WEBVTT
00:00:00.980 --> 00:00:10.720
In this video, we will learn how to solve applications of systems of inequalities by translating each condition into an inequality.
00:00:11.740 --> 00:00:19.700
A system of linear inequalities is a set of two or more linear inequalities in several variables.
00:00:20.570 --> 00:00:36.440
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.
00:00:37.250 --> 00:00:41.860
Letβs begin by looking in more detail how we can represent these.
00:00:42.880 --> 00:01:03.120
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.
00:01:03.940 --> 00:01:12.520
The equation π₯ equals two can be represented by a vertical line passing through two on the π₯-axis.
00:01:13.310 --> 00:01:21.950
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.
00:01:22.690 --> 00:01:27.200
We can therefore shade out anything to the left of this line.
00:01:28.360 --> 00:01:38.640
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.
00:01:39.540 --> 00:01:47.950
The equation π¦ is equal to four is represented by a horizontal line passing through the π¦-axis at four.
00:01:49.000 --> 00:01:56.510
As the inequality was π¦ is greater than or equal to four, the required area is above this line.
00:01:57.330 --> 00:02:01.080
And once again we can shade out the area below.
00:02:02.060 --> 00:02:08.780
Finally, we need to draw the straight line two π₯ plus three π¦ equals 24.
00:02:09.770 --> 00:02:18.210
This will intersect the π¦-axis when π₯ equals zero, and this occurs when π¦ is equal to eight.
00:02:19.060 --> 00:02:28.730
Likewise, the line will intersect the π₯-axis when π¦ is equal to zero, and this occurs when π₯ equals 12.
00:02:29.500 --> 00:02:37.660
The equation two π₯ plus three π¦ equals 24 can be represented on the graph as shown.
00:02:38.530 --> 00:02:46.420
To work out which side of this line we require, we can rearrange our inequality to make π¦ the subject.
00:02:47.220 --> 00:02:51.140
Firstly, we can subtract two π₯ from both sides.
00:02:51.800 --> 00:03:00.450
We can then divide through by three such that π¦ is less than or equal to negative two-thirds π₯ plus eight.
00:03:01.550 --> 00:03:09.770
As π¦ is less than or equal to negative two-thirds π₯ plus eight, the area required is below the line.
00:03:10.610 --> 00:03:14.490
We can therefore shade out the area above this line.
00:03:15.390 --> 00:03:21.860
This leaves us with a triangular region that satisfies the system of inequalities.
00:03:22.990 --> 00:03:33.950
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.
00:03:34.910 --> 00:03:43.750
This is because all the inequalities need to be satisfied and a strict inequality excludes it from the solution set.
00:03:44.490 --> 00:03:51.310
We will now consider some examples where we obtain a system of inequalities from a word problem.
00:03:52.410 --> 00:03:56.260
A shepherd wants to build a rectangular sheep barn.
00:03:56.960 --> 00:04:06.260
The length of the barn must be more than 88 meters, and its perimeter must be less than 253 meters.
00:04:07.250 --> 00:04:17.240
Derive the system of inequalities that describes the situation, denoting the length of the barn by π₯ and its width by π¦.
00:04:18.310 --> 00:04:28.320
In this question, we are told that a shepherd wants to build a rectangular sheep barn with length π₯ meters and width π¦ meters.
00:04:29.290 --> 00:04:34.940
We need to state a system of inequalities that satisfy the conditions given.
00:04:35.890 --> 00:04:40.930
We are told that the length of the barn must be more than 88 meters.
00:04:41.510 --> 00:04:44.960
Therefore, π₯ must be greater than 88.
00:04:45.800 --> 00:04:53.500
Since the width of the barn cannot be a negative value, we have π¦ is greater than or equal to zero.
00:04:54.420 --> 00:05:01.490
Weβre also told that the perimeter of the barn must be less than 253 meters.
00:05:02.310 --> 00:05:13.960
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¦.
00:05:14.750 --> 00:05:17.920
This simplifies to two π₯ plus two π¦.
00:05:18.870 --> 00:05:26.760
And factoring out a two, we have the perimeter of the barn is equal to two multiplied by π₯ plus π¦.
00:05:27.690 --> 00:05:32.600
This perimeter must be less than 253 meters.
00:05:33.310 --> 00:05:40.010
Therefore, two multiplied by π₯ plus π¦ is less than 253.
00:05:41.020 --> 00:05:46.660
We now have a system of three inequalities that describe the situation.
00:05:47.510 --> 00:05:50.960
The length π₯ must be greater than 88.
00:05:51.700 --> 00:05:55.410
The width π¦ must be greater than or equal to zero.
00:05:56.080 --> 00:06:02.750
And two multiplied by π₯ plus π¦ is less than 253.
00:06:03.680 --> 00:06:10.620
Whilst it is not required in this question or indeed this video, we could represent this graphically.
00:06:11.460 --> 00:06:24.690
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.
00:06:25.360 --> 00:06:35.490
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.
00:06:36.280 --> 00:06:43.460
As π¦ is greater than or equal to zero, the region required is above the π₯-axis.
00:06:44.140 --> 00:06:54.350
And finally, as two multiplied by π₯ plus π¦ is less than 253, we require the region below this line.
00:06:55.350 --> 00:07:01.940
This gives us a triangular region that satisfies all three of the inequalities.
00:07:02.770 --> 00:07:06.960
Letβs now consider another example in context.
00:07:08.220 --> 00:07:11.860
A carpenter wants to buy two types of nails.
00:07:12.290 --> 00:07:19.290
The first type costs six pounds per kilogram, and the second type costs nine pounds per kilogram.
00:07:20.460 --> 00:07:26.870
He needs at least five kilograms of the first type and at least seven kilograms of the second.
