In interference we know that
$\hspace20mm I_{max}=(\sqrt{I_1}+\sqrt{I_2})^2$
and $\hspace20mm I_{min}=(\sqrt{I_1}-\sqrt{I_2})^2$
Under normal conditions (when the widths of both the slits
are equal)
$\hspace20mm I_1=I_2=I$ (say)
$\therefore \hspace20mm I_{max}=4I \, \, and \, \, I_{min}=0$
When the width of one of the slits is increased. Intensity due
to that slit would increase, while that of the other will remain
same. So, let:
$\hspace25mm I_1=I$
and $\hspace25mm I_2=\eta I \hspace10mm (\eta > 1)$
Then, $\hspace20mm I_{max}=I(1+\sqrt {\eta})^2 > 4I$
and $\hspace20mm I_{min}=I(\sqrt {\eta}-1)^2 > 0$
$\therefore$ Intensity of both maxima and minima is increased.