Question 214496
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If the uniform width is *[tex \Large x] then the new length and width must be *[tex \Large 15 - 2x] and *[tex \Large 9 - 2x].  The area is the length times the width, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (15 - 2x)(9 - 2x) = 112]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 135 - 48x + 4x^2 = 112]


In standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2 - 48x + 23 = 0]


Solve the quadratic for *[tex \Large x].  You will get two roots.  One of them is larger than 9, so twice that, subtracted from either 9 or 15 is less than zero.  Exclude that root.  The other one is your answer.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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