00:07:27.380 --> 00:07:31.200
He can spend less than 55 pounds.
00:07:32.170 --> 00:07:44.250
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.
00:07:45.140 --> 00:07:56.260
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.
00:07:57.030 --> 00:08:04.860
We will let π₯ represent the amount of the first type and π¦ represent the amount of the second type.
00:08:05.470 --> 00:08:08.940
This will be the amount of nails in kilograms.
00:08:09.580 --> 00:08:17.510
Since he needs at least five kilograms of the first type, we know that π₯ is greater than or equal to five.
00:08:18.180 --> 00:08:25.930
He also needs at least seven kilograms of the second type, so π¦ is greater than or equal to seven.
00:08:26.820 --> 00:08:29.640
The other constraint here is the cost.
00:08:30.580 --> 00:08:34.670
We are told that the first type costs six pounds per kilogram.
00:08:35.280 --> 00:08:38.050
This is equivalent to six π₯.
00:08:38.710 --> 00:08:45.990
The second type of nail costs nine pounds per kilogram, and this is equivalent to nine π¦.
00:08:46.880 --> 00:08:56.420
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.
00:08:57.280 --> 00:09:14.470
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.
00:09:15.450 --> 00:09:22.520
In our next two questions, we will look at more complicated problems where there are more constraints.
00:09:23.640 --> 00:09:32.860
A teacher gave his students 100 minutes to solve a test that has two sections: section A and section B.
00:09:33.910 --> 00:09:46.710
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.
00:09:47.910 --> 00:10:05.750
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.
00:10:06.810 --> 00:10:16.340
Use π₯ to represent the number of questions answered from section A and π¦ to represent the number from section B.
00:10:17.090 --> 00:10:21.650
We are told that a test has two sections A and B.
00:10:22.660 --> 00:10:31.210
And we will let π₯ be the number of questions answered from section A and π¦ the number answered from section B.
00:10:32.060 --> 00:10:37.540
We are told that a student has to answer at least four questions from section A.
00:10:38.180 --> 00:10:42.190
Therefore, π₯ must be greater than or equal to four.
00:10:43.160 --> 00:10:47.540
They must also answer at least six questions from section B.
00:10:48.030 --> 00:10:51.990
So π¦ must be greater than or equal to six.
00:10:52.900 --> 00:11:03.170
As any student must also answer at least 11 questions in total, π₯ plus π¦ must be greater than or equal to 11.
00:11:04.110 --> 00:11:08.210
There is also a time constraint of 100 minutes.
00:11:08.930 --> 00:11:14.590
And we are told that a girl answered each question in section A in three minutes.
00:11:15.230 --> 00:11:20.220
She also answered each question in section B in six minutes.
00:11:21.090 --> 00:11:29.350
This means that the total time she spent answering questions can be written as the expression three π₯ plus six π¦.
00:11:30.080 --> 00:11:37.300
And as the total time for the test was 100 minutes, this must be less than or equal to 100.
00:11:38.180 --> 00:12:01.910
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.
00:12:03.120 --> 00:12:07.390
We will now consider one final question in this video.
00:12:08.540 --> 00:12:12.890
A baby food factory produces two types of baby food.
00:12:13.740 --> 00:12:21.030
The first type contains two units of vitamin A and three units of vitamin B per gram.
00:12:21.990 --> 00:12:29.320
The second type contains three units of vitamin A and two units of vitamin B per gram.
00:12:30.410 --> 00:12:48.620
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.
00:12:49.420 --> 00:13:01.680
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.
00:13:02.570 --> 00:13:07.880
In this question, we are told that a factory produces two types of baby food.
00:13:08.630 --> 00:13:17.720
We will let π₯ represent the mass of the first type of baby food and π¦ represent the mass of the second type.
00:13:18.560 --> 00:13:25.360
Since these are masses given in grams, we know that both π₯ and π¦ must be nonnegative.
00:13:26.220 --> 00:13:32.610
Therefore, π₯ is greater than or equal to zero, and π¦ is greater than or equal to zero.
00:13:33.400 --> 00:13:45.330
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.
00:13:46.200 --> 00:13:59.300
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.
00:14:00.070 --> 00:14:03.890
We can find a similar inequality for vitamin B.
00:14:04.930 --> 00:14:12.110
The first type of baby food contains three units, and the second type contains two units.
00:14:12.740 --> 00:14:25.480
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.
00:14:26.460 --> 00:14:35.030
We can therefore conclude that we have a system of four inequalities that describes the food that a baby must eat each day.
00:14:35.860 --> 00:14:52.670
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.
00:14:53.780 --> 00:14:57.500
We will now summarize the key points from this video.
00:14:58.630 --> 00:15:07.830
In a given situation, in order to state the system of inequalities, we should label each of the quantities π₯ or π¦.
00:15:08.690 --> 00:15:20.520
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.
00:15:21.420 --> 00:15:28.860
We may also be given other constraints for the quantities, such as a minimum or maximum value for each.
00:15:29.680 --> 00:15:37.760
For example, π₯ is greater than or equal to 10 or π¦ is less than or equal to 15.
00:15:38.480 --> 00:15:48.200
These inequalities could also be strict inequalities, such as π₯ is greater than 10 and π¦ is less than 15.
00:15:49.290 --> 00:16:00.260
Additional linear inequalities can be translated from constraints given for the total combination of quantities, such as time and cost.
00:16:01.060 --> 00:16:08.030
These can be written in the form two π₯ plus three π¦ is greater than or equal to 15.
00:16:08.900 --> 00:16:13.930
We can represent any of these systems of inequalities graphically.
00:16:14.820 --> 00:16:22.690
And whilst it is outside of the scope of this video, we could also solve them to find optimal solutions.