
Find the points of intersection of the pair of curves.
y  x2  10 x  9; y  x  9
SOLUTION
The graphs of the two equations can be seen to intersect in the following graph. We can
use this graph to help us to know whether our final answer is correct.
100
80
60
40
20
0
-5
0
5
10
15
-20
-40
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 37 of 78
Graphs of Intersecting Functions
CONTINUED
To determine the intersection points, set the equations equal to each other, since they
both equal the same thing: y.
x2  10 x  9  x  9
Now we solve the equation for x using the quadratic formula.
x2  10 x  9  x  9
This is the equation to solve.
x2  11x  9  9
Subtract x from both sides.
x2  11x  18  0
Add 9 to both sides.
Here, a = 1, b = −11, and c =18.
x
  11 
x
 11
21
11  121  72
2
© 2010 Pearson Education Inc.
2
 4118
Use the quadratic formula.
Simplify.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 38 of 78
Graphs of Intersecting Functions
CONTINUED
x
11  49
2
Simplify.
11 7
2
11  7 11  7
x
,
2
2
x
Simplify.
Rewrite.
x  9, 2
Simplify.
We now find the corresponding y-coordinates for x = 9 and x = 2. We can use either of
the original equations. Let’s use y = x – 9.
x9
x2
y  x 9
y 99
y  x 9
y  29
y0
y  7
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 39 of 78
Graphs of Intersecting Functions
CONTINUED
Therefore the solutions are (9, 0) and (2, −7). This seems consistent with the two
intersection points on the graph. A zoomed in version of the graph follows.
10
5
0
0
2
4
6
8
10
-5
-10
-15
-20
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 40 of 78
Factoring
EXAMPLE
Factor the following quadratic polynomial.
6 x  2 x3
SOLUTION
6 x  2 x3
2 x3  x 2 

2
2 x 3  x2
This is the given polynomial.
Factor 2x out of each term.

2
Rewrite 3 as 3 .
Now I can use the factorization pattern: a2 – b2 = (a – b)(a + b).

2x 3  x

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3x

Rewrite
2
3  x 2 as

3x


3x .
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 41 of 78
Factoring
EXAMPLE
Solve the equation for x.
1
SOLUTION
5 6
 2
x x
5 6 
x 2 1    2  x 2
x x 
1
5 2 6 2
x  2 x
x
x
x 2  5x  6
x2 
x 2  5x  6  0
x  1x  6  0
x 1  0
x  1
x60
x6
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5 6

x x2
This is the given equation.
Multiply everything by the LCD: x2.
Distribute.
Multiply.
Subtract 5x + 6 from both sides.
Factor.
Set each factor equal to zero.
Solve.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 42 of 78
Exponents
Definition
Example
b n  b
 b



b
53  5  5  5
n times
1
n
b  b
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n
1
3
5 3 5
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 43 of 78
Exponents
Definition
m
n
b  b 
b

m
n
n
1
m
Example
 b
n
1
 m n m 
b
bn
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3
4
5  5 
m
1
 b
n
m
3
4
4
 5
4
1
5  3 4 3 
5
54

1
3
3
1
 5
4
3
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 44 of 78
Exponents
Definition
b b  b
r
s
1
b r  r
b
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r s
Example
1
3
2
3
6 6  6
1
2
1 2

3 3
3
3
 6  61  6
1 1
4  1 

4 2
42

1
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 45 of 78
Exponents
Definition
Example
br
r s

b
bs
b 
r s
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b
rs
7
7
4
3
1
3
7
4 1

3 3
3
3
 7  71  7
5
8
45
4
1

 
 9   95 8  98  92  9  3
 
4
5
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 46 of 78
Exponents
Definition
Example
125  27 
1/ 3
ab
r
 a r  br
r
a a
   r
b b
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r
 1251 / 3  271 / 3
 3 125  3 27  5  3  15
4
10  10 
4




2
 16


4
5
5
4
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 47 of 78
Applications of Exponents
EXAMPLE
Use the laws of exponents to simplify the algebraic expression.
 27 x 
5 2/ 3
3
x
SOLUTION
 27 x 
5 2/ 3
3
x
This is the given expression.
 x 5 
x
 27 
2/ 3
2/ 3
3
 27 
2/ 3
 x 5 
2/ 3
1/ 3
x
 27 
2/ 3
 x10 / 3
1/ 3

3
abr  a r  b r
x
2
 27  x10 / 3
x1 / 3

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1
n
b n b
b 
r s
m
n
 b rs
b  n bm 
 b
n
m
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 48 of 78
Applications of Exponents
CONTINUED
 3
2
 x10 / 3
1/ 3
3
 27  3
x
10 / 3
9x
x1 / 3
 32  9
9x
br
 br s
s
b
9 x9 / 3
Subtract.
9x3
Divide.
10 / 31 / 3
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 49 of 78
Geometric Problems
EXAMPLE
Consider a rectangular corral with two partitions, as shown below. Assign letters to the
outside dimensions of the corral. Write an equation expressing the fact that the corral has
a total area of 2500 square feet. Write an expression for the amount of fencing needed to
construct the corral (including both partitions).
SOLUTION
First we will assign letters to represent the dimensions of the corral.
y
x
x
x
x
y
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 50 of 78
Geometric Problems
CONTINUED
Now we write an equation expressing the fact that the corral has a total area of 2500
square feet. Since the corral is a rectangle with outside dimensions x and y, the area of
the corral is represented by:
A  xy
Now we write an expression for the amount of fencing needed to construct the corral
(including both partitions). To determine how much fencing will be needed, we add
together the lengths of all the sides of the corral (including the partitions). This is
represented by:
F  xxxx y y
F  4x  2 y
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 51 of 78
Surface Area
EXAMPLE
Assign letters to the dimensions of the geometric box and then determine an expression
representing the volume and the surface area of the box.
SOLUTION
First we assign letters to represent the dimensions of the box.
z
x
y
Therefore, an expression that represents the volume is:
V = xyz.
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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 52 of 78
Surface Area
CONTINUED
z
y
x
Now we determine an expression for the surface area of the box. Note, the box has 5
sides which we will call Left (L), Right (R), Front (F), Back (B), and Bottom (Bo). We
will find the area of each side, one at a time, and then add them all up.
L: yz
R: yz
F: xz
B: xz
Bo: xy
Therefore, an expression that represents the surface area of the box is:
S = yz + yz + xz + xz + xy = 2yz + 2xz + xy.
© 2010 Pearson Education Inc.
Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 53 of 78