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Re: More Calculus Fun
Originally Posted by martindcx1e
Originally Posted by spoonitnow
Well we need an equation for area first and foremost. The area of a rectangle is length times width, so let's define those. If we let the left side of the fence be length x, then the opposite side is also length x, and the top side is 100 - 2x (since we have 100 feet total of fencing). Now we have an equation for area: A = width * length = x(100-2x).
We want to maximize area here, so how would we do that?
thank you spoon. i guess my original statement comes off as lazy. sorry about that everyone. i assure you that's not the case.
ok so to find the max area of your example you would distribute the x, differentiate, set it equal to 0, solve for x then plug that into 100 - 2x to find the long side. so x = 25 and 100 - 2x = 50 and the max area is 1250. right?
That's right. Now look at this one:
1) If 1200 cm^2 of material is available to make a box with a square base and an open top, find the largest possible volume of the box.
I know the volume of an open box is (l-2x)(w-2x)(x). I just don't know what to do with the 1200 cm^2
Square base, open top. Draw a picture of what the box looks like laid out flat and you'll probably get somewhere on this one pretty quickly. Here, I'll get you started:
Letting x be the length of the base and h be the height.
Once you have the correct equation these problems are cake. So, we want to maximize volume which will be V = length * width * height, or V = x * x * h. We need everything to be in terms of x so we can take the derivative, so we need a way to write h in terms of x (kind of like when we wrote the length in terms of x in the fence problem and came up with the length being 100 - 2x).
So see if you can find a way to write h in terms of x and finish the problem. (Hint: "I just don't know what to do with the 1200 cm^2.")
I'm going to go see Saw 4 later so if you post before I leave I'll get back at you sooner :P
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