Shakira plans to build a 4 side-by-side rectangular enclosure whenever/wherever she wants around her garden. SHE DECIDES TO USE HER HOUSE AS ONE SIDE OF THE GARDEN. She has 1800 ft of fence. What would be the length and width of the garden for maximum AREA? What is the max area?
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OpenStudy (anonymous):
calc or no? doesn't make a difference, just asking
OpenStudy (amenah8):
yep. calculus :)
OpenStudy (danjs):
draw it, and total the perimeter using a variable for length and width
OpenStudy (anonymous):
not needed but we can do it that way
OpenStudy (anonymous):
first off, although i am not supposed to give a direct answer, lets answer it first
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OpenStudy (anonymous):
use half of the fence for the side opposite the house, split the remaining half for the other two sides
there, i said it, now lets find it
OpenStudy (anonymous):
|dw:1453254624175:dw|
OpenStudy (amenah8):
right
OpenStudy (anonymous):
how much is left for the part of the fence not labelled?
OpenStudy (amenah8):
A=lw so A=(1800-2x)(x) ??
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OpenStudy (anonymous):
bingo
OpenStudy (anonymous):
now for the lets not use calculus part, because it is unnecessary \[A(x)=1800x-2x^2\] is a parabola that opens down
the max is at the vertex use \(-\frac{b}{2a}\) to find it
OpenStudy (danjs):
i thought the thing was 4 rectangles side by side p=5x+y,, maybe not
OpenStudy (anonymous):
or if you are a slave to calc, take the derivative, set it equal to zero and solve
you will still get \(-\frac{b}{2a}\)
OpenStudy (anonymous):
oh crap i should learn how to read !!!
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OpenStudy (anonymous):
ignore all i said, @DanJS pointed out the correct question
OpenStudy (danjs):
not sure, you may be right, enclosure is singular, not 4 of em..
OpenStudy (amenah8):
x-450
OpenStudy (amenah8):
sorry, x=450
OpenStudy (danjs):
either way, same process to solve , sorry
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OpenStudy (anonymous):
i am not sure what a "4 side by side" rectangular enclosure is
maybe it is this |dw:1453255024187:dw|
OpenStudy (anonymous):
no i am sorry, i didn't read carefully enough
OpenStudy (amenah8):
right. that's what i was thinking
OpenStudy (anonymous):
then it is \(A=x(1800-5x)\) and the solution is different
OpenStudy (amenah8):
so x is not 450?
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OpenStudy (anonymous):
not any more
OpenStudy (anonymous):
now \[A=1800x-5x^2\] so it is different
OpenStudy (anonymous):
i mean the work is the same, but the answer is different
OpenStudy (amenah8):
ok, one moment while i try to solve for x
OpenStudy (anonymous):
ok, take the derivative first of course, then set that equal to zero
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OpenStudy (amenah8):
take the derivative of 1800x-5x^2?
OpenStudy (anonymous):
yes
OpenStudy (amenah8):
A(x)= 1800x - 5x^2 --> 1st D= 1800-10x
OpenStudy (amenah8):
x= 180
OpenStudy (anonymous):
yes
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OpenStudy (amenah8):
so now, A=(180)(w)
OpenStudy (anonymous):
\[
A=180(1800-5\times 180)\]
OpenStudy (amenah8):
ah, so plug in 180 for x into x(1800-5x)
OpenStudy (anonymous):
yes
OpenStudy (amenah8):
-1296000?
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