Each of the rectangles shown has an area of 12 square units. Notice that as the length increases, the width decreases. As the length decreases, the width increases. However, their product stays the same.
This is an example of an inverse variation. We say that y varies inversely
as x. This means that as x increases in value, y decreases
in value, or as y decreases in value, x increases in value.
The graphs below show how the graph of a direct variation
differs from the graph of an inverse variation.
Example
Construction Link
1. The number of carpenters needed to frame a house varies inversely as the number
of days needed to complete the project. Suppose 5 carpenters can frame a house
in 16 days. How many days will 8 carpenters take to frame the house? Assume
that they all work at the same rate.
Alternative Solutions :
Explore
You know that it takes 16 days
for 5 carpenters to
frame the house. You need to know how many days
it will take 8 carpenters to frame the house.
Plan Solve the problem by using
inverse variation.
Solve Let x the number of carpenters. Let y the number of
days. First, find the value of k.
xy = k Definition
of inverse variation
(5)(16) = k Replace x with 5 and y with 16.
80 = k The constant of variation is 80.
Next, find the number of days for 8 carpenters to
Frame the house.
A crew of 8 carpenters can frame the house in 10 days.
Examine
It takes 16 days for 5
carpenters to frame a house. It
makes sense that 8 carpenters could do the work in less
time.
10 days is a reasonable solution.
Just as with direct variation, you can use a proportion to
solve problems involving inverse variation. The proportion 
is only one of
several that can be formed. Can you name others?
Example
2. Suppose y varies inversely as x and y = –6 when x = –2. Find y when x = 3.
Alternative Solutions :
Inverse variations can be used to solve rate problems.
Example
Sports Link
3. In the formula d = rt, the time t varies
inversely as the rate r. A race car traveling 125 miles per hour
completed one lap around a race track in 1.2 minutes. How fast was the car
traveling if it completed the next lap in 0.8 minute?
Alternative Solutions :
First, solve for the
distance d, the constant of variation.
d =
rt Formula
for distance
= (125)
(1.2) Replace
r with 125 and t with 1.2.
= 150 The
constant of variation is 150
Next,
find the rate if one lap was completed at 0.8 minute.
Sumber
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