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anonymous

  • one year ago

"Explain the statement: Of all rectangles with a fixed perimeter, the square measures out to the largest area." How would you go about proving this to be true?

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  1. anonymous
    • one year ago
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    I guess I cant use the diagonal.. they're not consistent across different base & height.

  2. anonymous
    • one year ago
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    GIVEN area = base * height perimeter = 2base + 2height EQUATIONS area[x,y] = xy perimeter[x,y] = 2x+2y

  3. anonymous
    • one year ago
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    So maybe I need to prove that the argument of the statement is that base/height=1 will yield the highest value for base*height

  4. perl
    • one year ago
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    |dw:1434755324547:dw|

  5. perl
    • one year ago
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    Let define the fixed perimeter , call it C C = 2x + 2y We want to maximize the area of the rectangle. Area (A) of a rectangle is : A = x * y Area is now a function of two variables x and y. It would be easier to maximize the area if we could eliminate one of the variables. How about eliminating y. Use the first equation to solve for y in terms of x. C - 2x = 2y (C - 2x ) / 2 = y Now substitute. A = x * ( C - 2x)/2

  6. perl
    • one year ago
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    With me up to here?

  7. anonymous
    • one year ago
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    I think so.. was just thinking wow...

  8. anonymous
    • one year ago
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    Im not sure where to take it form here though..

  9. perl
    • one year ago
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    At this point you can you use calculus or use your knowledge of parabolas

  10. anonymous
    • one year ago
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    so Im guessing then a derivative ties into this somehow ..

  11. anonymous
    • one year ago
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    wow again..

  12. anonymous
    • one year ago
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    do we need to do the same thing for y?

  13. perl
    • one year ago
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    We have a quadratic function A = x * ( C - 2x)/2 bring the 1/2 in front A = x/2 * (C - 2x) distribute A = Cx/2 - 2x^2/2 A = Cx/2 - x^2 A = -x^2 + Cx/2 dA/dx = -2x + C/2 According to Fermat's theorem the extrema occurs when dA/dx = 0 dA /dx = 0 -2x + C/2 = 0 -2x = -C/2 2x = C/2 x = C/4 Plug this into C = 2x + 2y, and solve for y C = 2(C/4) + 2y C = C/2 + 2y C - C/2 = 2y C/2 = 2y 1/2 * C/2 = y C/4 = y So far we have x = C/4 , and y = C/4 We must have a square

  14. anonymous
    • one year ago
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    wow.. thats amazing, I dont think I was going to work this out on my own.. thank you.

  15. perl
    • one year ago
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    Here is a problem that is similar. Show that of all isosceles triangles of fixed perimeter, the triangle that has maximum area is an equilateral triangle.

  16. perl
    • one year ago
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    |dw:1434759096755:dw|

  17. perl
    • one year ago
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    Perimeter = x + 2y Area = 1/2 * x * height You can get the height using pythagorean theorem

  18. perl
    • one year ago
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    once you have a solution for x that minimizes the area, you can back substitute to find the y value

  19. anonymous
    • one year ago
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    oh cool, yeah that makes sense.

  20. perl
    • one year ago
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    We assumed that x was the length and y was the width (or vice a versa) in our drawing. Our work shows that the maximize area has length C/4 and width C/4 The only rectangle that has equal length and width is a square.

  21. perl
    • one year ago
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    where C was the fixed perimeter

  22. anonymous
    • one year ago
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    o to summarize .. 1. We redefined the perimeter as a constant C 2. Then defined area as a function of x and y, area = xy . .. the plan here is to find the derivative and look for the maxima by Fermat's Theorem. 3. Before looking for the maxima of area, we simplify the area function to make it a function of x with the constant C, I'm guessing so we dont have to deal with a partial derivative maybe, and we have area[x]= .... 4. We distribute and simplify our area function 5. Then we work out the derivative and find area'[x] 6. Then we Solve for area'[x] ==0 and apply Fermats Theorem. 6a ... perhaps we might add here a test for 2nd derivative <0 to prove we have a maxima? 7. Then plug y back into C = 2x + 2y and solve for y 8. Finally we get two values that are equal for x and y 9. we might be able to say that as a square has sides that are b/h = 1 then because x/y=1 we have a square.

  23. anonymous
    • one year ago
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    ah, yes, only square rect's have x=y

  24. perl
    • one year ago
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    right by definition

  25. anonymous
    • one year ago
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    well I'm good to go .. thank you.

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