TITLE: Minimize $\int_0^1 A(x)dx$ given $\int_0^1 \frac{dx}{A(x)}\le \text{constant}$
QUESTION [5 upvotes]: The following problem is derived from a Finite element problem.
We have a beam which is subject tot a constant axial force. With out loss of generality the beam can have a variable section area $A(x)$. The beam cannot exceed a certain strain value, that is, it cannot be shorten by more than a given percentage. We want the beam to be with the smallest possible volume.  
I have no idea how to approach the following optimization problem. Let $Volume=\int_a^bA(x)dx$. The constraint is given by $\int_a^b \frac{1}{A(x)}dx\leq const$. 
Hence:
 $$\min_{A} \{Volume\} \\S.T. \int_a^b \frac{1}{A(x)}dx\leq const$$

How do I approach such a problem if at all possible?
If not, is there a way to solve this, given the relaxation of $A(x)$ being a polynomial function lower than the 4th (inertia momentum) degree?

REPLY [4 votes]: This is not a general approach, but it will lead you to the final answer quickly.
Let's say $\int_a^b \frac{dx}{A} \le K$ for some given constant $K$, then by Cauchy Schwarz
$$K \int_a^b A(x) dx \ge \int_a^b \frac{dx}{A(x)} \int_a^b A(x) dx \stackrel{CS}{\ge} \left(\int_a^b dx\right)^2 = (b-a)^2$$
This  leads to following lower bound for the integral $$\int_a^b A(x) dx \ge \frac{(b-a)^2}{K}$$
Since this lower bound is achieved by the constant function $A(x) = \frac{b-a}{K}$, the minimum volume is $\frac{(b-a)^2}{K}$.

For a general approach, you need to set this up as a calculus of variation problem with constraint.
The constrainted Lagrange has the form
$$\mathcal{L}(x) = A(x) + \lambda\left(\frac{1}{A(x)} - k\right)$$ 
for unknown constant $\lambda$ and $k$.
You proceed to solve the associated Euler-Lagrange equation
$$\frac{\partial \mathcal{L}}{\partial A} - \frac{d}{dx}\left(\frac{\partial \mathcal{L}}{\partial A'}\right) = 0$$
In this case, the EL-equation tell us
$$1 - \frac{\lambda}{A(x)^2} = 0$$
This means the solution which minimize the functional $\int_a^b A(x) dx$ subject to constraint $\int_a^b \left( \frac{1}{A(x)}  - k\right) dx =  0$ is probably a constant function. Finally, you need to verify by suitable choice of constant value of $A(x)$, you will get the actual minimum.
To make this precise, it probably take one or two chapters from a book. I will just stop here.