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Litre. How wooden many fence SI litres 5 are high needed? and 52 in length The stain for the (2 fence encorks)s 'm per...

Question

Litre. How wooden many fence SI litres 5 are high needed? and 52 in length The stain for the (2 fence encorks)s 'm per

litre. How wooden many fence SI litres 5 are high needed? and 52 in length The stain for the (2 fence encorks)s 'm per



Answers

An equestrian buys a 5 -acre rectangular parcel (approximately 200,000 square feet) and is going to fence in the entire property and then divide the parcel into two halves with a fence. If 2200 linear feet of fencing is required, what are the dimensions of the parcel?

Here. We have a sketch off, Um, some fencing that has been put up and it creates a rectangular shape. And then, Ah, center fencing is put in two. Separate that into two equal sections. We are told the total area is 216 meters squared. Okay, so we need to find out two things. What are the dimensions off the outside rectangle that we're going to give us the least amount of fencing needed. Okay. And how much fencing do we need? So let's label some things here. So this is going to be the width. And from here, the whole way over here is gonna be my length. All right. So we know our formula for area is going to be length, times width. Okay. And what I want to dio right now is go ahead and solve this for l. Well, we know the areas to 16. We'll go ahead and put that in, and then l is going to be to 16. Divided by w not come into play here in a minute. Okay? The formula for the perimeter or for how much fencing we need or length of fencing. The perimeter is going to be to L Because we do have to side length here and then. Plus, we need three of the W lengths. OK, um, so what we want to do is go ahead and substitute in that l that we came up with. So this is gonna be to 16 w plus three w. So now we only have two variables to work with. All right, So I'm gonna go ahead and make that a w to the negative first power because we are going to take the derivative here, and that just makes it a little easier. So we have negative for 32 w to the negative. Second power plus three. All right. And what we want to dio to find the minimum fencing that we need. We know we need to set that equal to zero. All right, So I'm gonna go ahead and move the w term to the other side, so we're gonna have three is equal to, and I'm gonna go ahead and put that w back in the denominator, so it's gonna be w swear. So we're gonna have three. W squared is equal to 4 32 W squared is equal to 1 44 and then W is equal to 12. So I know the width of my fancying has to be 12. All right, so we're gonna just go right back to here for l. And I know L is going to be now to 16 divided by 12 which is 18. Okay, so my dimensions off my fencing. All right, I'm just gonna put it up here. W is equal to 12 and l is equal to 18. And those are meters. Okay, Now we want to know how much fencing we need. Well, all we have to do there is Go ahead to our perimeter formula. We're gonna have l is 18 and w is 12. You go ahead and calculate that out. We are going to need 72 meters off fencing.

Hello. Everyone in this problem has been said that What index is there? Okay, which is Ah, off dimension and feet by eight feet. Okay, so you have to stain that, wouldn't it? Okay. And, ah, the stains which you bring will be in measures off court. Okay, so each word off Steyn can hold. Ah, 50 feet squared off steam. Okay, So even hugs that how much worse you would require or how many courses do require toe complete the staining off the wooden deck. Okay, so let me first find the ideal daughter ending. So that area is length multiplied by week. So which is Stan Freed Money played by eight feet. So that would give us 80 for you to square. Okay, so the number off quirks would be brutal. Area that is 80 divided by area covered with each court. Okay, so this 80 by 50 it is eight by five. Or even write it one whole to be by five Quirk's. So the answer is one whole three by five works. They're so thank you

We're asked to maximize thie area so we know we're going to need to maximize area of a basically like a fence. So the best thing is going to be to go ahead and draw yourself a picture. So we know we are fencing in some rectangular thing. So we know we have some amount of fencing here, which is some excelling some X linked here, a fencing, some wiling fear, a fencing in some whiling pura fencing. So we know we're trying to maximize the area of this shape, and the other thing we know is that we only have five hundred meters of fencing. So what we know from this five hundred meters offense is that two Why plus two X because that is the perimeter. So to why plus two x equals five hundred. We know that because we need some amount. Why x y next? We also aren't limited to the fact of it being a break time tangle. It just tells us in the question, um that we have some field we need defense it. So we have to I post two X is equal to five hundred. We also know even maximized the area an area is equal to X times. Why so keeping this information in mind? We have two equations. We have a conditional equation which says we need this condition to be true. We also know that we are maximising the area. It will tow X Y and we can call this the perimeter. Okay, so now we know what we're maximizing. And this is kind of like step three, which is we have to figure out what we're maximizing and then make it a function of just one variable. So to do that, we're going to solve the perimeter function. So we know that too. Why is equal to Well, before we even do that, you can divide everything by two. So you have X class wide. It was two. Fifty. So why he holes two fifteen minus x. So we know we have this and this comes from the P. We also know that that means the area will be X times two fifty minus x. With some simplification, we get to fifty X minus X squared. Okay, so now we've kind of completed Step three, which is to get the function we're maximizing into just one variable. The next part we need to do Now that we know what we need to maximize is to fifty x minus X squared. We need to find critical points. So a prime is two fifty minus two x and we need to find critical points. So to do that, I'll do it in green. You need to set zero equal two two fifty minus two X. If you solve this, you get that X is one twenty five. Okay, so now we have a critical point. Now we need to know. We need to answer this question. Is this a max? Is this a maximum? I My advice would be to use the second derivative test is to figure out if it's a maximum. Is this the maximum? So to do this, we're going to take the second derivative. The second derivative is negative too, which is less than zero. That implies there is a max. Okay, a little bit of explanation behind this. A negative second derivative implies you have a Khan cave down, and that is definitely a max. I think of it as a negative sign Your sad. So this is a frowny face. So you will have a maximum, so that's just a quick little bit of information. So now we have that X is equal to one. Twenty five is a max. If we recall her original equation, which is that the perimeter has equal to X Plus two why? And that has to equal five hundred. We can plug in the one twenty five for X, and we get to fifty post two. Why equals five hundred two? Why equals two fifty? Why also equals one hundred and twenty five, And that's going to be your final answer. So the maximum area So the max area is when why equals one twenty five and eggs a horse one twenty five and there you have it. That should be the answer to question number ten.

So we know the question Bought a five acre parcel. Um, and they're going to fence the property and divide it in half. So, uh, if 2200 linear feet are acquired, we want to know the dimensions. So we're gonna have to create. And our equation one's going to include area, and one is going to include, um, one is going to include area, and everyone's going to include the amount of fencing required. I'm in terms of the perimeter. So, um, we will have x Will have. Ah, there are. If we let, um, the the vertical lines we call those why will have free why and then if the long horizontal ones are acts that we plus two X um, apple equal 2200. And then we will also have the area B x times y, and thats going equal 200,000. Um, if that's the case, we can zoom out and see that there's one area of intersection and that tells us that Oh, sorry. We need Teoh. This would be, uh, too acts right here. We have a couple areas of intersection. Um, one of them is 603 133.2 repeating and 504 100 so we can consider the dimensions of the parcel to be 500 feet by 400 feet.


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