Question

This question has been bugging me all night, its a realtive extrema problem. y = x^2 sqrt(16-x)

Answer 1

find the derivative of this

\[\frac{dy}{dx}=\frac{d}{dx}[x^2(16-x)^{1/2}]\]

Answer 2

use the product rule

\[uv'+vu'\]

\[x^2(\frac{1}{2})(16-x)^{-1/2}(-1)+(16-x)^{1/2}(2x)\]

Answer 3

\[\frac{-1}{2}(\frac{x^2}{(16-x)^{1/2}})+2x(16-x)^{1/2}\]

Answer 4

now you just have to find out when

\[-1(x^2)+2x(16-x)^{1/2}=0\] with hte constraint that \[16-x<0\]

Answer 5

oh alright, what about the local max?

Answer 6

ahh yes there is other zeroes hmm we have to figure out some more zeros for this equation

Answer 7

when
\[x^2=2x(16-x)^{1/2}\] is our other zero where this hits i'm not quite sure

Answer 8

lets try simplifying the derivative to get something more easier to deal with

Answer 9

needs to find x from this mess