Question 977643
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The volume of the box is *[tex \Large lwh\ =\ 1440], but you know that the length is *[tex \Large 20] so the product of the width and the height must be *[tex \Large 1440] divided by *[tex \Large 20].  But you also know that two times the height is the width, so *[tex \Large wh\ =\ 2h^2].


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2h^2\ =\ 72]


Solve for the positive value of *[tex \Large h].  Then calculate *[tex \Large w\ =\ 2h]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \