00:01
So in the following problem, there is a farmer that has a total area of 300 acres.
00:06
It has $70 ,500 to spend on planting and $12 ,000 to spend on fertilizer.
00:15
She wants to plant cauliflower and cabbage.
00:19
And we know from the information given that one acre of cauliflower costs $70 for planting and $25 for fertilizer.
00:26
And the same area, one acre for cabbage, cost $35 for planting and $55 for fertilizer.
00:35
So we want to do a system of inequalities that will satisfy all of this information given.
00:42
So to start off, we know i'm going to say that x, it's going to be the cauliflower.
00:49
So it's going to be cauliflower.
00:54
And why is going to be my cabbage? now, we need to take a look at all of these equations.
01:02
So all of this information is in real numbers.
01:06
So we, first of all, we want to delimit our x values, greater than or equal to zero, and our y values greater than or equal to zero.
01:15
Because there is no sense in saying that we have negative cauliflowers and negative colorflowers and negative cabbages.
01:24
So next up, we need to group all of this information into similar groups.
01:29
So first of all, we're going to group everything that has to do with money, money, and everything that has to do with land or area.
01:44
So in this case, we have the total area that must be 300, and we know that we have to plant all of this, sorry.
01:57
So in this case, we have the fertilizer.
02:00
So in this case, we have our total fertilizer for one acre, which is cauliflower in area.
02:09
All of this is in area.
02:11
So for one acre of cauliflower, we have $70.
02:17
So for every acre that we plant, we're going to multiply it by $70 to obtain the total amount of money.
02:24
Plus the total area of our acres, of our cabbage, sorry, which is going to be y, multiply by its costs.
02:31
So $35 ,000.
02:32
$35, sorry.
02:34
And that should be equal to $70 ,500.
02:44
In this case, this is the maximum amount of money that the farmer can spend.
02:50
So this left -hand side should be less than or equal to this amount of money.
02:54
We cannot go above it, but we can go below it.
03:00
So the next step will be to take into account the fertilizer.
03:04
So for every acre of cauliflower, which is eggs, we need to spend $25 in fertilizer.
03:13
So $25 for fertilizer.
03:15
Plus for every acre of cabbage, we need to spend $55 .5.
03:21
So in this case, 55y must be less than or equal again to $12 ,000.
03:30
So in this case, we have covered all of this part of the information, as well as this one, and this total again.
03:38
So now we're left with the total area.
03:41
So for our total area, we know that for every acre of cauliflower and every acre of cabbage, where that should be less than or equal to 300 acres.
03:51
Because that's the maximum amount of land that the farmer has to plant.
03:57
So all of this information will give us the following system of inequalities.
04:02
So we have that x plus y must be less than or equal to 300.
04:08
70x plus 35y must be less than or equal to $70 ,500.
04:16
And 25x plus 55y must be less than or equal to $12 ,000.
04:24
We also know that x must be positive and y must be positive as well.
04:32
So now we're going to graph the system of inequalities.
04:34
I have summarized all of this previous information over here.
04:38
So if x is greater than zero and y is greater than zero, then we're looking only on the first quadrant, which is this one right here.
04:51
So now i'm going to take red color to graph this line.
04:57
So i'm going to assume that this is just a simple equation and say that x plus y is equal to 300.
05:05
I'm going to solve for y and see that 300 minus x is the line that we're going to be graphing on.
05:12
So this line has a y intercept of 300 over here and an x intercept of 300 as well.
05:21
In these two points with a solid line, we're including the values on this line.
05:28
And we're going to test with some point over here if that satisfies the inequality.
05:35
I'm going to use the point 100, 100.
05:38
So test with 100, comma, 100.
05:45
So in that case, we'll have that 100 plus 100 must be less than or equal to 300.
05:52
So 200 is less than or equal to 300, which is true.
05:57
So we're going to be shading this side of the graph.
06:02
I'm not drawing it, but you can imagine that there are two lines, one up on the y -axis and one on the other on the x -axis, that it's limiting our area just to this first quadrant.
06:14
And that comes from the fact that we have these two inequalities.
06:19
So the second one is 70x plus 35y.
06:25
Again, i'm just going to assume it's a simple equation, just $70 ,500.
06:32
And i'm going to start to solve for y again.
06:36
So we'll have 35y is equal to $70 ,500, $70 ,500 minus $70x.
06:43
And we're going to divide this everything by $35.
06:47
So $7 ,500 divided by $35 is just $500, and $70 divided by $35 is just 2x.
06:56
So in this case, we have a y intercept at 500 and an x intercept at 250, which is about here.
07:05
Again, we're going to be joining these two points with a solid line.
07:08
Oh, that's not an accurate.
07:11
Line, that's so much better.
07:16
And we're going to test again with one point.
07:18
I'm going to use a point again.
07:20
I'm going to use the point 0 .0.
07:24
So in this case, test with 0 .0, which is part of our domain.
07:33
So we'll have that 70 times 0 plus 35 times 0 is less than, oh, sorry, less than, or equal to 70 ,500, 70 ,500.
07:47
So we'll have zero is less than equal to $70 ,500.
07:53
We know this is true, so we'll be shading this side of the graph, which also includes this shaded region of the red line.
08:02
Finally, i'm going to do the last equation.
08:06
25x plus 55y is less than equal, sorry.
08:12
Again, we're going to assume that this is just an equal sign to $12 ,000.
08:18
So i'm going to solve for y again.
08:20
55y is equal to 12 ,000 minus 25x.
08:27
We're gonna divide by 55 from both sides and that will give us 218 .88 minus 0 .45x.
08:40
So in this case we have a y intercept at 218, which is about here, and we're gonna solve for the x intercept.
08:48
So in this case, we're gonna assume that y is equal to 0, then we'll have 218 minus 0 .45x we're going to add 0 .45x on both sides so 0 .5x is equal to 218 .8 we're going to divide by 0 .45 and we'll get that x is just equal to 480.
09:12
So our x intercept is at 480 .0.
09:19
So for 80 it's going to be a point outside of my line outside of the board, and we can just join the graph like that.
09:30
And i want to test with the same point, 0 .0.
09:33
Test with 0 .0.
09:37
And we'll have that 0 is less than equal to $12 ,000.
09:41
And that is true.
09:43
If that's true, then we're going to shade all of the region below the line.
09:50
The solution to the system of inequality, since we have delimited our domain for x and y to be only positive numbers, it's going to be this region right here.
10:05
The region i'm shading in black is a solution for this system of inequalities.
10:09
And that means that if you choose any point inside this region, all of these five inequalities are going to be true.
10:16
No matter what the point is, all of these points will give a positive solution for the system of inequalities.
10:25
So in this case, we have one vertis over here and that is the intersection of the blue line.
10:31
So that will be the point 0 .2 .2.
10:36
18 .8.
10:39
That is one of our vertices...