When the dimensions of the shape, such as radius, height, or length, change, both surface area and volume also change. However, the volume of the object always changes MORE than the surface area for the same change in dimensions.

Step-by-step explanation:

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3 years ago

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Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

nalin [4]

First we have to find the derivative of the function:
f ` ( x ) = ( ln x / x^6 ) ` =
= $\frac{ \frac{1}{x} x^{6}- 6 x^{5}ln x }{ x^{12} } = \\ \frac{ x^{5}(1-6lnx) }{ x^{12} }= \\ \frac{1-6lnx}{ x^{7} }$
The critical point is when f ` ( x ) = 0
1 - 6 ln x = 0
6 ln x = 1
ln x = 1/6
$x = \sqrt[6]{e}$
f ( e^(1/6)) = e^(-1) * 1/6 = 1 / (6 e)
Answer:
$( \sqrt[6]{e}; \frac{1}{6e} )$

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4 years ago