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# A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. No fencing is needed along the barn, and the fencing along the west side of the plot is shared with a neighbor who will split the cost of that portion of the fence. If the fencing costs \$20 per linear foot to install and the farmer is not willing to spend more than \$ 5000, find the dimensions for the plot that would enclose the most area.

## 10416.6$\overline{6} f t^{2}$

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Aniela W.

April 23, 2021

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I think I got all the details here. We have the farmer's barn and on the north wall he's gonna build a pastor or a field or whatever. Uh huh. Okay. And he doesn't have to put fence on the where the barn is. Okay. And then on one fence his neighbor is going to pay for part of that. I guess I should put that at the end of it barn. Maybe his barn is right on the edge of the property or something that. Okay so that's the blue one, his neighbor is going to pay for half of that. And then um The Green one, he has to pay all of it. And so what and the most he will spend is $5,000. And the fence costs$20 per foot. And so we want to maximize the area of the um pasture field or I forgot what it was. So we'll just call it area. So we want to maximize area equals. Well what's the equation for the area of a rectangle? Well it's like times with but we need to give our specific names, let's call this one eggs because it's horizontal. And then these two will be linked Y. So that's the length. So we're going to maximize X times Y. Subject to, okay, the total cost has to be less than 5000. So if it costs $20 per foot and he's paying for all of it and he has to pay for Y. Feet. That's 20 Y. Plus, he has to pay for exhibit 20 x. Plus, he only has to pay for half of this. So instead of being$20 per square foot, it will be \$10 per square foot and that's why. Okay, I can't stand that. Why I got to fix it. Okay, 20 Why? So the first one is the cost for this part. Right here. 2nd 1 20 X costs for this third one. His cost for this His neighbors have to pay the rest and we need that to be less than or equal to 5000. So let's just say equal to. All right. So, if I saw if I simplify that a little bit is 30 Y plus 20 X equals 5000. All right. Any time to find a maximum. You're gonna take the derivative of the thing and then set it equal to zero. But the thing has too many variables. That's what the constraint is for. This is the constraint restrains the size of things depending on how much they cost. Okay, So, we need to take out either X or Y. So you can decide you want to solve this for X or Y. I don't think it's gonna matter. Let's just solve it for why 30 Y. Is 5000 minus 20 X. Yeah, I would have got 2000. If I would have solved drags, I wouldn't have fractions on both of these because the zeros cancel and we get 503 -2/3 x. So area is x times y. Which is 503 -2/3 x. Which is 503 X -2/3 x squared. All right now we're gonna take the derivative with respect to X. So we'll call it a prime derivative of a constant times X is the constant minus the constant times the derivative of X squared which is two X. We're gonna set that equal to zero. We're going to multiply this whole equation by three. So there are no fractions 500 -4 x equals zero. So for x equals 500 X equals 500 over four, which is 1 25. And then remember it wants to know the dimension. So it wants to know why also and it was 503 minus two thirds X. Is that right? Minus two thirds X. So that's 503 -2/3 times 1 25. That's 500 -2/53. So 253 Or uh 83 and 3rd. So he should make it um North wall, North fence, sorry for what we're talking about. North fence, ft. And then the west and east vince 83 and a third feet each

Oklahoma State University

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