a rectangular lot is to have an area of 1600 sq. m. find the least amount of fence that could be used to enclose the area?

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a rectangular lot is to have an area of 1600 sq. m. find the least amount of fence that could be used to enclose the area?

Mathematics
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|dw:1433649074172:dw|
Let A = xy 2x + 2y = 1600
Since we are looking for the "least" amount of fencing, we have to differentiate (or even twice) to see that the outcome is a minimum

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why does it became 2x +2y sir?
2x + 2y is the perimeter of the rectangle. So lets make 'x' the subject hence -> x = 1600 - 2y 2 Now we resub the x value into A = xy -> sub that in and start differentiating -> Double differentiate if you have to find that the value is indeed a minimum.
rectangle with given area but with minimum perimeter: |dw:1433659554258:dw|
A=xy since x=y A=y*y 1600=y^2 y=40=x
how do you know y=x will give maximum? are you allowed to assume that?
from the perimeter: 2x+2y=P 2(40)+2(40)=P P=160 therefore P=160 is the answer
that seems to be correct but are you allowed to assume the max is given by y=x are you do you have to show that?
the min*
by differentiating you can get y=x
ok your answer seems to be fine as long as you also show in your paper how you came to the y=x part
1 Attachment
just see the attached file
I know I would have done this: \[xy=1600 \\ \text{ so that means we can write } y=\frac{1600}{x} \\ \text{ and we also have \to minmize } P=2x+2y \\ P=2x+2(\frac{1600}{x}) \\ P=2x+\frac{3200}{x} \\ \text{ then I would differentiate and set } P'=0 \text{ and solve for } x \]
and then once I found x from that equation plug into xy=1600 to find y
yeah you can also do that
we came up with the same answer
actually i am the one who asked this question because this is our assignment in school. unfortunately i didnt understand the explanation said by the first person. that's why I searched along the net. and luckily got the right answer :-)
@freckles thanks for your clear explanation
hmm it looks like he got the area and perimeter mixed up
yep that is exactly what happened
oooh yeah..
http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx I really like paul's notes he does other things besides optimization problems
ooh thank you for this sir.. it will be a great help for me
also that way you showed is an awesome way I have never done optimization problem the way it was done in that pdf
ooh thanks
doing Paul's first problem like that: we have: \[A'=y+xy' \\ 0=1+2y' \\ \text{ so } y'=\frac{-1}{2} \\ \text{ so } A'=y+x(\frac{-1}{2}) \\ \text{ but } 2y=500-x \\ \text{ and multiplying } \frac{1}{2} \text{ on both sides gives } y=\frac{500}{2}-\frac{1}{2}x \\ \text{ so } A'=\frac{500}{2}-\frac{1}{2}x+x(\frac{-1}{2}) \\ A'=\frac{500}{2}-x \\ A'=0 \text{ when } x=\frac{500}{2}=250 \]
that second equation came from differentiating 500=x+2y
w.r.t. x
anyways neat
oooh yes i got it

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