### Video Transcript

In this video, we will learn how to
solve applications of systems of inequalities by translating each condition into an
inequality. A system of linear inequalities is
a set of two or more linear inequalities in several variables. These are often used when a problem
requires a range of solutions and there is more than one constraint on the
solutions, for example, a shop trying to buy stock with a given budget. Letβs begin by looking in more
detail how we can represent these.

For example, if we had the system
of inequalities π₯ is greater than or equal to two, π¦ is greater than or equal to
four, and two π₯ plus three π¦ is less than or equal to 24, we can represent this on
the two-dimensional coordinate plane. The equation π₯ equals two can be
represented by a vertical line passing through two on the π₯-axis. As the given inequality is π₯ is
greater than or equal to two, the area that we require is to the right of this
line. We can therefore shade out anything
to the left of this line.

It is important to note that had we
had a strict inequality, such as π₯ is greater than two, then this would be
represented by a dashed line. The equation π¦ is equal to four is
represented by a horizontal line passing through the π¦-axis at four. As the inequality was π¦ is greater
than or equal to four, the required area is above this line. And once again we can shade out the
area below. Finally, we need to draw the
straight line two π₯ plus three π¦ equals 24. This will intersect the π¦-axis
when π₯ equals zero, and this occurs when π¦ is equal to eight. Likewise, the line will intersect
the π₯-axis when π¦ is equal to zero, and this occurs when π₯ equals 12. The equation two π₯ plus three π¦
equals 24 can be represented on the graph as shown.

To work out which side of this line
we require, we can rearrange our inequality to make π¦ the subject. Firstly, we can subtract two π₯
from both sides. We can then divide through by three
such that π¦ is less than or equal to negative two-thirds π₯ plus eight. As π¦ is less than or equal to
negative two-thirds π₯ plus eight, the area required is below the line. We can therefore shade out the area
above this line. This leaves us with a triangular
region that satisfies the system of inequalities.

Once again, it is important to note
that we only include the values at the edges of intersection of the regions if there
is a solid line on both. This is because all the
inequalities need to be satisfied and a strict inequality excludes it from the
solution set.

We will now consider some examples
where we obtain a system of inequalities from a word problem.

A shepherd wants to build a
rectangular sheep barn. The length of the barn must be more
than 88 meters, and its perimeter must be less than 253 meters. Derive the system of inequalities
that describes the situation, denoting the length of the barn by π₯ and its width by
π¦.

In this question, we are told that
a shepherd wants to build a rectangular sheep barn with length π₯ meters and width
π¦ meters. We need to state a system of
inequalities that satisfy the conditions given. We are told that the length of the
barn must be more than 88 meters. Therefore, π₯ must be greater than
88. Since the width of the barn cannot
be a negative value, we have π¦ is greater than or equal to zero.

Weβre also told that the perimeter
of the barn must be less than 253 meters. Recalling that the perimeter of any
two-dimensional shape is the distance around the outside, then this is equal to π₯
plus π¦ plus π₯ plus π¦. This simplifies to two π₯ plus two
π¦. And factoring out a two, we have
the perimeter of the barn is equal to two multiplied by π₯ plus π¦. This perimeter must be less than
253 meters. Therefore, two multiplied by π₯
plus π¦ is less than 253. We now have a system of three
inequalities that describe the situation. The length π₯ must be greater than
88. The width π¦ must be greater than
or equal to zero. And two multiplied by π₯ plus π¦ is
less than 253.

Whilst it is not required in this
question or indeed this video, we could represent this graphically. We can draw a solid line at π¦
equals zero and dashed lines at π₯ equals 88 and two multiplied by π₯ plus π¦ equals
253. As π₯ is greater than 88, we can
shade out the region to the left of this, as we require the region to the right. As π¦ is greater than or equal to
zero, the region required is above the π₯-axis. And finally, as two multiplied by
π₯ plus π¦ is less than 253, we require the region below this line. This gives us a triangular region
that satisfies all three of the inequalities.

Letβs now consider another example
in context.

A carpenter wants to buy two types
of nails. The first type costs six pounds per
kilogram, and the second type costs nine pounds per kilogram. He needs at least five kilograms of
the first type and at least seven kilograms of the second. He can spend less than 55
pounds. Using π₯ to represent the amount of
the first type and π¦ to represent the second type, state the system of inequalities
that represents this situation.

