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A piece of wire $ 10 m $ long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximum? (b) A minimum?

(a) For maximum area, all the wire should be used to make the square(b) For minimum area, 4.35 $\mathrm{m}$ should be used for square and 5.65 $\mathrm{m}$ should be used for the equilateral triangle

07:46

Wen Z.

02:04

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Missouri State University

Baylor University

University of Michigan - Ann Arbor

Idaho State University

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So if we have this 10 m piece of wire and we're going to cut it somewhere and use one side to create a square and one to create a equal I'll triangle. And we want to maximize and minimize the areas that will be within these. The first thing we're going to want to do is actually just come up with an area equation. So over here, um, I'll call this just X and then I'll call this why? Just for right now. So for us to get the area over here, well, we would have to kind of, like, been this into shape. So the area over here would be Well, it should be equally partitioned for the perimeter. So each side, if we were to kind of draw, this would be for X over four. And then the area of this is going to be just where we square it. So area is going to be X squared over 16. We have that power and over here for our equal lateral triangle. So we would need to cut this into three or bend it into three equal pieces. So if you were to kind of draw what our triangle. Looks like one of these sides are all the sides are going to be. Why, over three, right? And then the area of a equal lateral triangle is supposed to be three route four times the side length squared like that, right? And so now that we have this information together, we could go ahead and add these up. So let's see what we actually get. So total area is going to be equal to so X squared over 16, plus Route three over or and then that would be why squared over nine. So actually, we can rewrite. This is 36. I'll just make that one fraction now. Yeah, so that's gonna be a total area. But now knows we have this where it depends on two variables. So what we want to do now is come up with some kind of constraint equation, so we can just kind of plug it in for the other and go from there. And our constraint is going to be well, how much we have for X and how much we have for why should add up to be 10. So we have X plus y is equal to 10 and then we can solve for either of these. Um, so I'm just gonna solve for X and doing that. Give us X is equal to 10 minus. Why? So now we can take this and plug it in over here, and that's going to give us. So 10 minus y squared over 16 and then plus route 3/36. Why squared? And since this now gives us our area equation and Onley terms of why we can go ahead and take the derivative of this Now, um, so let's go ahead and do that. So a prime why is going to be so for this first one, we would need to use chain rule, So it'll be two times 10 minus why? And then the directive on the inside would just be negative one, then all over 16, and then over here, it would be plus, um, we would just use power rule, so it would be Route 3/18. Why now? We want to set this equal to zero, and then we can kind of solve her there. Um, so let's go ahead and distribute. Actually, this and this would simplify down to eight And so now let's go ahead and distribute the negative. A swell is that eight. So that is going to give us why, Over eight, minus 10/8, which is going to be five force on. Then we have plus route 3/18 y and so this is equal to zero. So now, from thes to weaken factor out of why So that would be why won over eight, plus Route 3/18. And we can add the 1/5 over by force over. And now we can just go ahead and divide that over. So that's gonna be why is equal to 5/4? Um, actually, that's off on the side. Clean this up a little bit, so we can, like, cross multiply. So would be 18 plus eight, Group three and then, um, 18 times eight is 1 44. So be 1 44. So, actually, this is like I multiplied by 1 44 and then divided by 18 plus eight. Route three. Okay, on. And to simplify these, um, four and 1 44 gives us 36 then 36 times five is 1 80. So he end up with. Why being 1 80/18 plus eight thirds eso This is meters. And if we want an approximation, let's see what we get. So 18 plus eight, Route three. So this gives us something around 5.65 m. Um, and now, if we were to go ahead and do this for X Well, we know X plus why was supposed to be equal to 10 or very yet X is supposed to be 10 minus. Why? So we could just go ahead and plug this quantity into here? You could use the approximation. But I mean, since we have the exact answer, we might as well just kind of keep that there for the time being. So be 10 minus 1 80 over 1800 plus eight, Route three and then getting a common denominator. 1 80 just 18. That would give 1 80 plus eight. Route three minus 1 80 all over 18 plus eight. Route three. So the one eighties cancel out with each other and we end up with where X is supposed to be, um eight. Route three over 18 plus eight, Route three. And I mean you could like, divide and kind of simplify this little bit, but I'm just leave it like that, Um, and then this would be approximately 4.35 m. Okay, so let's actually think about what we just found here. So we said that the square is going to need to be 4.35 And then we found that the equilateral triangle, the total much the total amount we're gonna cut off this. 5.65 So that's right, this often side. So we have square about 4.35 m and triangle about 5.65 m. But now they actually asked us to different questions, and we only ended up with one critical point. So there's something else we kind of need to think about. What? We're going through this, um So if we come back up, Thio our total area here, the first equation we came up with, there's too extreme cases that we're going to need to consider. What if Excuse me? What if we let the entire thing be the square or the entire thing be the rectangle or the triangle? So while we do need to check these two here so again, this was X. Why, Um, we will need to check the edge cases as well. So this is possibly a men, possibly a max. Or it could just be like some critical, Um, some like a saddle point or something like that. So let's come over here and see. So we have the case of when X is equal to zero and why is equal to 10. So if we do that our area is going to be will be zero plus o route 3/10 squares. They'll be 100 over 36 which is going to be approximately eso 100 divided by 36 times Route three. So this is somewhere around an area 4.81 m squared. Now if we have where it's flipped, we have X is equal to tend and why is equal to zero? Our area is going to now be, um so would be 100 over 16 plus zero and then 100 divided by 110 divided by 16. This is equal to 6.25 m squared. Okay, and now we'll need to come up here and plug both of these into our area. equation as well. So and we just hope that the rounding didn't throw anything off too much eso That's 4.35 Why is, um, 5.65 and then we just plug those it So 4.35 squared s actually just start rounding this little bit S o B 18.92 to 5. Very. I'll just leave it like that and then plus Route 3/36. And then if we do, why squared as we have 5.65 square to That would be 31.9225 and this is approximately so now just multiply everything together. Also times Route three, divided by 36. That's about 1.5 and then plus 18.92 to 5, divided by 16. And that is going to give us something around. So 2.71 or 2.72 m squared after we round. So we have the case of having a max when excess 10 and Y zero, and then we have a men at this critical point that we were working with before. So what was the order for these again? Let me check the question so the maximum So the max area will be when all 10 m is a square. And then to minimize our area, we will need about about, ah, 4.35 m or square and about 5.65 m, or are equal lateral triangle. And so I guess this is a good question to kind of like bring up the idea of its when we're taking these derivatives is always good to check to see if this is a max or men using, like the first or second derivative tests. Um, because if this just said like, okay, find them in or find the max. And we just did that and then didn't really think about Oh, there's multiple questions going on. We may have just blindly said, Oh, this is a Max or Min when in reality it would have been neither, um, or it could have been neither in this case, Um, yeah, so just kind of keep in mind, always using like the first or second degree of the test is a good thing for these kinds of problems.

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