Question 755438
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From the formula for the perimeter we derive:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l\ =\ 72\ -\ 2w]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 36\ -\ w]


Multiplying length times width to get area:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 36w\ -\ w^2\ =\ 315]


Put the quadratic into standard form, factor, and solve.  The two roots will actually be the two dimensions, and by convention you would generally assign the smaller one to the designation "width".


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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