Lesson Video: Applications on Systems of Inequalities Mathematics

In this video, we will learn how to solve applications of systems of inequalities by translating each condition into an inequality.

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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. Let’s 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, 𝑦 is greater than or equal to four, and two π‘₯ plus three 𝑦 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 𝑦 is equal to four is represented by a horizontal line passing through the 𝑦-axis at four. As the inequality was 𝑦 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 𝑦 equals 24. This will intersect the 𝑦-axis when π‘₯ equals zero, and this occurs when 𝑦 is equal to eight. Likewise, the line will intersect the π‘₯-axis when 𝑦 is equal to zero, and this occurs when π‘₯ equals 12. The equation two π‘₯ plus three 𝑦 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 𝑦 the subject. Firstly, we can subtract two π‘₯ from both sides. We can then divide through by three such that 𝑦 is less than or equal to negative two-thirds π‘₯ plus eight. As 𝑦 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 𝑦.

In this question, we are told that a shepherd wants to build a rectangular sheep barn with length π‘₯ meters and width 𝑦 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 𝑦 is greater than or equal to zero.

We’re 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 𝑦 plus π‘₯ plus 𝑦. This simplifies to two π‘₯ plus two 𝑦. And factoring out a two, we have the perimeter of the barn is equal to two multiplied by π‘₯ plus 𝑦. This perimeter must be less than 253 meters. Therefore, two multiplied by π‘₯ plus 𝑦 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 𝑦 must be greater than or equal to zero. And two multiplied by π‘₯ plus 𝑦 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 𝑦 equals zero and dashed lines at π‘₯ equals 88 and two multiplied by π‘₯ plus 𝑦 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 𝑦 is greater than or equal to zero, the region required is above the π‘₯-axis. And finally, as two multiplied by π‘₯ plus 𝑦 is less than 253, we require the region below this line. This gives us a triangular region that satisfies all three of the inequalities.

Let’s 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 𝑦 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 𝑦 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 𝑦 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 𝑦. 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, 𝑦 is greater than or equal to seven, and six π‘₯ plus nine 𝑦 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 𝑦 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 𝑦 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 𝑦 must be greater than or equal to six. As any student must also answer at least 11 questions in total, π‘₯ plus 𝑦 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 𝑦. 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, 𝑦 is greater than or equal to six, π‘₯ plus 𝑦 is greater than or equal to 11, and three π‘₯ plus six 𝑦 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 𝑦 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 𝑦 represent the mass of the second type. Since these are masses given in grams, we know that both π‘₯ and 𝑦 must be nonnegative. Therefore, π‘₯ is greater than or equal to zero, and 𝑦 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 we’re also told that a baby needs at least 100 units of vitamin A per day, we know that two π‘₯ plus three 𝑦 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 𝑦 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, 𝑦 is greater than or equal to zero, two π‘₯ plus three 𝑦 is greater than or equal to 100, and three π‘₯ plus two 𝑦 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 𝑦. If the quantities are values that can never be negative, then we always start with π‘₯ is greater than or equal to zero and 𝑦 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 𝑦 is less than or equal to 15. These inequalities could also be strict inequalities, such as π‘₯ is greater than 10 and 𝑦 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 𝑦 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.

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