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A farmer has 2400 feet of wire to build a rectangular or square pen that encloses the maximum area: If the farmer wants to place two (2) lines of wire on each side,...

Question

A farmer has 2400 feet of wire to build a rectangular or square pen that encloses the maximum area: If the farmer wants to place two (2) lines of wire on each side, the dimensions of the pen that generate the maximum area are:Select one: a. 600 ft * 1200 ftbThe correct answer is notc. 1200 ft * 1200 ftd. 300 ft * 600 fte 600 ft x 600 ft

A farmer has 2400 feet of wire to build a rectangular or square pen that encloses the maximum area: If the farmer wants to place two (2) lines of wire on each side, the dimensions of the pen that generate the maximum area are: Select one: a. 600 ft * 1200 ft b The correct answer is not c. 1200 ft * 1200 ft d. 300 ft * 600 ft e 600 ft x 600 ft



Answers

Area A farmer is constructing a rectangular pen with one additional fence across its width. Find the maximum area that can be enclosed with 2400 m of fencing.

So in this problem we are given that a pen for a pet is under construction. It's rectangular and it is using 100 ft offense. And in part for us to uh huh. Find the dimensions that would give the pen 576 square feet. So we have when we draw this out it's just a rectangle with besides Ellen W. Length and width. Mhm. Lengthen with. And so 100 ft is the perimeter. So that would be to L plus two W. Is the addition of all the sides. That's equal to 100. And the area is 5 76. Or that's what we want. So L. Times W. Which is the area is 5 76. Okay, so now we can just solve these out to find the dimensions that would give 576 square feet. So first let's take this to L. Plus two W equals 100. Um And let's solve for L. In terms of W. So first we can divide by two on each term. We get L plus W equals 50. Okay, so then we get subtract W. L equals 15 minus W. Okay well why don't we plug this into this equation? So we get 15 minus W. Which is equal to L. A. Times W equals 576. And now we can distribute the W. We get 50 W minus W square equals 576 And subtract or well add these terms, will subtract 50 W and add W square to each side and I get zero equals W squared minus 50 W plus 5 76. And that is a quadratic equation. So you can solve it any way you like? I did graphically. So I used as most graphing utility and you just plug it in and you look for where it crosses the X axis and those are going to be our solutions, we get 18 and 32. So either of those are viable values for w let's say W equals 18. Well then W plus L is equal to 50 as we see here. So therefore that would be equal to 32. Okay, so those are dimensions to get 576 square feet. Now part B asks us to maximize the area of the power. So why don't we make a function and look for its maximum? Let's say we have F. W. And going back to our equation for area right here, 50 my S W times W equals 576 So this 15 minus W times W will solve for the area. And in our case here we were just trying to find what value of W would get 5 76. So now we just want to kind of this as a function and find its maximum, so 50 minus W times W is equal to F W. And so we distribute the w. We get 50 W minus W squared. And now we just want to find where this function f f W reaches its maximum value. So what the value for W is there? Okay, well I plug this into a graphing calculator and so right now we're not looking for the zeros were just looking for where this function is at its highest value. And you see here it reaches its maximum at X is equal to 25. W equals 25 and it gives an area of 625 square feet. So therefore W equals 25 is the maximum and therefore since W plus L is 50 also, L will equal 25. And those are the dimensions that give the maximum square footage, which is 625 square feet. And that is all we need for this problem.

If a farmer wishes to enclose a rectangular region bordering a river. So here's the river with fencing. As shown here in this picture? I'm drawing? Suppose that X. Represents the length of each side of the three parallel pieces of fencing. There's actually a line down the middle here. We've got X. X. And X. And they've got 600 ft of fencing available. What is the length of the remaining piece of fencing in terms of acts? So this piece right here we're going to call why for a moment. And if the whole thing we get three X plus Y. Because I've got +123 exes Equals 600 because that's how much fencing they have available. Subtract three X. From both sides And we get y equals 600 -3 x. I'm going to erase that and plug in the 600 minus three X. There. Our second piece wants us to determine a function A that represents a total area of the enclosed region. We're going to find any restrictions. So the area is going to be X. This length here Times this length here, 600 -3 x. We're going to distribute. We had X times 600. 600 X. X. Times negative three X. Is negative three X squared. Want to rewrite that salts in standard form, the area is negative three X squared plus 600 X. What are the restrictions? What we said? We've got that 600 And we have three exes. So 600 divided by three is 200. So X. has to be less than or equal to or less than 200. So my restrictions are that access between zero and 200. What are the dimensions for the total enclosed region? That would give it an area of 22,500 ft. So we said area was this guy here negative three X squared plus 600 X. And we wanted to be 22,500 ft. I'm going to subtract 22 500 from both sides. We get zero equals negative three X squared plus 600 X minus 22,500. We now have this in the quadratic form so we can use the quadratic equation X equals negative B plus or minus the square root of B squared minus four. A. See All of that divided by two A. X. is negative 600 plus or minus square root of 600 squared minus four times negative. Three times negative 22,500 is 90,000 over negative six And continue over here, X equals negative 600 plus or minus. The score of 90,000 is 300 over negative six. So we get X equals negative 600 plus 300 over negative +64 X equals negative 600 minus 300 over negative six. Excuse me, X equals 50 ft. Or x equals 150 ft. What is the maximum area that can be enclosed? So determine the maximum area. Get D over dx Equals 600 -6 x. I'm gonna set that equal to zero. I'm gonna say Track 600 from both sides I get negative six X equals negative 600 Divide both sides by -6. And we get x equals 100 feet. We said at the beginning when I put y equals 600 -3 x. So why equal 600 -3 times 100. We're substituting this 100 and Y equals 600 -300 And y equals 300. See the area. Therefore remember the area is negative three X square Plus 600 x. Area equals negative 300 times 100 square plus 600 times 100 Area equals negative. three times 10,000 plus 60,000. Area equals negative 30,000 plus 60,000. The area is 30,000 feet square

