Maximizing Area with a Curve Length Constraint

[jacobi1]

Given:

\(\displaystyle y=b-ax^2, \ a>0, \ b>0, \ y \geq 0\)

Find the values of a and b that maximize the area, subject to the constraint that the length of the curve above the x-axis is 10.

We integrate over

\(\displaystyle \left [- \sqrt{\frac{b}{a}}, \sqrt{\frac{b}{a}} \right ] \),

which gives the area as \(\displaystyle \frac{4}{3}b \sqrt{\frac{b}{a}}\).
Setting up and evaluating the arc length integral, we have

\(\displaystyle 2 \sqrt{4a^2 b^2 + ab} + \operatorname{arcsinh} 2 \sqrt{ab}=10a\)

as the constraint on a and b.
I used the method of Langrange multipliers to maximize this.
Set \(\displaystyle \Lambda=f+\lambda g\), where f is the area and g is the constraint set equal to zero.
We now solve the system

\(\displaystyle \nabla_{a,b,\lambda} \Lambda=0\).

Using Wolfram Alpha for the partial derivatives, we get

\(\displaystyle \nabla_{\lambda} \Lambda=0 \implies 2 \sqrt{4a^2 b^2 + ab} + \operatorname{arcsinh} 2 \sqrt{ab}=10a\) (the original constraint)

\(\displaystyle \nabla_{a} \Lambda=0 \implies -\frac{2b \sqrt{ab}+3 \lambda \operatorname{arcsinh} 2 \sqrt{ab}}{3a^2}=0\)

and

\(\displaystyle \nabla_{b} \Lambda=0 \implies \frac{2b+2 \lambda \sqrt{4ab+1}}{\sqrt{ab}}=0\)
as our system.
Wolfram times out on it, but plugging it into this link gives the solution \(\displaystyle a=b= \lambda =0\).
What has happened?

 

[Opalg]

Not trusting Wolfram Alpha, I did this by hand. I prefer to use the logarithm rather than the inverse hyperbolic function, and I found that the equation for the length is $\sqrt{ab + 4a^2b^2} + \frac12\ln\bigl(2\sqrt{ab} + \sqrt{1 + 4ab}\bigr) = 10a.$

Looking at that, it seems convenient to use $\sqrt{ab}$ instead of $b$ for one of the parameters. So let $c = \sqrt{ab}$, and then the equation for the length is $c\sqrt{1+4c^2} + \frac12\ln\bigl(2c + \sqrt{1 + 4c^2}\bigr) = 10a.$ In terms of $a$ and $c$, the expression for the area is $\frac43c^3a^{-2}.$

Now let $f(a,c,\lambda) = \frac43c^3a^{-2} + \lambda \bigl(c\sqrt{1+4c^2} + \frac12\ln\bigl(2c + \sqrt{1 + 4c^2}\bigr) - 10a\bigr)$. The equations from the Lagrange multiplier method are $$\nabla_a\,f = -\tfrac83 c^3a^{-3} - 10\lambda = 0,$$ $$ \nabla_c\,f = 4c^2a^{-2} + \lambda\left( \sqrt{1+4c^2} + \frac{4c^2}{\sqrt{1+4c^2}} + \frac{2 + \frac{4c}{\sqrt{1+4c^2}}}{2\bigl(2c+ \sqrt{1+4c^2}\bigr)} \right) = 0.$$ The $\lambda$-equation is just the expression for the length. The $c$-equation looks horrendous, but the expression inside the large parentheses simplifies drastically, and the equation becomes $2c^2a^{-2} + \lambda\sqrt{1+4c^2} = 0.$ Eliminating $\lambda$ from the $a$- and $c$-equations leads to the equation $a = \frac2{15}c\sqrt{1+4c^2}.$ When you substitute that into the equation for the length, you get this equation for $c$: $$2c\sqrt{1+4c^2} - 3\ln\bigl(2c + \sqrt{1+4c^2}\bigr) = 0.$$ My graphing calculator gives the positive root of that equation as $c\approx 0.973$, from which $a\approx 0.284$ and $b\approx 3.335.$ The graph of $y = b - ax^2$ then looks like this, and it appears that the length is indeed about $10$:

[GRAPH]rsk9zojftw[/GRAPH]

 

[jacobi1]

Oh, I see...I missed a factor of two in the integral. (Blush)
Also, instead of subtracting 10a before doing the partial derivative, I divided by a and then subtracted 10 before taking the partial derivative for some reason.
Anyway, thanks for explaining it!