In this question, we need to state
the system of inequalities that satisfy the conditions for a carpenter who wants to
purchase two types of nails. We will let π₯ represent the amount
of the first type and π¦ represent the amount of the second type. This will be the amount of nails in
kilograms. Since he needs at least five
kilograms of the first type, we know that π₯ is greater than or equal to five. He also needs at least seven
kilograms of the second type, so π¦ is greater than or equal to seven.

The other constraint here is the
cost. We are told that the first type
costs six pounds per kilogram. This is equivalent to six π₯. The second type of nail costs nine
pounds per kilogram, and this is equivalent to nine π¦. As the total amount he can spend
must be less than 55 pounds, we know that the sum of these must be less than 55. The system of inequalities that
represents the situation are π₯ is greater than or equal to five, π¦ is greater than
or equal to seven, and six π₯ plus nine π¦ is less than 55.

In our next two questions, we will
look at more complicated problems where there are more constraints.

A teacher gave his students 100
minutes to solve a test that has two sections: section A and section B. The students had to answer at least
four questions from section A and at least six questions from section B and answer
at least 11 questions in total. If a girl answered each question in
section A in three minutes and each question in section B in six minutes, derive the
system of inequalities that would help to know how many questions she tried to solve
in each section. Use π₯ to represent the number of
questions answered from section A and π¦ to represent the number from section B.

We are told that a test has two
sections A and B. And we will let π₯ be the number of
questions answered from section A and π¦ the number answered from section B. We are told that a student has to
answer at least four questions from section A. Therefore, π₯ must be greater than
or equal to four. They must also answer at least six
questions from section B. So π¦ must be greater than or equal
to six. As any student must also answer at
least 11 questions in total, π₯ plus π¦ must be greater than or equal to 11.

There is also a time constraint of
100 minutes. And we are told that a girl
answered each question in section A in three minutes. She also answered each question in
section B in six minutes. This means that the total time she
spent answering questions can be written as the expression three π₯ plus six π¦. And as the total time for the test
was 100 minutes, this must be less than or equal to 100. The system of inequalities that
would help to know how many questions the girl tried to solve in each section is π₯
is greater than or equal to four, π¦ is greater than or equal to six, π₯ plus π¦ is
greater than or equal to 11, and three π₯ plus six π¦ is less than or equal to
100.

We will now consider one final
question in this video.

A baby food factory produces two
types of baby food. The first type contains two units
of vitamin A and three units of vitamin B per gram. The second type contains three
units of vitamin A and two units of vitamin B per gram. If a baby needs at least 100 units
of vitamin A and 120 units of vitamin B per day, state the system of inequalities
that describes the food that the baby must eat each day to meet these
requirements. Use π₯ to represent the mass of the
first type of baby food in grams and π¦ to represent the mass of the second type of
baby food in grams.

In this question, we are told that
a factory produces two types of baby food. We will let π₯ represent the mass
of the first type of baby food and π¦ represent the mass of the second type. Since these are masses given in
grams, we know that both π₯ and π¦ must be nonnegative. Therefore, π₯ is greater than or
equal to zero, and π¦ is greater than or equal to zero. We know that the first type of baby
food contains two units of vitamin A per gram and the second type contains three
units of vitamin A per gram. As weβre also told that a baby
needs at least 100 units of vitamin A per day, we know that two π₯ plus three π¦
must be greater than or equal to 100.

We can find a similar inequality
for vitamin B. The first type of baby food
contains three units, and the second type contains two units. As a baby requires 120 units of
vitamin B per day, we have three π₯ plus two π¦ is greater than or equal to 120. We can therefore conclude that we
have a system of four inequalities that describes the food that a baby must eat each
day. π₯ is greater than or equal to
zero, π¦ is greater than or equal to zero, two π₯ plus three π¦ is greater than or
equal to 100, and three π₯ plus two π¦ is greater than or equal to 120.

We will now summarize the key
points from this video. In a given situation, in order to
state the system of inequalities, we should label each of the quantities π₯ or
π¦. If the quantities are values that
can never be negative, then we always start with π₯ is greater than or equal to zero
and π¦ is greater than or equal to zero. We may also be given other
constraints for the quantities, such as a minimum or maximum value for each. For example, π₯ is greater than or
equal to 10 or π¦ is less than or equal to 15. These inequalities could also be
strict inequalities, such as π₯ is greater than 10 and π¦ is less than 15.

Additional linear inequalities can
be translated from constraints given for the total combination of quantities, such
as time and cost. These can be written in the form
two π₯ plus three π¦ is greater than or equal to 15. We can represent any of these
systems of inequalities graphically. And whilst it is outside of the
scope of this video, we could also solve them to find optimal solutions.