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Calculus

Name: Anonymous 2007-08-10 6:45 ID:QFt2Y3wa

I cant math for poop, so I need your help.

How would one find the maximum area of a Right Angle Triange with a Hypotenuse of 21cm using calculus. 

Im stumped, I can do it with a rectangle but thats as far as i go.

Thanks :X

Name: Anonymous 2007-08-10 7:31 ID:kK4sxctx

Optimization Problem

two equations:

b^2+h^2=21^2

A=0.5*b*h

sub the first into the second, and find the maxima by taking the first derivative, and setting to 0, then solve for the variable.

Name: Anonymous 2007-08-10 10:20 ID:Heaven

You're on a rolle

Name: Anonymous 2007-08-10 10:32 ID:Heaven

>>3
you're on a rick roll'd

Name: Anonymous 2007-08-10 12:04 ID:6tZL/kgV

GILGAMESH DID IT

Name: Anonymous 2007-08-11 0:40 ID:wRdo1dEr

>>2
...

Thank you so damn much, so very very very very much.

I love you.

Name: Anonymous 2007-08-11 1:14 ID:bKYBOalH

isn't this a multivariable calculus problem?

Name: Anonymous 2007-08-11 1:36 ID:K1M7KTiQ

I suppose it would be isoceles, then the side length would be square root of (121/2) and the area would be (121/4)

Name: Anonymous 2007-08-11 1:40 ID:VETa1ZXJ

Let x be one leg of the triangle. Then the other leg of the triangle has length y = sqrt(21^2-x^2). Let the area be A = x*y/2. Substitute accordingly, differentiate, and get the fucking maximum.

Name: Anonymous 2007-08-11 16:24 ID:K1M7KTiQ

>>8
oh damn, change 121 to 441, oops

Name: Anonymous 2007-08-11 16:26 ID:HfZBu+D7

>>7
No, because you are only dealing with A(theta) or A(x).
Don't change these.
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