Question 1121065
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Width of base:  *[tex \Large w]


Length of base:   *[tex \Large 2w]


Height of prism:  *[tex \Large h]


Volume of prism:  *[tex \Large V(w,h)\ =\ 2w^2h\ =\ 84]


Surface Area:  *[tex \Large S(w,h)\ =\ 4w^2\ +\ 4wh\ +\ 2wh\ =\ 4w^2\ +\ 6wh]


But then:  *[tex \Large h\ =\ \frac{42}{w^2}]


Hence the surface area as a function of the width is *[tex \Large S(w)\ =\ 4w^2\ +\ \frac{252}{w}]


And then the first derivative of the surface area function is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ S'(w)\ =\ 8w\ -\ \frac{252}{w^2}]


Set the derivative equal to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8w\ -\ \frac{252}{w^2}\ =\ 0]


Solve for *[tex \Large w], then calculate *[tex \Large 2w] and *[tex \Large \frac{42}{w}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
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