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anonymous
 one year ago
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
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433649074172:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let A = xy 2x + 2y = 1600

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since we are looking for the "least" amount of fencing, we have to differentiate (or even twice) to see that the outcome is a minimum

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why does it became 2x +2y sir?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02x + 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
 one year ago
Best ResponseYou've already chosen the best response.0rectangle with given area but with minimum perimeter: dw:1433659554258:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A=xy since x=y A=y*y 1600=y^2 y=40=x

freckles
 one year ago
Best ResponseYou've already chosen the best response.2how do you know y=x will give maximum? are you allowed to assume that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from the perimeter: 2x+2y=P 2(40)+2(40)=P P=160 therefore P=160 is the answer

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0by differentiating you can get y=x

freckles
 one year ago
Best ResponseYou've already chosen the best response.2ok your answer seems to be fine as long as you also show in your paper how you came to the y=x part

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just see the attached file

freckles
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.2and then once I found x from that equation plug into xy=1600 to find y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah you can also do that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we came up with the same answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually 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
 one year ago
Best ResponseYou've already chosen the best response.0@freckles thanks for your clear explanation

freckles
 one year ago
Best ResponseYou've already chosen the best response.2hmm it looks like he got the area and perimeter mixed up

freckles
 one year ago
Best ResponseYou've already chosen the best response.2yep that is exactly what happened

freckles
 one year ago
Best ResponseYou've already chosen the best response.2http://tutorial.math.lamar.edu/Classes/CalcI/Optimization.aspx I really like paul's notes he does other things besides optimization problems

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh thank you for this sir.. it will be a great help for me

freckles
 one year ago
Best ResponseYou've already chosen the best response.2also that way you showed is an awesome way I have never done optimization problem the way it was done in that pdf

freckles
 one year ago
Best ResponseYou've already chosen the best response.2doing Paul's first problem like that: we have: \[A'=y+xy' \\ 0=1+2y' \\ \text{ so } y'=\frac{1}{2} \\ \text{ so } A'=y+x(\frac{1}{2}) \\ \text{ but } 2y=500x \\ \text{ and multiplying } \frac{1}{2} \text{ on both sides gives } y=\frac{500}{2}\frac{1}{2}x \\ \text{ so } A'=\frac{500}{2}\frac{1}{2}x+x(\frac{1}{2}) \\ A'=\frac{500}{2}x \\ A'=0 \text{ when } x=\frac{500}{2}=250 \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.2that second equation came from differentiating 500=x+2y
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