|
A child wants to build a tunnel using three equal boards, each 4 feet wide, one for the top and one for each side as shown. The two sides are to be tilted at equal
angles theta to the floor. What is the maximum cross sectional area A that can be achieved?
Ok here's what I did: drew a diagram.
http://img392.imageshack.us/my.php?image=trapezoidsg2.jpg
I basically get a trapezoid. The top part of the trapezoid is 4 feet. The bottom angles, which are equal, are "theta". The two sides of the trapezoid are also 4 feet.
How I approached this problem:
I divided the trapezoid into three pieces. There is now a right triangle, a rectangle/square, and another right triangle.
Since we are trying to find the maximum area, I tried to find an equation for the Area,A of this trapezoid.
Area of the Triangle 1= (0.5*b*h)
Area of the Square= (b*h)
Area of the Triangle 2= (0.5*b*h)
In other words if I substitute everything (like x, y, and 4), my equation for the Area is A = ((2x+4+4)*y)/(2)
Now for theta/angle (i'll label it @), it is tan(@) = y/x
Multiply x to both sides and get x*tan(@)=y. Sub in 4 for x and y = 4sin(@) and x = 4cos(@)
Now I am confused what to do here.
I think that since we're looking for the maximum value of the area, we need to find the derivative of da / dx or something like that.
But for the A, my calculates were the following:
x=4cos(@)
y=4sin(@)
A = ( (2x+4+4)*y ) / (2) as stated above, or
A = 1/2 * (2x+8)*y
susbsitute x and y and simplify to get:
A=4xcos(@) + 4 + 2sin(@)
Am I on the right track? Do I find the derivative of the Area after that?
Help greatly appreciated!
/back to studying MATLAB ;;.
|
|
|
you should get A=16sin@ + 16sin@cos@
differentiate and solve for A'=0. Remember to restrict 0<=@<=90 (working in degrees not radians). Check solutions for second derivative.
|
On November 06 2008 13:12 megastarcraft20 wrote:
the answer is 42
You're useless, actually.
You're definitely on the right track. Assuming you got all the math you stated above correctly, take the derivative of the function in theta and find the maximum (where theta equals zero). You might have multiple answers and make sure that a) you're in radians b) the radian measure is between zero and pi/2, c) they're maxes, not minimums. After that, you should only end up with one 'legit' answer.
Edit:
You take the derivative of dA/d@. It's the change of A with respect to @, so it's dA/d@.
|
ew. wtf are you doing. max area of fences ftw, this is ftl. find the number, number^2 = max area.
|
actually no its not. That's the maximum area for a given perimeter. Here that's not an issue. Your final answer should be 12*sqrt(3), with @=60.
|
On November 06 2008 13:36 Raithed wrote:
ew. wtf are you doing. max area of fences ftw, this is ftl. find the number, number^2 = max area.
LOL you fucking fail so much at calc this is ridiculous
|
I got the answer.
the answer is 20.7
thx.
|
16935 Posts
On November 06 2008 13:36 Raithed wrote:
ew. wtf are you doing. max area of fences ftw, this is ftl. find the number, number^2 = max area.
I know you're trying to be helpful, but please post help when you're absolutely sure it's the correct response.
But hey, it's the thought that counts!
|
What's with people and using TL.net for homework.
*thinks*
I should do that from now on.
|
On November 06 2008 14:37 Empyrean wrote:Show nested quote +On November 06 2008 13:36 Raithed wrote:
ew. wtf are you doing. max area of fences ftw, this is ftl. find the number, number^2 = max area. I know you're trying to be helpful, but please post help when you're absolutely sure it's the correct response. But hey, it's the thought that counts!
no its not raithed is a fucking idiot who fails at calc who shouldnt be making his dumbass calc help threads, much less trying to "contribute" to those of others
|
On November 06 2008 15:42 SiegeTanksandBlueGoo wrote:
What's with people and using TL.net for homework.
*thinks*
I should do that from now on.
Using? Dude, it's not like I posted the problem, and said "hey do it"
I gave my work in, asked them what I did wrong, and asked what I should do next.
Never asked for answers.
|
clzziquai u would've been a very good student if you just went to office hour w/ all the works u've shown. 
|
On November 06 2008 17:25 evanthebouncy! wrote:clzziquai u would've been a very good student if you just went to office hour w/ all the works u've shown.
Haha really? :o
I never go to office hours >_>
|
office hours is like insta-A on homework
|
^ Doubt it lol.
Because our TA does the grading.
My teacher doesn't grade them -_-, and he's the one with office hours.
Whatever, not that big of a deal ^^;;
|
On November 06 2008 13:36 Raithed wrote:
ew. wtf are you doing. max area of fences ftw, this is ftl. find the number, number^2 = max area.
raithed why are you such a novice at school?
|
So, you gotta do this junk too huh? I hate workshops... When do you have your recitation?
|
lol yea.
where do you go to school??
|
Rutgers New Brunswick. You?
|
|
|
|
|
|
EPT
