Max Beam Deflection: Find $x$ Value for $D$

[karush]

The deflection D of a beam of length L is
$D=2x^4-5Lx^3+3L^2 x^2$ where $x$ is the distance from on end of the beam.
Find the value of $x$ that yields the maximum deflection.

this is from a unit on inflection points so assume this is $\frac{d^2}{dx^2}$ and also looks like an implicit derivative and quadradic formula from ans. but I can't seem to go in the right direction with this. thnx ahead for suggestions.

Ans $x=\frac{15-√33}{16}L\approx 0.578L$

 

[MarkFL]

Try differentiating D with respect to x, and equate this in factored form to zero to get 3 critical values (using the quadratic formula for the quadratic factor), then use the first derivative test to determine only 1 of the 3 extrema is a maximum. Also, you must have:

$\displaystyle 0\le x\le L$

You will then find the intended value.

 

[karush]

ok this seems to where I get ?

$\frac{d}{dx} 2 x^4-5 L x^3+3 L^2 x^2 = x (6 L^2-15 L x+8 x^2)$

so the quadradic has L and x as varibles so how do get zero's from this
I asssume the \frac{d}{dx} is not correct

R

 

[MarkFL]

L is a constant, representing the length of the beam which does not change. So, for the quadratic factor, we have:

$\displaystyle 8x^2-15Lx+6L^2=0$

Use the quadratic formula where:

$\displaystyle a=8,\,b=-15L,\,c=6L^2$