need help on calculus problem

Bump! I'm absolutely stuck

Well if we label the two dimensions of the yard as x and y, then 20,000 = xy and we are trying to minimize 2x+y (assuming y is the side opposite the building). Then x=20000/y and you try to minimize 2*(20000/y)+y by taking its derivative and setting it to zero. That's your answer.

You need to find two equations, one a function of x that you are going to minimize. That will be the perimeter function, which is very easy to describe. You should be able to work that out. Then, you need another equation relating the two variables so that you can substitute one for the other. That is where the area comes in, namely the area function is a constant 20,000. Minimize the perimeter function in either x or y, and then use the value you get there in the area equation to get the other.

Assuming the building wall is much larger than whatever distance you need?

Let the side of the building be "x" in length. Then you can express the total length of the wall in terms of "x" and the total area "20,000 sq ft", and minimize the length with respect to x. You can do the math yourself, figuring out how to set up the problem is always the hard part. :)

alright well I've found the zero which is the sq. root of 20,000 (141.2). I can't figure out what the domain is though. Because I need to plug in 141.2 plus both of the domain values into original equation to find y then plug both into the equation of the perimeter to see which uses less fencing

So here's my next question. What's the domain of it?

Its been a while, but do you really need calc to solve this one?

The shape that would require the least fencing is a square, so find the sq root of 20,000 which is 141.6

multiply by 3 to give you 424.8 ft of fence.

Or is there more that Im missing?

Tommy2Nuts wrote:

Its been a while, but do you really need calc to solve this one?

The shape that would require the least fencing is a square, so find the sq root of 20,000 which is 141.6

multiply by 3 to give you 424.8 ft of fence.

Or is there more that Im missing?

Well, that's the answer I got, but here's the problem. The maximum would forcibly have to be different than the minimum, and if 424.8 is the minimum, then what is the maximum. My point by saying that is even if 424.8 is the correct answer I'll get points marked off if I can't prove why, which is why I need somebody to tell me what the domain is. Once I know the domain, I'll be able to plug in other values to see what the min and max really is.

XCMiler wrote:alright well I've found the zero which is the sq. root of 20,000 (141.2).

You solved for the zero incorrectly. The procedure as described by slowkid is correct and the solution uses less than 424 feet of fence.

There is a nifty geometric method to get the answer as well.

Yes, you're missing something -- I'm guessing you don't need fence when you use the building as a wall. So it makes sense to make it a lot longer in that direction.

not guessing, it says it in the original quesiton :)

That makes absolutely no sense. It clearly says that the yard should be rectangular, and the length of the wall is fixed as the length of the rectangle. So why would you want that to be longer?

XCMiler wrote:

Well, that\'s the answer I got, but here\'s the problem. The maximum would forcibly have to be different than the minimum, and if 424.8 is the minimum, then what is the maximum. My point by saying that is even if 424.8 is the correct answer I\'ll get points marked off if I can\'t prove why, which is why I need somebody to tell me what the domain is. Once I know the domain, I\'ll be able to plug in other values to see what the min and max really is.

This may or may not help, but I\'ll try anyway. The previous poster is right in how to calculate the answer. On your point here, there is no maximum. If you could have a really skinny box (large perimeter with same area), and in theory you can always have a skinnier box (infinite perimeter).

Okay so if I did it the way slowkid says I get: 400

If I do it the other way its 424

Okay, now post who's right slowkid or the other guy. I'm torn between the two

Tommy2Nuts wrote:

That makes absolutely no sense. It clearly says that the yard should be rectangular, and the length of the wall is fixed as the length of the rectangle. So why would you want that to be longer?

This is the problem with calculus homework. It doesn't say the length of the wall, it just says "large store." If they did give the length of the wall you would have to figure out if it matter. For this problem you just assume that the "large store" has a side wall that is really, really, really long so you will not use the whole wall for the fenced-in area. In the "real world," your solution would often be correct, but you probably would not use calculus to get there.

why would I ever think of using calculus in the real world?

It's just a loop I have to get through to get into a good college

slowkid

XCMiler wrote:

It's just a loop I have to get through to get into a good college

If you're still undecided, I'd recommend a major from the wonderful world of liberal arts for someone of your intellectual pedigree. Oh, and Slowkid's method is correct.

Come on, this problem is peanuts. Slowkid is right.

You assume the wall is as long as you'll ever possibly need it to be (the problem could have been a bit clearer about this). Your constraint is that xy=20000. Draw a diagram of the building, with two sides of fence coming out of it, and a third side of fence connecting those two. Label the two sides "x" and the third "y" (the variables you use don't matter, just be clear about which correspond to the double or the single side). f, which is the length of fence you'll use, is equal to 2x + y. Solve your constraint for y.

y=20000/x

plug into f

f=2x + 20000/x

take derivative

f'=2 - 20000/x^2

set equal to zero, solve for x

2=20000/x^2

x^2=10000

x=100

You can have a fence that's barely there at all or that goes on forever. Domain: 0