00:01
So we have a circular piece of sheet metal that has a diameter of 20 inches.
00:07
So let's draw that out.
00:09
You have a diameter of 20 inches.
00:18
The edges are to be cut off to form a rectangular.
00:22
Okay, so we have a rectangle.
00:25
Okay, of area of 160 inches.
00:37
So we have an area of 160 inches, which is cute.
00:42
Okay, it's great.
00:43
What are the dimensions of the rectangle? okay, so what we need is length and the width, or it's not called the width and the length of the rectangle.
01:00
That's the dimensions of that.
01:03
So we know the equation of a circle is x squared plus y squared is equal to r squared, right? and we have the diameter.
01:15
Well, the diameter is this equal to.
01:19
2 times the radius.
01:22
So our radius is thus, 20 is equal to 2 times r, so r is equal to 10.
01:34
So we can plug that into our r squared, so we go x squared plus y squared is equal to 100 squared, 10 squared, which is 100.
01:47
Okay, and we also have the equation for an area of a rectangle, which is this area is equal to, length times width we have our area which is 160 link to length times with okay so we have different variables x y length and width and perhaps there's a way to relate them so if you recall if we are on a coordinate system you can have a point here which is is negative x comma y and x comma y similarly we have a point here which is x comma negative y and this is negative x comma negative y right since the x and y points are or variables are points that lie on this circle and the rectangles touch these points at four points.
03:12
Okay, so how can we get the length and the width out of these four points? what we recall we have, we know the distance formula, which is t is equal to the square root of x1 minus x2 squared plus x or y1 minus y2 squared.
03:34
So let's get our width.
03:37
What is that? oh, what's x1? let's call it.
03:44
Negative x minus x2 which is x x2 plus y minus y 2.
04:03
That gives us square root 2 x or negative 2 x 2 of 2 plus 0.
04:34
Actually let me swap this around so our system better.
04:41
Let me rewrite this a quick.
04:45
So let's have this point the x2.
04:50
So let's have x1 comma x or y 1 equal x comma 1, y and x2 comma y2 equal to negative x comma y2 and we're looking for our width.
05:10
Okay, so playing that into our distance formula, i need to page.
05:20
And...
05:21
Okay, distance is equal to square root x1 minus x2 squared, plus y1 minus y2, squared.
05:50
Plugging that in, x1, if this is x minus x2, which is negative of x, x to the power of 2 plus light 1 which is y i just y 2 which is y which is y just 2 okay that gives us the root of 2x to the part 2 plus y nice y which is 2 so we just get 2 x is our width width the distance of our width and now we're going to do the same we find our length.
06:26
So length is going to be decided.
06:33
Okay.
06:34
So in order to find our length, what we need is x1, x1, is equal to x comma y, and x2, 2 is equal to x comma negative y.
06:58
So this is all for the distance of our length.
07:02
Plugging that in, we get x minus x to the bar of two plus y minus nirutri of two.
07:17
X minus x is zero plus y minus negative y which is 2 y to provide 2.
07:26
That gives us 2 y and that's our length.
07:32
So just to know, it doesn't matter which side we choose, but i just chose sides that would make the formula or the equations easier to handle.
07:44
But our width here and here are the same, and length here and here are the same because we have a rectangle.
07:51
Okay, so now that we have the width and the length in terms of x and y, and our circle in terms of x and y, let's rewrite that.
07:59
So our system of equations is x squared plus y squared is equal to 100.
08:13
So we have, what do we have? length times width is equal to 160.
08:21
So 160 is equal to length, which is 2y times with, which is 2x.
08:33
Okay, so let's simplify this one a bit.
08:36
We get 4xy is equal to 160, and we have x squared plus y squared is equal to 100.
08:54
Okay.
08:55
So what can we do? let's solve for...
09:00
Let's isolate y and plug y into here.
09:04
We solve for x.
09:05
So we get y is equal to 160 over 4x with 160 over 4.
09:16
But that's this 40 over x.
09:20
So we have y is equal to 40 x.
09:24
Now let's plug that into our first equation.
09:27
X squared plus y squared is equal to 100 so we get x squared plus y squared but y is 40 over x 2 is equal to 100 okay simplifying this event with x squared 40 times 40 is 1 ,600 over x squared is equal to 100 and multiplying x squared and we'll say it to give it a fraction here we get x zero power of four plus one thousand six hundred is equal to a hundred x squared and moving off our variables to one side just to check 100 x squared minus hundred x squared we get x two five four minus a hundred x two plus one thousand six hundred is equal to zero.
