anonymous
  • anonymous
"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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
I guess I cant use the diagonal.. they're not consistent across different base & height.
anonymous
  • anonymous
GIVEN area = base * height perimeter = 2base + 2height EQUATIONS area[x,y] = xy perimeter[x,y] = 2x+2y
anonymous
  • anonymous
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

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perl
  • perl
|dw:1434755324547:dw|
perl
  • perl
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
perl
  • perl
With me up to here?
anonymous
  • anonymous
I think so.. was just thinking wow...
anonymous
  • anonymous
Im not sure where to take it form here though..
perl
  • perl
At this point you can you use calculus or use your knowledge of parabolas
anonymous
  • anonymous
so Im guessing then a derivative ties into this somehow ..
anonymous
  • anonymous
wow again..
anonymous
  • anonymous
do we need to do the same thing for y?
perl
  • perl
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
anonymous
  • anonymous
wow.. thats amazing, I dont think I was going to work this out on my own.. thank you.
perl
  • perl
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.
perl
  • perl
|dw:1434759096755:dw|
perl
  • perl
Perimeter = x + 2y Area = 1/2 * x * height You can get the height using pythagorean theorem
perl
  • perl
once you have a solution for x that minimizes the area, you can back substitute to find the y value
anonymous
  • anonymous
oh cool, yeah that makes sense.
perl
  • perl
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.
perl
  • perl
where C was the fixed perimeter
anonymous
  • anonymous
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.
anonymous
  • anonymous
ah, yes, only square rect's have x=y
perl
  • perl
right by definition
anonymous
  • anonymous
well I'm good to go .. thank you.

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