Help Bobmuhthol do a simple calculus problem by plotting data with a TI-83+.

[Bobmuhthol]

I have a calculus problem I need to do. I think I know the process in which to do it; but I do not know how to execute that process.

I need to find the dimensions of the rectangle that gives the most area in a region that is bounded by the parabola 16 - x^2 and the x-axis. It looks like the dimensions are going to be approximately 5 x 9.75, but I need an exact number so I would have to plot the data in a scatter plot, find the curve/line that best passes through the data, and find the maximum.


In short, I need to know how to make scatter plots with a TI-83+ because I forget how. The last time I did it was in Algebra II two years ago.

[Edited on 9-12-2005 by Bobmuhthol]

[AnticorRifling]

I'd hit the RTFM button and then use the search glossary feature n3wb.

[Xandalf]

Buy yourself a real calculator.. a TI-89.

(In other words, I don't have a clue on an 86.. sorry)

[Sean of the Thread]

I bet Warclaidtard could do it!

[DeV]

I sold my TI-83 on ebay about 3 years ago. I'm of no help.

Edited because I meant TI-83! Oopsie.

[Edited on 9-12-2005 by DeV]

[Bobmuhthol]

Does anyone at least know how on a TI-83? I think I'm going to make that my primary calculator starting right now.

[AnticorRifling]

See my post above, the same RTFM feature is integrated in the TI-83 as well. :cool:

[Bobmuhthol]

Okay I know how to make a scatter plot but I fucked this entire problem up completely. I need help very hard.

"Find the maximum area of the rectangle described below by using a numerical method and also by using a graphical method.

A rectangle is bounded by the x-axis and by the portion of the parabola y = 16 - x^2 that is on or above the x=axis."

[Janarth]

U2U me, and I'll send you CAD, screw calculators, MathCAD can solve your 20 variable equations :p As for the problem...maybe I can do it by hand...waiting for my boss to get off the phone.

Lets see...any rectangle bounded by the x axis and the parabola y = 16-x^2 would have an area of area = 2x*(16-x^2), or 32x-x^3. darea = 32-3x^2...crit point is at the square root of 10.6 repeating, obviously a max, so...x= (32/3)^1/2, so the length would be twice that, and the height whatever 16-(that value)^2 is, no?

[Janarth]

Oh...do you -have- to do it with a calculator? I just looked at your post again and it sorta seems you -have- to use a calculator, in which case I suck, and I apologize. I wasn't allowed to use a calculator in college :(

[Latrinsorm]

Originally posted by Janarth
Lets seeHe can't use derivatives.

[Wezas]

Originally posted by Janarth
I wasn't allowed to use a calculator in college :(

<insert abacus joke here>

[Janarth]

Originally posted by Latrinsorm

Originally posted by Janarth
Lets seeHe can't use derivatives.

Damn...okay...the area of the rectangle would still be 2x*(16-x^2)...where the domain is 0 <= x <= 4. So...hit the y= key on your calculator to go into the equations mode for graphing, type in y = 2x*(16-x^2), then hit graph, when the graph pops up, hit...calc I think, then ask it to find the max between 0 and 4, the answer is half the with, and the height is the y cordinate. yah?

[Bobmuhthol]

I decided to go the "fuck calculators" route, and Latrinsorm spent about an hour teaching me derivatives, etc., especially pertaining to this problem.

I now have a beautiful array of equations.

y = 16 - x ^ 2
x = x
A = 2xy
A = 32x - 2x ^ 3
dA / dx = 1 * 32 * x ^ 0 + 3 * -2 * x ^ 2
dA / dx = 32 - 6x ^ 2
0 = 32 - 6x ^ 2
6x ^ 2 = 32
3x ^ 2 = 16
x * sqrt(3) = 4
x = +- (4 / sqrt(3))
x = +- (4 * sqrt(3) / 3)
x = +- 2.309
y = 16 - 2.309 ^ 2
y = 16 - (5 + 1 / 3)
y = 10 + 2 / 3
A = 2(2.309)(10 + 2 / 3)
A = 49.258666...

Hooray. Thanks, Latrinsorm.

[Parkbandit]

Originally posted by Bobmuhthol
I have a calculus problem I need to do. I think I know the process in which to do it; but I do not know how to execute that process.

I need to find the dimensions of the rectangle that gives the most area in a region that is bounded by the parabola 16 - x^2 and the x-axis. It looks like the dimensions are going to be approximately 5 x 9.75, but I need an exact number so I would have to plot the data in a scatter plot, find the curve/line that best passes through the data, and find the maximum.


In short, I need to know how to make scatter plots with a TI-83+ because I forget how. The last time I did it was in Algebra II two years ago.

[Edited on 9-12-2005 by Bobmuhthol]

[Keller]

I don't remember any of this shit. It's sad.

Good luck on your problem, Alex.

[Jayvn]

You know... you actually aren't going to use that in real life alex... Unless you get one of those 'jobs'... and no one wants to do that..

[Ylena]

So fucking glad that I was a liberal arts major...

[Bobmuhthol]

I laugh at liberal arts majors.

No offense.

[longshot]

Enginerds are a dime a dozen. Don't kid yourself...

[Ylena]

Go ahead and laugh, I'm a liberal arts major with a Big Oil job, and I've got all the enginerds begging me for my help with the process network.

Fuck calculus in the ear.

[Doyle Hargraves]

Fuck calculus in the ear.

With a high powered dildo coated in sandpaper.

[Bobmuhthol]

<<Go ahead and laugh, I'm a liberal arts major with a Big Oil job, and I've got all the enginerds begging me for my help with the process network.

Fuck calculus in the ear.>>

I'm still laughing. :)

[Dwarven Empath]

Damn they still teach Calc...I did that in 1982.

And I still havent used it since.

[Bobmuhthol]

Surprisingly, they still teach English, too.

And math doesn't change.

[Latrinsorm]

Thanks, Latrinsorm. Welcome. Calculus f-f-f-for life.