
#1
Feb2708, 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 



#2
Feb2708, 10:55 PM

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P: 25,165

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.




#3
Feb2708, 11:05 PM

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 =( 



#4
Feb2708, 11:31 PM

P: 16

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



#5
Feb2708, 11:43 PM

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P: 25,165

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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