# Optimization help!

1. Apr 26, 2010

### doublek1229

The problem states: The trough in the figure is to be made to the dimensions shown. Only the angle theta can be varied. What value of theta will maximize the troughs volume?

http://img81.imageshack.us/img81/5963/24ni3.jpg [Broken] (There is an image of the problem)

I know the height in terms of theta is h/1 = h

So the cross sectional area is

1/2(h)(1)sin(theta) = 1/2 cos(theta)sin(theta)

This is as far as I have gotten. I don't even know if this is right. Please help!

Last edited by a moderator: May 4, 2017
2. Apr 27, 2010

### Rajini

Hi,
I dont know the procedure but i can say this:
Area of a triangle = (1/2)h(x). You have two triangles so Area = h(x). 'x' is the base of triangle and h is height. Now you have some portion in-between the two triangles and its area = h.
So final area of the cross section is = $$h(x+1)$$.
Consider one triangle and we have by Pythagoras law:
$$x^2+h^2=1$$ and $$\sin\theta=x$$ so $$h^2=1-\sin^2\theta$$.
Now Area becomes = $$h(1+\sin\theta)=(1+\sin\theta)\sqrt{(1-\sin^2\theta)}$$. [Note: in this relation you have only angle].
Finally you get:
Area = $$\cos\theta(1+\sin\theta)$$.
This is all what i got. I think may be you have to solve for the area such that area becomes maximum.
Actually i don't know how to solve using formula..but by plotting i guess it is << answer deleted by berkeman >>
hope this helps.

Last edited by a moderator: Apr 27, 2010
3. Apr 27, 2010

### Staff: Mentor

Rajini, please take care not to do too much of the OP's work for them, and please do not post answers. Thanks.