Maximize Volume of Trough: Find Theta Value

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In summary, the problem states that 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?
  • #1
doublek1229
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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!
 
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  • #2
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 = [tex]h(x+1)[/tex].
Consider one triangle and we have by Pythagoras law:
[tex]x^2+h^2=1[/tex] and [tex]\sin\theta=x[/tex] so [tex]h^2=1-\sin^2\theta[/tex].
Now Area becomes = [tex]h(1+\sin\theta)=(1+\sin\theta)\sqrt{(1-\sin^2\theta)}[/tex]. [Note: in this relation you have only angle].
Finally you get:
Area = [tex]\cos\theta(1+\sin\theta)[/tex].
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.
 
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  • #3
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 = [tex]h(x+1)[/tex].
Consider one triangle and we have by Pythagoras law:
[tex]x^2+h^2=1[/tex] and [tex]\sin\theta=x[/tex] so [tex]h^2=1-\sin^2\theta[/tex].
Now Area becomes = [tex]h(1+\sin\theta)=(1+\sin\theta)\sqrt{(1-\sin^2\theta)}[/tex]. [Note: in this relation you have only angle].
Finally you get:
Area = [tex]\cos\theta(1+\sin\theta)[/tex].
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.
 

What is the purpose of maximizing the volume of a trough?

The purpose of maximizing the volume of a trough is to find the optimal shape and size of the trough that can hold the maximum amount of liquid. This can be useful in various industries such as agriculture, manufacturing, and construction.

What factors affect the volume of a trough?

The volume of a trough is affected by its dimensions, such as length, width, and height. The material used to construct the trough can also impact its volume, as well as any additional features such as handles or spouts.

How is the theta value determined in maximizing the volume of a trough?

The theta value is determined through mathematical calculations and optimization techniques. It involves finding the optimal angle at which the trough should be constructed in order to maximize its volume.

What is the relationship between the theta value and the volume of a trough?

The theta value and the volume of a trough have an inverse relationship. This means that as the theta value increases, the volume of the trough decreases, and vice versa. The goal is to find the optimal theta value that will result in the maximum volume of the trough.

Can the volume of a trough be maximized without changing the theta value?

No, the theta value is a crucial factor in maximizing the volume of a trough. Changing the theta value allows for different angles and dimensions to be considered, which can result in a larger volume. Therefore, in order to achieve the maximum volume of a trough, the theta value must be optimized.

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