GRE question: maximize $\int_{0}^{3}f(x)dx$ where $f$ is a differentiable function such that $f(3)=7$ and $f'(x) \geq x$ for all $x>0$ Let $f$ be a differentiable real valued function such that $f(3)=7$ and $f'(x) \geq x$ for all  positive $x$. What is the maximum possible value of $$\int_{0}^{3}f(x)dx$$
The answer is 12.. so far what I have done is:
$$F(x)=\int_{0}^{x}f(x)dx = f'(x) - f'(0) \geq x - f'(0)$$ 
so to maximize $F(x)dx$ we take the derivative and set it equal to 0 :
$$\frac{d}{dx} \int_{0}^{x} f(x) dx = f(x) - f(0) = 0 $$
so $f(x) = f(0)$
I have no idea what to do next or if I'm even doing it correctly. Thanks in advance.

Between $x=0$ and $x=3$, $f'(x)>0$ so $f$ increases.  So we've got an increasing function on $[0,3]$ and we want to maximize its integral. Since the function is fixed to go through $(3,7)$, the steeper it is the lower the value of the integral. (As an example, see $f_1$ and $f_2$ below, $f_2$ is steeper so it has a smaller integral on $[0,3]$.)

We get the best results if $f'$ is small, so we want strict equality: $f'(x) = x$.  This means $f(x) = x^2 /2 + C$.  To have $f(3) = 7$ we need $C=5/2$.  Then you get your answer:
$$ \int_0^3 \frac{x^2}{2} + \frac{5}{2} dx = 12.$$
