Question 399146
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Pick any value you like less than 24 (and greater than zero, of course).  Divide your selected value by 2, and call that the length.  Take the value you originally chose and subtract it from 24 then divide that result by 2.  Call that result the width.  Multiply the length times the width to get the area.  Follow these steps three times with different choices for the first step.  If you choose 12 for the first step, you will get the largest possible area.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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