4. Given the three sides of a triangle $a, b, c$ are integers, and $a+b+c=11$. Then when the product $abc$ takes the minimum value, the area of the triangle is $\qquad$ .
4. Since $a, b, c$ are the lengths of the three sides of a triangle, when $a=1$, $b=c=5$, $abc=25$; when $a=2$, $b=4$, $c=5$, $abc=40$; when $a=3$, $b=3$, $c=5$, $abc=45$, or $b=c=4$, $abc=48$. It is evident that the product $abc=25$ is the minimum value. In the isosceles $\triangle ABC$, the height from vertex $A$ to the base $BC$ is $h=\sqrt{5^{2}-\left(\frac{1}{2}\right)^{2}}=\frac{3 \sqrt{11}}{2}$, thus the area is $S=\frac{3 \sqrt{11}}{4}$.