Effect of enlarging one of the slits in Young's Double Slit experiment

I'm trying to figure out what happens to the maximum intensity of a fringe pattern in the Fraunhoffer regime when one of the slits in the Young's double slit experiment is widened. The other slit is kept at a constant (finite) slit width, and the separation between the center of the slits is also kept constant.

So far, I've attempted numerical solutions by plotting the slit function in MATLAB and taking the FFT of the slit function to get the corresponding fringe pattern. It seems that the maximum intensity just climbs linearly with the ratio of the two slit widths.

I haven't attempted an analytical solution and I don't quite trust my programming enough to accept the linear answer I got above. What is the theoretical outcome of this experiment?

P.S: Here's my MATLAB code if you want to check my numerical solution:

clear

for hw = 1:100 %hw is half width of slit with variable area
hw0 = 20; %other slit kept at 20 units half width

X = [-300:300];
Xamp = 0; %amplitude of slit function
Xamp(201-hw0:200+hw0) = 1; %left slit (unchanged)
Xamp(401-hw:400+hw) = 1; %right slit (dependent on hw)
Xamp(401+hw:601) = 0;

ff = fft(Xamp); %fft to give interference pattern
rightf = ff(1:300);
leftf = ff(301:601);
trunf = [leftf,rightf]; %truncation to center about 0

trunfval = abs(trunf);

relC(hw) = max(trunfval);
ratio(hw) = hw/hw0; %finds ratio of hole areas
end

plot(ratio, relC)
title('Maximum intensity of pattern VS ratio of area of holes')


In this limit, all the light interferes constructively at $\theta = 0$, the direction perpendicular to the slits. The amplitude is therefore linear in the combined width of the two slits (IE the amplitude is proportional to the sum of the widths). The peak intensity is the square of this.