Maximize Volume of Trough: Find Theta Value

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SUMMARY

The discussion focuses on maximizing the volume of a trough by determining the optimal angle theta. The cross-sectional area is derived as Area = cos(theta)(1 + sin(theta)), based on the relationship between height and angle using Pythagorean theorem. Participants suggest that solving for the maximum area involves calculus or graphical methods. The conversation emphasizes the need for a clear understanding of trigonometric relationships in this context.

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  • Understanding of trigonometric functions, specifically sine and cosine.
  • Familiarity with the Pythagorean theorem.
  • Basic knowledge of calculus, particularly optimization techniques.
  • Ability to interpret geometric shapes and their properties.
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  • Explore graphical methods for visualizing area maximization problems.
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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 (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:
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Hi,
I don't 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:
Rajini said:
Hi,
I don't 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.

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

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