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

by chaosblack
Tags: anaylze, calculus, derivatives, function, models, simple, word
 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
 HW Helper Sci Advisor Thanks P: 24,454 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.
 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 $$A = 8*D - \frac{D^2}{2} - \frac{pi*D^2}{4} + \frac{pi*D^2}{8}$$ 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 =(
P: 16

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

 Quote by Dick 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 by RyanSchw 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 $$A = 8*D - \frac{D^2}{2} - \frac{pi*D^2}{4} + \frac{pi*D^2}{8}$$ 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
 HW Helper Sci Advisor Thanks P: 24,454 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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