Minimize $\int_0^1 A(x)dx$ given $\int_0^1 \frac{dx}{A(x)}\le \text{constant}$ 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? 

 A: 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.
A: First I will show that any minimum value of the integral you are considering can be attained by a constant function $A(x).$ Then, once you assume $A(x)$ is constant, the answer falls into your lap.
For the sake of simplicity, I will assume that we are integrating over the interval $[0,1]$ instead of $[a,b]$. I will also assume that $A(x)\le 1$ for all $x\in [0,1]$, which I think is a reasonable assumption because "beams" are typically longer than they are thick.
So, to restate the question, we want to find a function $A(x)$ that minimizes
$$I=\int_0^1 A(x)dx$$
with the constraint that
$$\int_0^1 \frac{dx}{A(x)}\le C$$
where $C$ is some given constant.
To solve this problem, we will make use of the following fact, which can be verified using integration by substitution: if $f$ is an integrable function on $[0,1]$, then
$$\int_0^1 f(x)dx=\int_0^1 \frac{f(\frac{x}{2})+f(\frac{x+1}{2})}{2}dx$$
Now suppose that $A_m(x)$ is a "well-behaved" function that minimizes $I$. Then, by the above identity, $I$ takes on the same value if we use the function 
$$\frac{A_m(\frac{x}{2})+A_m(\frac{x+1}{2})}{2}$$
instead. Further, notice that
$$A_m(x)\le 1 \implies \frac{2}{A_m(\frac{x}{2})+A_m(\frac{x+1}{2})}\le \frac{1}{2} \bigg(\frac{1}{A_m(\frac{x}{2})}+\frac{1}{A_m(\frac{x+1}{2})}\bigg)$$
Which implies that
$$\int_0^1 \frac{2dx}{A_m(\frac{x}{2})+A_m(\frac{x+1}{2})}\le \int_0^1 \frac{dx}{A_m(x)}\le C$$
This is all that we need! This demonstrates that if $A_m(x)$ is a "minimizing function," then the function
$$\frac{A_m(\frac{x}{2})+A_m(\frac{x+1}{2})}{2}$$
is also a "minimizing function." If we repeat the same process with this function, we see that
$$\frac{A_m(\frac{x}{4})+A_m(\frac{x+1}{4})+A_m(\frac{x+2}{4})+A_m(\frac{x+3}{4})}{4}$$
is also a "minimizing function." In fact, repeating this process infinitely many times shows that the constant function function
$$\int_0^1 A_m(t)dt$$
is also a "minimizing function." This proves that for any function minimizing the given integral, there exists a constant function that also minimizes it; for this reason, you may assume that your function $A(x)$ is constant. 
So let $A(x)=A$ be constant. Then we have
$$\int_0^1 \frac{dx}{A}=\frac{1}{A}\le C$$
and so $A\ge 1/C$, implying that $I\ge 1/C$ and $I=1/C$ is the minimum value of your integral. Done!
