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 Fun Calculus Problem
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Dreadnought




PostPosted: Mon May 27, 2013 10:22 pm   Post subject: Re: Fun Calculus Problem

Well it's not bad to learn more. In my opinion the best teachers I had in high school were the ones that gave me the most work. My calculus teacher, for example taught integration near the end of the course.

Nevertheless, I still feel that question was evil for a high school test.
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Panphobia




PostPosted: Sat Jun 08, 2013 4:39 pm   Post subject: RE:Fun Calculus Problem

I just had another test in calculus and this is my last before the exam, it is in optimization, I found my thinking question too easy so I asked another class for theirs and this is what it was
code:
You are given 60 cm of wire, some of this wire is to be used for an equilateral triangle and the rest should be used for a circle, what are the dimensions fof both the circle and the rectangle such that their combined area is a minimum?
I got the radius = 0.454 cm and one side of the triangle = 19.05, did I get it right?
Dreadnought




PostPosted: Sat Jun 08, 2013 5:47 pm   Post subject: Re: Fun Calculus Problem

I'm pretty sure the answer is x = 12.4642 and r =3.5981 where x is the length of one side of the equilateral triangle and r is the radius of the circle.
Panphobia




PostPosted: Sat Jun 08, 2013 7:32 pm   Post subject: Re: Fun Calculus Problem

How did you get that? This is what I did
code:
Since we know the wire is 60cm that is one constraint, so
60 = 2(PI)r + 3a
I am going to be using the equation for the area of a triangle that only needs the three sides so here that is
area of triangle = sqrt(p(p-a)^3)
p = perimiter /2 = 3a/2
area of triangle = sqrt((3a/2)(3a/2-a)^3) = sqrt((3a/2)(a^3/8)) = sqrt(3a^4/16)
area of circle = (PI)r^2
A = sqrt(3a^4/16)+(PI)r^2

from there just sub in and differentiate, please tell me if I did something wrong, and I know you can use Pythagorean theorem for the height of the triangle also, but I chose to do it this way, anyway if I did do anything wrong could you show me your solution?
Dreadnought




PostPosted: Sat Jun 08, 2013 8:53 pm   Post subject: Re: Fun Calculus Problem

The formula you give for the area is correct and it has a minimum at a = 12.4642.

What do you differentiate with respect to, a or r?

EDIT: If we substitute for r, here's a plot of the area (this url doesn't play nice with the tags apparently...)

http://www.wolframalpha.com/input/?i=plot+sqrt%283%29*%28x%5E2%29%2F4+%2B+pi*%5B%2860+-+3*x%29%2F%282*pi%29%5D%5E2
Panphobia




PostPosted: Sat Jun 08, 2013 9:18 pm   Post subject: RE:Fun Calculus Problem

yea I differentiated wrong, haha when it came around to it I didn't do it right, doh! I also have a similar worded question and I am not sure what the second part is saying so
code:
You are given 100 cm of wire to create a circle and a square, what are the dimensions of the circle and square such that their total area is a minimum? a maximum?
I got a minimum but when he said maximum, did he just mean to put all the wire into the circle or all the wire into the square? Oh and by the way I think I got like 7 cm as the radius and 14 cm as one side to the square, forgot, did this a while ago.
Dreadnought




PostPosted: Sat Jun 08, 2013 10:21 pm   Post subject: Re: Fun Calculus Problem

Panphobia wrote:
yea I differentiated wrong, haha when it came around to it I didn't do it right, doh!

Happens a lot, when I know my answer isn't right, I usually take a break then start again from scratch later and compare.
Panphobia wrote:

I got a minimum but when he said maximum, did he just mean to put all the wire into the circle or all the wire into the square?

You're right, the answer is to use all the wire for one shape, but which shape? The second part can be done with any differentiation if you know how to compute perimeters and areas of circles and squares.

EDIT: Also 7 and 14 looks good to me.
Panphobia




PostPosted: Sat Jun 08, 2013 10:33 pm   Post subject: RE:Fun Calculus Problem

Oh thank the lord, that was my test question, my other one was pretty damn easy it was basically
code:
two different lines
y' = -x/3 + 30
y = x
you want to find the maximum area box you can make where the width is along the y axis and the two other vertices touch the two lines
that was the best I could do without a picture but basically what I did was Area = xy and that the width is equal to y' - y sooooo A = x(-x/3 + 30 - x), is that correct at all?
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Dreadnought




PostPosted: Sat Jun 08, 2013 11:31 pm   Post subject: Re: Fun Calculus Problem

Well, if my understanding of how the box is formed is correct then that looks like it would be the area.
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