91Ó°ÊÓ

The perimeter of the triangle is given as 9 units, which means \(a + b + c= 9\).

Heron's formula states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle. The perimeter is 9 units, so \(2s=9 \Rightarrow s=\frac{9}{2}\). Now we can write down the area as \(A=\sqrt{\frac{9}{2}\left(\frac{9}{2}-a\right)\left(\frac{9}{2}-b\right)\left(\frac{9}{2}-c\right)}\).

To maximize the area, we will eliminate one variable from the formula for area by using the perimeter constraint. Let's express \(c\) in terms of \(a\) and \(b\): \(c = 9 - a - b\). Now, substitute the expression for c in the area formula: \(A=\sqrt{\frac{9}{2}\left(\frac{9}{2}-a\right)\left(\frac{9}{2}-b\right)\left(\frac{9}{2}-(9-a-b)\right)}\) Simplify: \(A=\sqrt{\frac{9}{2}\left(\frac{9}{2}-a\right)\left(\frac{9}{2}-b\right)\left(a+b-\frac{9}{2}\right)}\) Note that we have a constrained optimization problem, where we're working within the domain: \(a+b>0\), \(a+b<9\), \(a>0\), \(b>0\)(to have a valid triangle) We maximize the area by applying AM-GM inequality: \(\frac{(a+b)}{2}\geq\sqrt{ab}\). Apply inequality to the area formula: \(\frac{9}{2}\left(\frac{9}{2}-a\right)\left(\frac{9}{2}-b\right)\geq\left(\frac{9}{2}ab\right)^2\) Since equality holds in AM-GM when all components are equal, the maximum area occurs when: \(\frac{9}{2}-a=\frac{9}{2}-b=a+b-\frac{9}{2}\) This implies that \(a=b\), and hence \(c=a=b=\frac{9}{3}=3\).

The maximum area occurs when the triangle is equilateral with equal sides, which, in this case, have a length of 3 units. Therefore, the dimensions of the triangle are \(a=3, b=3,\) and \(c=3\).