Here we have another modeling problem, but this time it is about area. And so it says that a renter has 200 ft to enclose these two corrals here, right? And it gives us these units where excess this distance. It's also that distance. And it tells us that why is this distance here? So we know that the entirety of the fencing must equal 200 ft. So that means that four X plus three wide, equal to 200 I'm gonna solve for why? So we could get it in terms of X. So 200 minus four X and then we divide by three. So I've got 200 minus four x divided by three. Great. So now, part of the question asked us What's the area of the garage as a function of X? I'm gonna call that function a of X, and I'm gonna take this side length multiplied by this side length. So that's gonna be two x multiplied by why which we know is 200 minus four x over three. I'm gonna distribute that to multiply everything by the two X right. I've got 400 x minus eight X squared all over three. And that is our function. Now, if you want, you could simplify this in fact, her out and ate them doing so you're left with 50 minus X over three. Now, I'm gonna use this warm right, because that's our function, right? Either one of those will work apart, be asked us to, Um it asked us what dimensions produce a maximum enclosed area. So recall that when we have a function in this form a X squared plus V X plus c, you know that the maximum occurs at half of negative be over to a so we need to figure out mhm what negatively over to a is. Now, I'm gonna write this equation here just a little bit differently on a separate the three plus 400 X over three. And that's of course, FX. Now, this makes it easy to identify our bnr a. So let's figure out what be over to a negatively over today. So be in this case is 403rd. We're gonna divide that by negative to a So excuse me, this 400 years should be negative. We're gonna divide that by two A. That's two times negative 82 3rd. So that's 400 over three Negative, divided by a negative 16 3rd. I remember when you're dividing fractions like this, you're really just multiplying by the reciprocal. So I'm gonna make this a reciprocal you see easily now that the threes will cancel, and so will these negatives. So we're left with 400 divided by 16, which evaluates to 25. So that's the X dimension 25. So now we need to do is find out why dimension. Now we know that. Why is 200 minus four x over three? So I'm just gonna plug that in to this expression 200 minus four times 25 over three. And that gives us a 200 minus 14. 25 is 102 100 minus 100 is simply 100 over three. Right. And reading that as an improper fraction, I'm gonna write or a mixed number, Rather is gonna be 33 and one third

Here I have recreated figure 22. So a farmer was just to fence off three identical adjoining rectangular pens. So those three rectangles are all identical and they each have an area of 300 square feet. So let's go ahead and write that in. That's 300 square feet. So now the question is what should be with the length of each pen B. So that the least amount offensive required? So we're basically asked if I'd X. M. Live so that it maximizes the total amount of fencing needed. So let's write down a couple of equations. So what we know is that since this area here is 300 square feet, we know X times Y. Must be 300 so X. Y. Is equal to 300. Next we need to figure out how much fencing is actually required here. So let's label all the sides, this is also an X. And then the top three sides are all excess and then these are all wise. Okay, so we can see that the total amount of fencing required is going to be six X. And four Y. Okay so let's call that up for fencing is equal to six X plus four white. Next we are going to use our equation here to isolate one of the variables so that we can get um this function here in terms of only one variable. So if we were to isolate why we get Y. Is equal to 300 over X. So now we can say f is just a function of X. And this is going to be six X plus four times 300 over X. So that would be 1200 extra power of negative ones. Okay. And now because we want to minimize the amount of fencing required, this means that we're looking for the critical values. So we do that by finding F. Prime of X. And that's going to be six minus 1200 extra power of negative two. Okay, so we can see that when X zero F prime of X doesn't exist but X can't be zero because then we won't have an area of 300 right? Because then X times Y can't be 300 if X is equal to zero. So really our domain has to be X. Y greater than zero. Okay, So which means the only critical values that we're going to be looking for would be when F prime of X is equal to zero. Okay, so let's set that equal to zero and solve for X. Okay, So then we get 1200 extra part of negative two is really just over X squared is equal to six. Okay, So which means X squared is equal to 1200 divided by six, which is 200. So therefore X must be the positive square root of 200. So the reason I'm tossing out the negative number is because again, that's not in our domain, right? We can't have a negative blank. Okay. So now the last thing that we need to do is we need to double check and make sure that this critical value does actually give us a minimum and not a maximum. And you do that by doing the secondary test. Okay? So secondary of the test, that would be a double crime. So that's going to be negative 1200 times negative two. So that will be positive 2400 and this is extra power of negative three which is really over execute. So we can see that if we take exit we cubit that's going to be a positive number and then we have a positive divided by a positive. So therefore a double prime of this particular critical value is going to be greater than zero and that's all we care about. Whether it's greater than zero where less than zero were equal to zero. We don't actually need to know what the number is. So if it's greater than zero that means it's concave up at this at this critical value. And concave up means that here we have a minimum which is exactly what we're looking for. Okay, so therefore we found our X. Values. So let's go ahead and put this back into the original um like it's not original into this equation to find our why? Okay so if we put this back we get y must be equal to 300 over um the square root of 200. Which actually let's simplify this because 100 is a perfect square. So we can take that out. That's really 10 route to. So which means why is 300 over 10 route to multiply top and bottom by Ruutu over route to to rationalize it. So this will cancel, that's 30. So this is 30 route to over two, which is really 15 we took. Okay. So therefore our final answer is going to be X is equal to 10 route to, and don't forget our unit which is in feet and why is equal to 15 route to.


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