Maximizing y=kx³.ln(1/x) with Positive Constant k | Math Homework Help

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dan_fmx
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Hi I am new on the forums here. I am having troube with my maths assignment. It says to find the maximum value of y= kx³ . ln(1/x) where k is a positive constant. I am also told to find the value of x, in terms of e. I am really struggling and any help would be appreciated.

Thanks in advance

Dan
 

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If this is an assignment, I'm sure it relates to what you are currently covering. Does that, by any chance, have something to do with derivatives? Can you (perhaps looking back in your notes) tell us how, in general, you can find the x-coordinate where the function y(x) takes its maximum?
 


The maximum will be found when the gradient function = 0. I found the gradient function using the product rule as y'=-x²+3x².Ln(1/x). Then when 0=-x²+3x².Ln(1/x) x= 0.7165. How ever it says in terms of e and this is where i get really stuck. I am drawing a blank on this one
 


Okay, -x2+ 3x2ln(1/x)= x2(3 ln(1/x)- 1)= 0 so one possible solution is x= 0. The other is given by 3 ln(1/x)= 1 or ln(1/x)= 1/3. Surely you can find x without resorting to a calculator? If ln(x)= a, the x= ea by definition.
 
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Wow it all became so clear, thanks so much for all the help everyone. I found the maximum x value to be x=1/(e^(1/3)). Thanks everyone for helping