10:32
Okay.
10:35
So is there a way to factor this to find our x terms? well, let me rewrite this.
10:44
We can use our quadratic formula.
10:48
Since i'm not sure how to factor this, the numbers are really large.
10:53
So let's work with that.
10:56
So a is equal to 1, b is equal to negative 100.
11:00
And c is equal to 1 ,600.
11:07
So we have negative b, which is maybe 100, plus minus square roots of b squared.
11:17
1 ,100 squared minus 4 times a, time c, all over 2a, which is 1.
11:30
Okay, so what's 100 squared? it's 10 ,000 minus 4 times 1 ,600 which is 6 ,400.
11:46
Scurrude of that plus minus 100 over 2.
11:54
10 ,000 minus 6k 400.
12:01
It says 3 ,600 0 .000.
12:04
I'll just go rid of that.
12:05
Plus minus 100 over 2.
12:09
Square root, 6 .0.
12:15
That's the 60.
12:16
So we get 100 plus minus 60 over 2.
12:21
We get two answers.
12:24
100 plus 60 over 2.
12:26
And 100 plus or minus 60 over 2.
12:31
100 plus 60 divided by 2 is 80.
12:36
100 minus 60.
12:38
Excuse me, minus 60, but if at 2 is 20.
12:44
So we have two, x squared terms.
12:48
Because if you recall, we have x2 to 4 and x2 to 2, but x2 4 is x2 to 5 2 times 2.
12:54
So our term is x squared.
12:59
On the next page, we have x squared minus, or it's equal to 80, and x squared is equal to 20.
13:09
Okay, if you're writing that, we have.
13:12
X squared minus 80 and x squared minus 20 is equal to zoh okay so can we solve for this what is it is it square root of 80 x so if we take the square root of most signs we get x is equal to plus minus square root of 80 but 80 is just 8 times 10 yeah that's this 4 times 2 the times 10 times 2 which is equal to 2 times 2 times 2 times 10 so these 2 would come out of the square root so it would be 4 to the square root of 10 okay and the same thing goes for x squared is equal to 20 we would get x is equal to plus minus square root of 20 the 20 the 20 is equal to 2 times 10 which is equal 2 times 5 times 2 so we can take out a 2 so we get plus minus 2 square root of 5 okay so recall we're looking for our length and our width but our length and our width can't be negative so we can get rid of these negatives and we just have to look for our 2 terms which is x is equal to square root or 4 times square root of 10 and x is equal to 2 times square root of 5.
15:26
Okay, so we're going to plug that into one of our equations that we have.
15:31
Let's see.
15:34
Okay, so we have y is equal to 40 over x.
15:39
You have these two terms, but we manipulated the second to look like this.
15:47
So let's just solve for our y using that.
15:51
So we know y is equal to 40 over x is that correct yeah okay so using x is equal to 4 times square a bit of 10 let's plug that in y is equal to 40 over 4 times square root 10 40 is 4 times 10 and we can cancel it force so we're left with 10 over square root of 10 but that's this 10 square rate of 10 over 10 you can cancel the 10 so it's the square root 10 okay so we have y is equal to square root of 10 for one of them now just to check the better x is equal to square rate of 5 plugging that into our equation we get y is equal to 40 over 2 times square a 5.
16:56
But 40 is just 4 times 10 over 2 to square root of 5.
17:01
So 4 is just 2 times 2 times 10.
17:04
Over 2, square root of 5.
17:07
You can cancel one of the 2s.
17:10
We're left with 20 over square bit of 5.
17:14
Let's bring the square rate of 5 to the top of 20 over 5 over 5.
17:19
Over five but 20 is just five or 10 times two the time is just five times two so we can cancel one of our fives now and we're left with two times two times square root of five so it's four square width of five okay, so our other solution for x is equal 2 times square root of 5.
17:51
4 times square root of 5 is our y.
17:55
Okay, so it looks good.
17:57
So it seems like these are solutions, right? so a way to check is to plug it into the equations and check if it's correct.
18:08
Oh, hold on.
18:09
So you know how we have everything in terms of x and y right now, but those are the coordinate systems that we're working with, right? we're trying to look for the length and the width.
18:21
So if you recall, our width is equal to 2x and our length is equal to 2y.
18:29
So let's write that down.
18:32
Our length is equal to 2x.
18:36
Is that correct? no, that's our width...