1
0.6
Functions and Graphs in Application
2
Section Outline

• Geometric Problems
• Cost, Revenue, and Profit
• Surface Area
• Functions and Graphs

3
Geometric Problems
EXAMPLE
(Fencing a Rectangular Corral) 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
4
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
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
5
Cost Problems
EXAMPLE
(Cost of Fencing) Consider the corral of the
last example. Suppose the fencing for the
boundary of the corral costs 10 per foot and the
fencing for the inner partitions costs 8 per
foot. Write an expression for the total cost of
the fencing.
SOLUTION
This is the diagram we drew to represent the
corral.
y
x
x
x
x
y
Since the boundary of the fence is represented by
the red part of the diagram, the length of
fencing for this portion of the corral is x x
y y 2x 2y. Therefore the cost of fencing
the boundary of the fence is (2x 2y)(cost of
boundary fencing per foot) (2x 2y)(10) 20x
20y.
6
Cost Problems
CONTINUED
Since the inner partitions of the fence are
represented by the blue part of the diagram, the
length of fencing for this portion of the corral
is x x 2x. Therefore the cost of fencing the
inner partitions of the fence is (2x)(cost of
inner partition fencing per foot) (2x)(8) 16x.
Therefore, an expression for the total cost of
the fencing is
(cost of boundary fencing) (cost of inner
partition fencing)
(20x 20y) (16x)
36x 20y
7
Surface Area
EXAMPLE
Assign letters to the dimensions of the geometric
box and then determine an expression representing
the surface area of the box.
SOLUTION
First we assign letters to represent the
dimensions of the box.
z
y
x
8
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 yz yz xz xz
xy 2yz 2xz xy.
9
Cost, Revenue, Profit
EXAMPLE

• (Cost, Revenue, and Profit) An average sale at a
  small florist shop is 21, so the shops weekly
  revenue function is R(x) 21x where x is the
  number of sales in 1 week. The corresponding
  weekly cost is C(x) 9x 800 dollars.
• What is the florist shops weekly profit
  function?
• How much profit is made when sales are at 120
  per week?
• If the profit is 1000 for a week, what is the
  revenue for the week?

SOLUTION
(a) Since Profit Revenue Cost, the profit
function, P(x), would be
P(x) R(x) C(x)
P(x) 21x (9x 800)
P(x) 21x 9x - 800
P(x) 12x - 800
10
Cost, Revenue, Profit
CONTINUED
(b) Since x represents the number of sales in one
week, to determine how much profit is made when
sales are at 120 per week, we will replace x with
120 in the profit function and then evaluate.
P(120) 12(120) - 800
P(120) 1,440 - 800
P(120) 640
Therefore, when sales are at 120 per week, profit
is 640 for that week.
(c) To determine the revenue for the week when
the profit is 1000 for that week, we use an
equation that contains profit, namely the profit
function
P(x) 12x - 800
Now we replace P(x) with 1000 and solve for x.
1000 12x - 800
1800 12x
150 x
Therefore x, the number of units sold in a week,
is 150 when profit is 1000.
11
Cost, Revenue, Profit
CONTINUED
Now, to determine the corresponding revenue, we
replace x with 150 in the revenue function.
R(x) 21x
R(150) 21(150)
R(150) 3,150
Therefore, when profit is 1000 in a week, the
corresponding revenue is 3,150.
NOTE In order to determine the desired revenue
value in part (c), we needed to solve for R(x).
But in order to do that, we needed to have a
value for x to plug into the R(x) function. In
order to acquire that value for x, we needed to
use the given information profit is 1000.
12
Functions Graphs
EXAMPLE

• The function f (r) gives the cost (in cents) of
  constructing a 100-cubic-inch cylinder of radius
  r inches. The graph of f (r) is given.
• What is the cost of constructing a cylinder of
  radius 6 inches?
• Interpret the fact that the point (3, 162) is on
  the graph of the function.
• Interpret the fact that the point (3, 162) is
  the lowest point on the graph of the function.
  What does this say in terms of cost versus
  radius?

13
Functions Graphs
CONTINUED
SOLUTION
To determine the cost of constructing a cylinder
of radius 6 inches, we look on the graph where r
6. The corresponding y value will be the cost
we are seeking.
The red arrow is emphasizing the point in which
we are interested. The y value of that point is
270. Therefore, the cost of constructing a
cylinder of radius 6 inches is 270 cents or 2.70.
14
Functions Graphs
CONTINUED
(b) The fact that the point (3, 162) is on the
graph tells us that the cost to make
100-cubic-inch cylinders with a radius as small
as 3 inches is 162 cents or 1.62.
(c) The fact that the point (3, 162) is the
lowest point on the graph tells us that the least
expensive 100-cubic-inch cylinder that can be
made is a 3 inch cylinder at a cost of 1.62.
Therefore, the 3 inch cylinder is the most
cost-effective one that is offered.