Math help. Surface area

[Xenus]

A covered metal triangular trough is constructed as follows:
A square shaped sheet of metal which is 120 centimeters wide and long square is folded along the center. Next, two pieces of metal in the shape of isosceles triangles are are welded to the ends. (See picture above). Finally, a metal cover is attached to the top (not shown).

We want to find the smallest and largest surfaces area that a so constructed trough can have, and at what opening angle of the triangular pieces it is attained. Proceed as follows:

Find the surfaces area as a function of the angle. (Be sure to include all five sides of the trough).

I'm trying to find the formula for the surface area

 

[OpenGL guy]

The only thing that really needs computing is the area of the ends of the trough. Let h be the height of the trough (the line bisecting the angle 2w) and let b be the base of the small triangle. Then:

cos w = h/60
sin w = b/60

h = 60 cos w
b = 60 sin w

Thus area of an end is:

A = 2 * 1/2 * b * h = 60 * 60 * sin w * cos w = 3600 sin w cos w = 1800 sin 2w

Double the result and add in the large sides and you get the final answer:

120 * 60 * 2 + 2 * 1800 sin 2w = 14400 + 3600 sin 2w

Edit: This only gives the surface area of the outside. Double the result to add in the area of the interior as well.

 

[Xenus]

Got a problem there is supposed to be a lid on the thing so would that be 7200sin(w). but the program sees somehting as wrong. And the TA seemed to think we needed cos(2w) in the problem.

Nevermid got it atleast so far 14400 + 3600 sin (2w)+14400sin(w).

 

[Xenus]

Anybody get what it is trying to say.

On its domain has one stationary point . Although there is no explicit formula for the value of itself, it is possible to derive the exact value of the cosine of . Find it:

 

[OpenGL guy]

Got a problem there is supposed to be a lid on the thing so would that be 7200sin(w). but the program sees somehting as wrong. And the TA seemed to think we needed cos(2w) in the problem.

TA seemed to think? Any TA that can't solve this should be fired.

Nevermid got it atleast so far 14400 + 3600 sin (2w)+14400sin(w).

That sounds reasonable if you only want the area of the outside. I didn't realize it was a closed object

 

[Xenus]

Probably for the derivative

 

[DemoCoder]

You don't really need to go through with the derivatives to get the min/max. The solution is constant except for the sin 2w, and it is trivial to see what value of w produces the min and max for sin 2w.

 

[Xenus]

Oh it's already due I gave up because I couldn't find out what they were asking for. Here
Another part of the same problem.

On its domain has one stationary point . Although there is no explicit formula for the value of itself, it is possible to derive the exact value of the cosine of . Find it:

 

[KimB]

On its domain has one stationary point . Although there is no explicit formula for the value of itself, it is possible to derive the exact value of the cosine of . Find it:

This is what the problem states word for word? I can't make heads or tails of it, and I'm a graduate student in physics, so you'd think I could.

 

[Xenus]

Apperently it didn't copy and past quite properly

On its domain S(w) has one stationary point w=c . Although there is no explicit formula for the value of c itself, it is possible to derive the exact value of the cosine of c. Find it:

There we go the variables were missing.

 

[K.I.L.E.R]

I remember that question.
I had one just like it and I solved it with derivatives.

Quite easily done!

 

[KimB]

Apperently it didn't copy and past quite properly

On its domain S(w) has one stationary point w=c . Although there is no explicit formula for the value of c itself, it is possible to derive the exact value of the cosine of c. Find it:

Much better. But now I'm not quite sure what they mean by a stationary point in this context.

Deriving the cosine of an angle is trivial, however: it's just adjacent over hypotenuse. If you have a specific w, then just ask yourself what the hypotenuse is (always 60 in this case), and what the adjacent side is (the height of the trough).

 

[Xenus]

That was the problem they just gave you ther variable c and expected you to come up witha an answer. That bwas a number the didn't want't h/60.

 

[Basic]

You got the area right: 14400 + 3600 sin (2w)+14400sin(w)
(Times two for in and outside.)

To find min/max, find where the derivative is 0 for possible candidates. (That's what's called a "stationary point".) Then you'll have to prove that it's a maximum.

 

[Xenus]

Yeah talked to the TA today. He could of answered my email which would of got me it much quicker and actually been useful.

 

[KimB]

To find min/max, find where the derivative is 0 for possible candidates. (That's what's called a "stationary point".) Then you'll have to prove that it's a maximum.

Yeah, that's what I was thinking, but I wasn't sure because it seemed redundant.