Concavity and the 2nd Derivative Test: Finding Maximum Deflection

[karush]

this is a problem in the topic of Concavity and the second Derivative Test

The deflection \(\displaystyle D\) of a beam of length \(\displaystyle L\) is

\(\displaystyle D=2x^4-5Lx^3+3L^2x^2\),

where \(\displaystyle x\) is the distance from one end of the beam.

Find the value of \(\displaystyle x\) that yields the maximum deflection

The answer to this is \(\displaystyle x=\Bigg(\frac{15-\sqrt{33}}{16}\Bigg)L \approx 0.578L\)

well first of all this equation has 2 variables in x and L so not sure what to do perhaps implicit differentiation. also the answer looks it came from a quadratic formula

so not to sure what the first step is.

 

[MarkFL]

Re: beam deflection

The length of the beam $L$ is a constant.

So what you want to do is compute $D'(x)$ to determine the critical values in the range:

\(\displaystyle [0,L]\)

and then evaluate $D''(x)$ at these critical values, looking for negative values, i.e., concave down, and hence a maximum.

Can you proceed?

 

[mathworker]

Re: beam deflection

L is not a variable its a constant as length is constant and hence you can go for first derivative which is equal to slope.it is maximum when tangent comes parallel to x-axis so = '0' and proceed, you might have mistook L as variable

 

[karush]

Re: beam deflection

\(\displaystyle D'=8x^3-15Lx^2+6L^2x\)
\(\displaystyle =x(8x^2-15Lx+6)=0\)

so \(\displaystyle x=0\)
or \(\displaystyle \frac{15L\pm \sqrt{33}L}{16}\)
\(\displaystyle =\left(\frac{15 \pm \sqrt{33}}{16}\right) L\)
\(\displaystyle =0.5784L\) or \(\displaystyle 1.2965L\)

so \(\displaystyle D'' = 24x^2-30Lx+6L^2\)

\(\displaystyle D''(0.5784L)>0 \) so max is \(\displaystyle 0.5784L\)

 

[MarkFL]

Yes, good work! (Yes)

You can discard the critical value that is greater than $L$, since this is outside the relevant domain of \(\displaystyle x\in[0,L]\).