Question 312230
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First divide the volume by the given length, 24.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{4368}{24}\ =\ 182]


We now know that the product of the height and the width must be 182.


Since the height of the box is 1 inch more than the width, we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w(w\ +\ 1)\ =\ 182]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ w\ -\ 182\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (w\ +\ 14)(w\ -\ 13)\ =\ 0]


Discard the negative root:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ 13]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ w\ +\ 1\ =\ 14]


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