To find maximum possible value of this integral 
If $\int_{0}^{1} f dx=3$  and  $\int_{0}^{1} xf dx  =2$, then find the maximum value of $$\int_{0}^{1} f^2 dx.$$

What methods would apply to find this maximum value? I am not approaching the methods...
 A: For a comprehensive introduction to the method of calculus of variation, you may look at this reference.
We want to maximize $\int_{0}^1 f^2 dx$ subject to the conditions $\int_{0}^1 f dx=3,\ \int_{0}^1 xf dx=2$. We form the Lagrangian, $l(x,\lambda,\mu)=f^2(x)-\lambda f(x)-\mu xf(x)$. Then use the Euler-Lagrange equations to get $$\frac{\partial l}{\partial x}=0\implies 2f(x)f'(x)-\lambda f'(x)-\mu(f(x)+xf'(x))=0$$ Solve this differential equation for $f(x)$ and use the constraints to solve for $\lambda,\mu$ and that will give you an extremum.     
A: Let $f(x) = 6x + \lambda (x^2-x+{1 \over 6})$. It is easy to check that $f$
satisfies the conditions and
$\int_0^1 f^2 = 12+\lambda^2 {1 \over 180}$.
Hence we can choose $\lambda$ as large as we want and so there is no maximum.
Here is how I approached the problem: Let $\langle f, g \rangle = \int_0^1 f(x)g(x) dx$.
Then we are looking to maximise $\|f\|^2$ for  $f$ such that $\langle f, x \mapsto 1 \rangle = 3$,
$\langle f, x \mapsto x \rangle = 2$.
By choosing a function that is orthogonal to $x \mapsto 1, x \mapsto x$, we
can add multiples of this function without changing the value of the constraints. In this case, one such function is $x \mapsto x^2-x+{1 \over 6}$.
The rest is algebra.
