Optimization of a Trough

pts) https://webwork.math.ohio-state.edu/...045bebae81.png
A covered metal triangular trough is constructed as follows:

A square shaped sheet of metal which is 200 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 https://webwork.math.ohio-state.edu/...575ddb21c1.png that a so constructed trough can have, and at what opening angle https://webwork.math.ohio-state.edu/...9b8ccd6cf1.png 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).
https://webwork.math.ohio-state.edu/...5b662fc5f1.png
The natural domain of https://webwork.math.ohio-state.edu/...fbc39b2671.png is an interval with left endpoint https://webwork.math.ohio-state.edu/...4f6c767fe1.png and right endpoint https://webwork.math.ohio-state.edu/...4f6c767fe1.png.
On its domain https://webwork.math.ohio-state.edu/...eb479d0001.png has one stationary point https://webwork.math.ohio-state.edu/...7d610c7571.png. Although there is no explicit formula for the value of https://webwork.math.ohio-state.edu/...25e85dc621.png itself, it is possible to derive the exact value of the cosine of https://webwork.math.ohio-state.edu/...25e85dc621.png. Find it:
https://webwork.math.ohio-state.edu/...88d04e42c1.png
Find from this the surfaces area at the stationary point:
https://webwork.math.ohio-state.edu/...0e50ebde01.png Find also the global minimal value of https://webwork.math.ohio-state.edu/...575ddb21c1.png on its natural domain:
https://webwork.math.ohio-state.edu/...7a3e4ac791.png

Ok I have tried doing the first part and was pretty confident that I had the right answer but the computer was telling me that I was wrong. I know that 0 is the endpoint of the domain of w. Please Help

Could you show us what you've done?

ok i just now figured out the first part
but i know surface area is the area of all of the sides added to each other
i know that the square folded into a v has side lengths of 200 so A=lw so that Area=40,000
the area of a triangle is A=.5bh
h=100cos(w) and b=100sin(w)
there are 2 b for each triangle so the area of one of the triangles is A=.5*100cos(w)*2*100sin(w)=10000cos(w)sin(w)
there are two triangles so their area is A=20000sin(w)cos(w)
for the rectangle on top A=lw
the length is 200 and the width is 2b so A=200*2*100sin(w)=40000sin(w)
Add all these areas up and you get
Surface Area=40000+20000sin(w)cos(w)+40000sin(w)

i also know that the domain of w is from (0,pi/2) because it can't have an angle greater than 90 degrees
I am stuck on the next part where it is talking about a stationary value (which is where the first derivative=0) and cos(c)

Yes, find the derivative of S(w).

At w = c, S'(c) = 0.

i found the derivative to of S(w) to be S'(w)=20000(cos^2(w)-sin^2(w))+40000cos(w)
i set s'(w)=0 and got w=1.196062
i'm not sure what it wants when it asks for cos(c)=?

What about using the identity;



Choose the one which you think will lead you to the answer (Smile)

ok im confused which one to use
i have
cos^2(w)-sin^2(w)+2cos(w)=0

Quote:

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Originally Posted by rawkstar

ok im confused which one to use
i have
cos^2(w)-sin^2(w)+2cos(w)=0

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w is now c.

And use 2cos^2A - 1 to get everything in terms of cos.

so correct me if im wrong, you're telling me to change
cos^2(c)-sin^2(c)+2cos(c)=0
to
2cos^2(c)-1+2cos(c)=0 ?

Yes.

However, your problem seems you give you all sorts of 'odd' numbers. I'm afraid you'll have to express it into radicals, using the equadratic formula.