1. ## Optimization Help!

The problem says...

The trough in the figure is to be made to the dimension shown. Only the angle delta can be varied. What value of delta will maximize the trough's volume. (I attached a picture)

Can someone please show me and explain to me how I would do a problem like this!

2. Originally Posted by KarlosK
The problem says...

The trough in the figure is to be made to the dimension shown. Only the angle theta can be varied. What value of theta will maximize the trough's volume. (I attached a picture)

Can someone please show me and explain to me how I would do a problem like this!
Hi Karlos,

The volume of the trough can be subdivided into the central cube
and the two outside pyramids.

To calculate volume, use

$\displaystyle V=(cross-sectional\ area)(length)$

hence you must express the trough's height "h" in terms of $\displaystyle \theta$

$\displaystyle cos\theta=\frac{h}{1}=h$

The cross-sectional area of the right and left triangles are

$\displaystyle \frac{1}{2}(h)(1)sin\theta=\frac{1}{2}cos\theta\ sin\theta$

Hence the volume of the trough is

$\displaystyle V=20cos\theta+20(2)\frac{1}{2}cos\theta\ sin\theta=20cos\theta(1+sin\theta)$

Theta giving max volume is found by differentiating this and setting it to zero, using the product rule of differentiation

$\displaystyle \frac{d}{d\theta}\left(20cos\theta(1+sin\theta)\ri ght)$

$\displaystyle 20\left(cos\theta\ cos\theta+(1+sin\theta)(-sin\theta)\right)=0$

$\displaystyle \left(cos^2\theta-sin\theta-sin^2\theta\right)=0$

$\displaystyle 1-sin^2\theta-sin\theta-sin^2\theta=0$

$\displaystyle 2sin^2\theta+sin\theta-1=0$

$\displaystyle (2sin\theta-1)(sin\theta+1)=0$

For an acute angle use the first factor

$\displaystyle 2sin\theta=1$

$\displaystyle sin\theta=\frac{1}{2}$

$\displaystyle \theta=sin^{-1}\frac{1}{2}=30^o$

3. Originally Posted by KarlosK
The problem says...

The trough in the figure is to be made to the dimension shown. Only the angle delta can be varied. What value of delta will maximize the trough's volume. (I attached a picture)

Can someone please show me and explain to me how I would do a problem like this!
basic trig w/ the two congruent right triangles on each side ...

trapezoid height = $\displaystyle \cos{\theta}$

top trapezoid base = $\displaystyle 1 + 2\sin{\theta}$

cross-sectional area ...

$\displaystyle A = \frac{\cos{\theta}}{2}[(1+2\sin{\theta}) + 1]$

proceed to find the value of $\displaystyle \theta$ that will maximize the cross-sectional area, and hence, the volume.

4. Originally Posted by skeeter
basic trig w/ the two congruent right triangles on each side ...

trapezoid height = $\displaystyle \cos{\theta}$

top trapezoid base = $\displaystyle 1 + 2\sin{\theta}$

cross-sectional area ...

$\displaystyle A = \frac{\cos{\theta}}{2}[(1+2\sin{\theta}) + 1]$

proceed to find the value of $\displaystyle \theta$ that will maximize the cross-sectional area, and hence, the volume.
Skeeter I have Area= cos(theta)+1/2sin2(theta) Is that incorrect? I attached how I came to that...

5. Originally Posted by Archie Meade
Hi Karlos,

The volume of the trough can be subdivided into the central cube
and the two outside pyramids.

To calculate volume, use

$\displaystyle V=(cross-sectonal\ area)(length)$

hence you must express the trough's height "h" in terms of $\displaystyle \theta$

$\displaystyle cos\theta=\frac{h}{1}=h$

The cross-sectional area of the right and left triangles are

$\displaystyle \frac{1}{2}(h)(1)sin\theta=\frac{1}{2}cos\theta\ sin\theta$

Hence the volume of the trough is

$\displaystyle V=20cos\theta+20(2)\frac{1}{2}cos\theta\ \sin\theta=20cos\theta(1+sin\theta)$

Theta giving max volume is found by differentiating this and setting it to zero, using the product rule of differentiation

$\displaystyle \frac{d}{d\theta}\left[20cos\theta(1+sin\theta)\right]$

$\displaystyle 20[cos\theta\ cos\theta+(1+sin\theta)(-sin\theta)]=0$

$\displaystyle 20\left(cos^2\theta-sin\theta-sin^2\theta\right)=0$

$\displaystyle 1-sin^2\theta-sin\theta-sin^2\theta=0$

$\displaystyle 2sin^2\theta+sin\theta-1=0$

$\displaystyle (2sin\theta-1)(sin\theta+1)=0$

For an acute angle use the first factor

$\displaystyle 2sin\theta=1$

$\displaystyle sin\theta=\frac{1}{2}$

$\displaystyle \theta=sin^{-1}\frac{1}{2}=30^o$

How come you have sin^-1 instead of just sin at the end of this? Thanks for the help

6. $\displaystyle A = \frac{\cos{\theta}}{2}[(1+2\sin{\theta}) + 1]$

$\displaystyle A = \frac{\cos{\theta}}{2}[2+2\sin{\theta}]$

$\displaystyle A = \cos{\theta}(1 + \sin{\theta})$

7. Originally Posted by skeeter
$\displaystyle A = \frac{\cos{\theta}}{2}[(1+2\sin{\theta}) + 1]$

$\displaystyle A = \frac{\cos{\theta}}{2}[2+2\sin{\theta}]$

$\displaystyle A = \cos{\theta}(1 + \sin{\theta})$
So after I have the area what would I do? Differentiate it? Can you tell me what steps I need to complete and then I can try to solve it?

8. Originally Posted by KarlosK
So after I have the area what would I do? Differentiate it? Can you tell me what steps I need to complete and then I can try to solve it?
find $\displaystyle \frac{dA}{d\theta}$ and determine the value of $\displaystyle \theta$ that maximizes $\displaystyle A$.

9. Originally Posted by KarlosK
How come you have sin^-1 instead of just sin at the end of this? Thanks for the help
At the end, we have the sine of the angle..

$\displaystyle sin\theta=\frac{1}{2}$

the "inverse sine" retrieves the angle

$\displaystyle arcsin\left(\frac{1}{2}\right)=sin^{-1}\left(\frac{1}{2}\right)=30^o$

cross-sectional area $\displaystyle =cos\theta+\frac{1}{2}sin2\theta$

this is correct, as

$\displaystyle \frac{1}{2}sin2\theta=\frac{1}{2}2sin\theta\ cos\theta=sin\theta\ cos\theta$

Since the volume is 20 times the cross-sectional area,
you only need to differentiate the cross-sectional area and set the result to zero.
Then discover the value of theta that causes the derivative to be zero.

$\displaystyle \frac{d}{d\theta}\left[cos\theta(1+sin\theta)\right]=0$

Use the product rule as shown earlier

or

$\displaystyle \frac{d}{d\theta}\left(cos\theta+\frac{1}{2}sin2\t heta\right)=0$

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