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Simple Calculus Word Problem (using derivatives to anaylze function models)

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chaosblack
#1
Feb27-08, 09:42 PM
P: 16
Hello, new here, first post. Just need some help with homework.

Question One

1. The problem statement, all variables and given/known data

This norman window is made up of a semicircle and a rectangle. The total perimeter of the window is 16 cm. What is the maximum area?

**
* * <<< Semicircle
*****
| | <<< Rectangle
L | |
______
D
2. Relevant equations

P (total) = 2L + D + (pi * d)

A (total) = D * L + (pi(d/2)^2)/2)


3. The attempt at a solution

What I did was using this equation:
16 = 2L + D + ((pi * d)/2)
L = 8 - d/2 - ((pi * d)/4)

A = D (8 - d/2 - ((pi * d)/4)) + (pi (d/2)^2)/2)
A = 8d - (d^2)/2
A' = 8 - d
Let 0 = A' to find critical value
then 8 = d.

When I sub that back into the original equation, I get L as a value less than 8, which doesn't make sense. (I think it works out to be L = 4 - pi)

I'm pretty much lost, sorry if this is too messy to read, any help would be appreciated. Thanks
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Dick
#2
Feb27-08, 10:55 PM
Sci Advisor
HW Helper
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P: 25,228
You made a great start. But where did you get "16 = 2L + D + ((pi * d)/2)"?? The "/2" wasn't in your original expression for P. You are just making algebraic mistakes.
RyanSchw
#3
Feb27-08, 11:05 PM
RyanSchw's Avatar
P: 37
I think you're doing fine up until this point:

A = D (8 - d/2 - ((pi * d)/4)) + (pi (d/2)^2)/2)

Which should simplify into

[tex]

A = 8*D - \frac{D^2}{2} - \frac{pi*D^2}{4} + \frac{pi*D^2}{8}
[/tex]

You would then go on to take the derivate and then set it to zero and solve for your D value

I've been beaten =(

chaosblack
#4
Feb27-08, 11:31 PM
P: 16
Simple Calculus Word Problem (using derivatives to anaylze function models)

Quote Quote by Dick View Post
You made a great start. But where did you get "16 = 2L + D + ((pi * d)/2)"?? The "/2" wasn't in your original expression for P. You are just making algebraic mistakes.
ah sorry, its actually supposed to be "/2", that way its half the area, sorry the drawing didnt show up. its supposed to be a semi-circle connected to a rectangle.

Quote Quote by RyanSchw View Post
I think you're doing fine up until this point:

A = D (8 - d/2 - ((pi * d)/4)) + (pi (d/2)^2)/2)

Which should simplify into

[tex]

A = 8*D - \frac{D^2}{2} - \frac{pi*D^2}{4} + \frac{pi*D^2}{8}
[/tex]

You would then go on to take the derivate and then set it to zero and solve for your D value

I've been beaten =(
sorry, that was a typing error as well haha.

A = L * D + (pi*d)/2
which becomes

A = 8 - d/2 - ((pi * d)/4)

this still doesnt work...I think i'm using the wrong equations somehow
Dick
#5
Feb27-08, 11:43 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
Why isn't there a D in all of the terms of A? I think you understand this problem perfectly well and you are using the right equations. You are simply making typographical mistakes right and left. Get a clean sheet of paper, calm down and take a stress pill and you can do this.


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