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October 20, 2014

October 20, 2014

Posted by **Tommy** on Tuesday, November 1, 2011 at 12:08pm.

Because I can't draw on here, the perimeter I have is

6x + 2y = P

(3x for each of the squares sides, because the rectangle covers the last one, 2y for each of the rectangles sides, and the last 3x are the rectangles length, x being half the length of the rectangle, so 2x on top of the rectangle and 1x at the bottom, because the other x is the junction between the square and the rectangle.)

-----------|

| |-------|

|___|

For the area, I found:

p = 6x +2y

y = (P-6x)/2

so

A(x) = x^2 + 2x((P-6x)/2)

now, I know I have to derive A(x)

but what I'm not sure is if the derivative of P will be 0, or do I simply leave P as it is?

- Optimization (Math) -
**Tommy**, Tuesday, November 1, 2011 at 12:09pmsorry for the drawing, apparently it didn't work when I submitted the question!

- Optimization (Math) -
**Steve**, Tuesday, November 1, 2011 at 1:22pmA(x) looks good.

A(x) = x(P-5x) = Px - 5x^2

You can take the derivative if you want, getting

A' = P - 10x

so, A' = 0 when x = P/10

Or, you can see that the roots of A(x) are 0 and P/5, making the max halfway between, or at x = P/10

y = (P-6x)/2 = (P-P/10)/2 = 9P/20

- Optimization (Math) -
**Steve**, Tuesday, November 1, 2011 at 1:24pmRats. y = (P-6P/10)/2 = 4P/20 = P/5

But then, we knew that, since y=2x

- Optimization (Math) -
**Tommy**, Tuesday, November 1, 2011 at 1:38pmI understand now, thanks a lot steve!

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