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?

- anonymous

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- anonymous

|dw:1433649074172:dw|

- anonymous

Let A = xy
2x + 2y = 1600

- anonymous

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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## More answers

- anonymous

why does it became 2x +2y sir?

- anonymous

@ganeshie8

- anonymous

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.

- anonymous

rectangle with given area but with minimum perimeter:
|dw:1433659554258:dw|

- anonymous

A=xy
since x=y
A=y*y
1600=y^2
y=40=x

- freckles

how do you know y=x will give maximum?
are you allowed to assume that?

- anonymous

from the perimeter:
2x+2y=P
2(40)+2(40)=P
P=160
therefore P=160 is the answer

- freckles

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?

- freckles

the min*

- anonymous

by differentiating you can get y=x

- freckles

ok your answer seems to be fine as long as you also show in your paper how you came to the y=x part

- anonymous

##### 1 Attachment

- anonymous

just see the attached file

- freckles

I know I would have done this:
\[xy=1600 \\ \text{ so that means we can write } y=\frac{1600}{x} \\ \text{ and we also have \to minmize } P=2x+2y \\ P=2x+2(\frac{1600}{x}) \\ P=2x+\frac{3200}{x} \\ \text{ then I would differentiate and set } P'=0 \text{ and solve for } x \]

- freckles

and then once I found x from that equation plug into xy=1600 to find y

- anonymous

yeah you can also do that

- anonymous

we came up with the same answer

- anonymous

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 :-)

- anonymous

@freckles thanks for your clear explanation

- freckles

hmm it looks like he got the area and perimeter mixed up

- freckles

yep that is exactly what happened

- anonymous

oooh yeah..

- freckles

http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx
I really like paul's notes
he does other things besides optimization problems

- anonymous

ooh thank you for this sir.. it will be a great help for me

- freckles

also that way you showed is an awesome way
I have never done optimization problem the way it was done in that pdf

- anonymous

ooh thanks

- freckles

doing Paul's first problem like that: we have:
\[A'=y+xy' \\ 0=1+2y' \\ \text{ so } y'=\frac{-1}{2} \\ \text{ so } A'=y+x(\frac{-1}{2}) \\ \text{ but } 2y=500-x \\ \text{ and multiplying } \frac{1}{2} \text{ on both sides gives } y=\frac{500}{2}-\frac{1}{2}x \\ \text{ so } A'=\frac{500}{2}-\frac{1}{2}x+x(\frac{-1}{2}) \\ A'=\frac{500}{2}-x \\ A'=0 \text{ when } x=\frac{500}{2}=250 \]

- freckles

that second equation came from differentiating 500=x+2y

- freckles

w.r.t. x

- freckles

anyways neat

- anonymous

oooh yes i got it

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