A calculus question

[Weyoun]

Hi all

I recently came across this question from our 3U maths textbook. I've tried doing it over 4 times, each instance yielding a different result.

A window consists of a rectangle surmounted by a semi-circle having as its diameter the width of the rectangle. If the perimeter of the window is t metres, find the greatest possible area of the window.

OK, to start with, I drew something like this:

. . . /---\
. . ./ . . .\
. . / . . . .\
. .|----x---|
. .| . . . . .y
. .|--------|

I used the pronumerals x and y for the width and height of the rectangle respectively. Since:
x + 2y + PI*x/2 = t
y = (t - PI*x/2 - x)/2
. .= (2t - PI*x - 2x)/4

Therefore the area (A) of the figure is:
x(2t - PI*x - 2x)/4 + (PI*(x/2)^2)/2
(2tx -PI*x^2 - 2x^2)/4 + PI*x^2/8
(4tx - 2PI*x^2 - 4x^2 + PI*x^2)/8
(4tx + x^2(-4 - PI))/8

Finding the derivative...

dA/dx = (4t + 2x(-4 - PI))/8

Finding the turning point (maximum)
0 = 4t + 2x(-4 - PI)
x = -4t/-2(PI + 4) . . . . . (taking a factor of -1 out of the last part)
= 2t/(PI + 4)

So, the area at this point.....

A(x) = (4tx - x^2(PI + 4))/8
A(2t/(PI + 4))= (4t(2t(PI + 4)) - (2t(PI + 4))^2(PI + 4))/8
= (8t^2/(PI + 4) - 4t^2/(PI + 4))/8
= 4t^2/8(PI + 4)
= t^2/2(PI + 4)

which is the right answer....

I hate this world at times....

Thanks for... uhh... making me re-do it again