Lesson 5
Procedure
Examples
Section 1.4: Functional Models
January 24th, 2014
Lesson 5
Procedure
Examples
In this lesson we will talk about solving word problems. These
types of problem will be common in this course, and developing
a good strategy for solving them will be essential. We will start
by listing steps for solving word problems and finish with
several examples.
Lesson 5
Procedure
Steps for solving a story problem
Examples
1
Read the problem carefully.
2
Identify and label the variables.
3
Draw a picture (if necessary).
4
Write down relationships with the variables.
5
Write down the main equation.
6
Solve.
Lesson 5
Procedure
Examples
Example
The highway department is planning to build a picnic area for
motorists along a major highway. It is to be rectangular with
an area of 3000 square yards and is to be fenced off on the
three sides not adjacent to the highway. Express the number of
yards of fencing required as a function of the length of the
unfenced side.
1. First we make sure we read the problem carefully and
understand what is being asked of us.
2. Secondly, we identify and label any variables. The
problems asks us to write a function in terms the length of
the unfenced side, and so ”the length of the unfenced
side” should be one of our variables.
x = length of the unfenced side (in yards)
Lesson 5
Procedure
2. Considering that we are asked a question about the
perimeter of a rectangle, it is reasonable to left the side
perpendicular to the road also be given by a variable.
Examples
y = length of the perpendicular side (in yards)
3. Since this question is about geometry, a picture will most
likely be helpful. It is very important that we label the
picture as well.
y
x
Lesson 5
Procedure
Examples
4. We now write down any relationships that exist between
the variables and the given constants. We are told that
the area of the region is to be 3000 square yards. This will
allow us to relate x and y , since we have the formula
A = xy for the area of a rectangle.
xy = 3000
5. We are asked to write an equation for the number of yards
of fencing used, so this will be our main equation. This
region will be fenced in by three sides, two of which have
length y and one of which has length x. So our equation is
F = x + 2y
Lesson 5
Procedure
Examples
We don’t know much about solving equations with more than
one variable, so our next step is to turn the previous equation
into an equation with only one variable. Since x is the variable
we care about, we must get rid of the y . We do this by solving
for y in terms of x, based on the relationship we wrote down in
Step 4.
3000
xy = 3000 ⇒ y =
x
Therefore we can write F as a function of x.
3000
6000
F = x + 2y ⇒ F (x) = x + 2
⇒ F (x) = x +
x
x
Since all we are asked to do in this problem is write the
function down, Step 6 is unnecessary and we are done here.
Lesson 5
Example
Procedure
Examples
A closed cylindrical can has surface area 120π square inches.
Express the volume of the can as a function of it radius.
Here we will proceed without such a detailed description of our
steps, but we will still follow those steps.
r = radius of the base, h = height of the cylinder
h
r
Lesson 5
Procedure
Examples
The surface area of a closed cylinder is the sum of the areas of
the top and bottom (which are circles) and the are of the
vertical portion. Notice that the vertical portion is actually a
rectangle with side lengths the height of the cylinder and the
circumference of the base. Thus the surface area S of the
cylinder is
S = 2πrh + 2πr 2 .
In the problem, we are told that the surface area is 120π, and
so we have
120π = 2πrh + 2πr 2 .
Lesson 5
Procedure
The volume of a cylinder is V = πr 2 h. Again, we want this to
be a function of just one variable, in this case r .
Examples
120π = 2πrh + 2πr 2 ⇒ 60 = rh + r 2
⇒ rh = 60 − r 2
60
⇒h=
−r
r
Therefore
V (r ) = πr
2
60
−r
r
= 60πr − πr 3 .
Lesson 5
Example
Procedure
Examples
A closed box with a square base is to have a volume of 1700
cubic inches. Express the surface area as a function of the
length of its base.
x = length of the base, h = height of the box
h
x
x
Lesson 5
Procedure
The volume of the box is given by V = x 2 h (since the base is a
square). Thus we have the relation 1700 = x 2 h. The equation
of for the surface area of the box is
Examples
S = 2x 2 + 4xh.
This comes from the base and top each having area x 2 and
each side having area xh. We must solve for h in terms of x:
x 2 h = 1700 ⇒ h =
1700
x2
Therefore
2
S(x) = 2x + 4x
1700
x2
= 2x 2 +
6800
x