What is the relation between drag and weight?
If you allow to make a few simplifications, the answer is easy:


Friction drag is not affected by the angle of attack change. That means no onset of flow separation on the heavier aircraft.
The wing's optimum L/D is reached with the same flap setting in both cases.
Lift changes linearly with angle of attack, so the lift coefficient $c_L$ can be expressed by the product of lift curve slope $c_{L\alpha}$ and angle of attack $\alpha$.
We neglect the changed lift contribution of engine thrust when the angle of attack is increased.


Now the drag $D$ can be expressed by this equation:
$$D = \frac{\rho\cdot v^2}{2}\cdot S\cdot\left(c_{D0}+\frac{(c_{L\alpha}\cdot\alpha)^2}{\pi\cdot AR\cdot\epsilon}\right)$$
Per the definition above the term $c_{D0}$ is constant, so the change in drag between the lighter aircraft (index 1) and the heavier aircraft (index 2) will be
$$\Delta D = \frac{\rho\cdot v^2}{2}\cdot S\cdot\frac{c_{L\alpha}^2\cdot\left(\alpha_2^2-\alpha_1^2\right)}{\pi\cdot AR\cdot\epsilon}$$
In order to express this drag difference $\Delta D$ in terms of the aircraft's mass, write the lift coefficient $c_L$ as $\frac{2\cdot m\cdot g}{\rho\cdot v^2\cdot S}$:
$$\Delta D = \frac{g\cdot\left(m_2^2-m_1^2\right)}{\pi\cdot AR\cdot\epsilon}$$

The other symbols are:
$\kern{5mm} \rho\:\:\:\:\:$ air density
$\kern{5mm} v\:\:\:\:\:$ velocity
$\kern{5mm} S\:\:\:\:\:$ wing surface area
$\kern{5mm} c_{D0} \:$  zero-lift drag coefficient
$\kern{5mm} \pi \:\:\:\:\:$  3.14159$\dots$
$\kern{5mm} AR \:\:$  aspect ratio of the wing
$\kern{5mm} \epsilon \:\:\:\:\:\:$  the wing's Oswald factor
$\kern{5mm} g \:\:\:\:\;$  gravitational acceleration

Now we can answer your questions:


  How much more drag will it be?


The drag will increase with the square of the mass increase. The gradient of that increase depends on the span loading of the aircraft.


  what would be the exponent?


2


  Is the aerodynamics in the design process optimised for "half-loaded" aircraft?


No, always for the fully loaded aircraft, as lighter loads can be tolerated much better than higher loads. However, since fuel burn will cause a change in aircraft mass over time, the aerodynamics must work over a range of altitudes.


  how does this translate to fuel consumption, say on 40000 feet with the usual speeds?


Drag is compensated by thrust, so you need more thrust to overcome the higher drag. Fuel consumption goes up linearly, but since cruise at 40.000 ft means that the aircraft is light and the engines are running at close to their maximum sustained thrust, a mass increase by 30% is impossible. For a practical result, the heavier aircraft would fly at the same angle of attack and speed but at a lower altitude where both the increased lift and the increased engine thrust can be provided by the higher air density.

If you need to calculate the trip fuel at different weights: We had a similar question before, so please follow the link for an explanation.