Question 202324
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Let *[tex \Large x] represent the measure of the side of one of the cut-out squares.  Since each of the dimensions of the original piece of cardboard will be reduced by *[tex \Large 2x] to form the base of the box, the dimensions of the base of the box will be *[tex \Large 11 - 2x] and *[tex \Large 14 - 2x].  Since the area of a rectangle is length times width and we know the area of the base must be 80 square inches, we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (11 - 2x)(14 - 2x) = 80]


Multiply the binomials and collect like terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 154 - 50x + 4x^2 = 80]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4x^2 - 50x + 74 = 0]


This does not factor conveniently, so use the quadratic formula:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{-(-50) \pm sqrt{(-50)^2 - 4(4)(74)}}{2(4)} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{50 \pm sqrt{1316}}{8} ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x = \frac{25 \pm sqrt{329}}{4} ]


Both roots are real and positive, but one of them is way too large, namely *[tex \Large \frac{25 + sqrt{329}}{4} \approx 10.8] meaning that the cutout squares would overlap and you would have no box at all.  So exclude this root as extraneous having been introduced by the act of squaring the variable.  The other root, *[tex \Large \frac{25 - sqrt{329}}{4} \approx 1.7] is the correct answer.



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