Need help doing this math...

[RESmonkey]

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part B. I tried finding the equation for area, hoping to derive it, set it to zero, and test critical and endpoints. Well, the area forumla ended up containing more than one variable.

What should I do?

 

[warmodder]

I hear there's a soda bottle that does the work for you...

 

[George P Burdell]

Sorry kid, I have my own homework to do.

 

[RESmonkey]

^...

I managed to do it, I think. I set x = w. I'll find out tomorrow if it's right or not, unless someone actually bothers to do it (NOTE: It's not hard, but it may confuse).

 

[TecHNooB]

nm, those answers were wrong T__T ill try again.

 

[RESmonkey]

I think I'm wrong. I got w = 0, or the initial point, as the minimum value.

I wrote this answer (in pen): "? According to the Mean Value Theorem & some poor math skills, I'm going to conclude that A(w) has a minimum @ w = 0."

 

[chuckywang]

Originally posted by: RESmonkey
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part B. I tried finding the equation for area, hoping to derive it, set it to zero, and test critical and endpoints. Well, the area forumla ended up containing more than one variable.

What should I do?

A line with slope of -2w and a point at (w,6-w^2).

(y+w^2-6) = -2w*(x-w)

When x = 0, y = w^2 + 6
When y = 0, x = 3/w + w/2.

The area of the triangle is

1/2*(6+w^2)*(3/w + w/2).

a) When w = 1, A(1) = 49/4.

 

[RESmonkey]

Yeah, I understand/successfully did part A. (Thanks, though. It confirms my answer is right )

 

[hypn0tik]

Part (b) is an optimization problem. Use the function for the area you found in part (a), i.e. A(w) = ....

Find A'(w) and set it equal to zero to find the critical points.

Find A"(w) and test the critical points to check if they are maxima or minima.

 

[chuckywang]

If you expand the expression, you get 18/w + 3w + 3w + w^3/2. = 18/w + 6w + w^3/2.

The derivative is -18/w^2 + 6 + 3w^2/2, so w^4 + 4w^2 - 12 = 0.

Therefore (w^2 +6)(w^2 -2) = 0. The only critical point in the range is w = sqrt(2).

 

[RESmonkey]

My A(w) is funky. The one single critical point I found was 0, because A'(w) = Sqrt.rt(6)/2 * w (it's probably wrong). That crit. point is also an endpoint...

 

[hypn0tik]

Originally posted by: RESmonkey
My A(w) is funky. The one single critical point I found was 0, because A'(w) = Sqrt.rt(6)/2 * w (it's probably wrong). That crit. point is also an endpoint...

Show your work. We might be able to spot the mistake (if there is any).

 

[RESmonkey]

Originally posted by: chuckywang
If you expand the expression, you get 18/w + 3w + 3w + w^3/2. = 18/w + 6w + w^3/2.

The derivative is -18/w^2 + 6 + 3w^2/2, so w^4 + 4w^2 - 12 = 0.

Therefore (w^2 +6)(w^2 -2) = 0. The only critical point in the range is w = sqrt(2).

Expand which expression?

 

[hypn0tik]

Originally posted by: RESmonkey

Originally posted by: chuckywang
If you expand the expression, you get 18/w + 3w + 3w + w^3/2. = 18/w + 6w + w^3/2.

The derivative is -18/w^2 + 6 + 3w^2/2, so w^4 + 4w^2 - 12 = 0.

Therefore (w^2 +6)(w^2 -2) = 0. The only critical point in the range is w = sqrt(2).

Expand which expression?

The expression for A(w) in one of his earlier posts.

 

[RESmonkey]

Okay:

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I can't make a clean link at all

 

[RESmonkey]

Holy jesus. I was on the right track (after looking at chunkywangs post and analyzing it). I didn't use x and y at the same time, though (that's prolly why I messed it up). I used 0 and y, and x and 0.

I'm going to test root 2 in a sec with the MVT.

 

[RESmonkey]

Yes! A(w) has a minimum at w = root 2, as proven by the mean value theorem!

Hehe...

Thanks EPICALLY, you guys make math fun

 

[Saint Michael]

I got an extremely close answer (although not precise because I did some rounding and what not) with an extremely complex series of equations that took over an hour. I think I need some more knowledge.

 

[bonkers325]

w = 1.414

find the slope of the tangent line in terms of w. you will get an equation that looks like y = w^# + # and x = #*w + #/w

area of a triangle = .5 * b * h, which in your case, is .5 * x * y

then optimize area, the answer is +/- 1.414 and since your range is b/w 0 and 6, your only answer is 1.414

 

[RESmonkey]

Uh...scan? I'd like to know how you did it.

 

[RESmonkey]

I meant Saint Michael (who claims to have done it a harder way).

Yeah (bonkers), I understand the way to do it now. My
My eyes hurt now...off to bed.

Thanks all

 

[bonkers325]

Originally posted by: RESmonkey
Uh...scan? I'd like to know how you did it.

i did it in mathcad and then i closed it.

f'(w) = -2w

y - y0 = m(x - x0)

m = -2w
x0 = w
y0 = 6 - w^2

plug that into point slope forumla and u will eventually get y = w^2 - w2x + 6, you know your boundaries which are x = 0 and y = 0, so u can find x and y respectively

you get y = w^2 + 6 and x = .5w + 3/w

plug into A = .5 x y, find A' and solve for the roots of A'

 

[Saint Michael]

Originally posted by: RESmonkey
I meant Saint Michael (who claims to have done it a harder way).

Yeah (bonkers), I understand the way to do it now. My
My eyes hurt now...off to bed.

Thanks all

It's not so much that the ultimate process I ended up with was different, it was that I was going down a bunch of false paths and rearranging my equations over and over again trying to figure it out. The simple answer was eluding me.