Wire of length 12 m is divided into two pieces and the pieces are bent into a square and a circle

Germin R.

A piece of wire 10 m long is to be cut into two pieces. One piece is to be bentinto a square and the other is to be bent into a circle. How should the wire be cut so

that the total area enclosed is a minimum

A wire of length 12 is divided into two pieces and the pieces are bent into a square and a circle. How should this be done in order to minimize the sum of their areas?

Chapter 4

APPLICATIONS OF THE DERIVATIVE

Section 7

Applied Optimization

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Video Transcript

Wire link 12 m and we're gonna cut it into two pieces and bend it into a square in a circle. And we want to know how should this be done to minimize enclosed area. So the trick to this problem is really getting the labels right, Okay, here we have a square and here we have a circle. And so with the circle we need both the circumference and the area Well, and same for the other one too. Okay. And both of those have pie in them. Okay, So we're gonna get a kind of goofed up answer. But first, let's decide what would you like to call the radiance here. How about X? Okay, then the circumference is two pi x. And the area is hi X square. Now, what would you like to call one of the sides of the squares? We can't call it X. Let's just call it. Why for now then the perimeter is for Y. And the area Y squared. All right. So what do we need the perimeter for? Well, we need the perimeter because we know that we have 12 m of wire that we have to make this square and the circle out of. Okay, so here we go minimize the area, which is equal to pi ax squared plus Y square subject to the constraint. Really? I'm 12 m of wire. So far y Plus two Pi X has to equal 12. Okay. Now, to minimize something, you take its derivative and you said that equal to zero. And as you can see, we have both an X and A Y in our area formula. So we have to get rid of one of them and that's what the constraint is for. Okay, So would you rather solve it for Y or X? I'm saying why I'm gonna solve for Y for y. Is 12 minus two pi X. Why is 12th minus two pi X over four? Um Mhm I'm trying to decide if I want to simplify that into. I do I'm gonna make it 12/4 which is three minus 2/4 which is one half pipe X. So pi X over two. Okay, so now my area is pi X squared plus three minus pi over two X quantity square take the derivative with respect to X. So I'll call it a prime two pi X plus two times three minus pi over two X. To the one times the derivative of three minus pi over two X. Which is minus pi over two Saturday equal to zero. Okay now let's do some simplifying. I have two pi x minus because of this minus Right here this too. And this two are going to cancel I think I will put the pie right here. Three minus pi over two X. Yeah so two pi x minus three pi plus pi squared over two X equals zero, yikes fractions multiply by 24 by x minus six pi plus pi squared X equals zero. Now let's solve for X. So I'm gonna move the six pi over to the right hand side. I'm going to factor the X out of the other two. For pi plus pi squared equals six pi X equals six pi over four pi plus pi squared six pi By four plus pi 6/4 plus pi. And then Why Remember? was 3 -5 or 2? X. Okay surprise two X. Which equals three minus pi over two times 6/4 plus pi three minus three pi over four plus pi. Let's get a common denominator 12 plus three pi minus three pi over four plus pi 12/4 plus pine. So how much is that much? How much was ext um Let's see six divided by parentheses. E for plus pi parentheses. E equals .84 on this one. And then this one is just twice it 1.68. So it's saying if we make the square 1.68 by 1.68 And the circle with radius .84 then we've either found the maximum or minimum area that we could possibly make out of these two. Let me go ahead and calculate what it is. So that's squared plus 0.84 plus parentheses. E High Times. .84 sq Parentheses. E Yeah looks five square feet about. Well how are we gonna tell if we found the maximum or minimum? We'll just like we always do we're going to draw a number line. We're gonna put um 6/4 plus pi there and zero here because we know we can't Have less than zero. Okay. Pick a number bigger than 6/4 plus pi. How about 10? Where's the derivative? Okay. Here it is. So I'm gonna plug in 10 for X. and I get 20 pie minus pi times three minus five by 20 pi minus three pi plus five pi square 17 Pi Plus five Pi scored. That's definitely positive. And then this other side it's going to be negative. So that is the minimum. So I believe this is the answer. Make the circle. Have this uh make the circle have this radius, make the square have this side that will give you the minimum. Uh huh. Area.