The perimeter of a triangle is \(10 \mathrm{~cm}\). If one of its side is \(4 \mathrm{~cm}\), then remaining sides of the triangle, when area of triangle is maximum are

Let \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) be the sides of the triangle.
The perimeter of triangle \((2 \mathrm{~s})=\mathrm{a}+\mathrm{b}+\mathrm{c}\)
Let \(\mathrm{a}=4\) and \(2 \mathrm{~s}=10\) i.e. \(\mathrm{s}=5\)
\(\therefore 10=4+b+c \Rightarrow b=6-c\)
Now, area of triangle \(=\Delta=\sqrt{s(s-a)(s-b)(s-c)}=\) \(\sqrt{5(5-4)(5-6+c)(5-c)}\)
\(\Delta=\sqrt{5(1)(\mathrm{c}-1)(5-\mathrm{c})} \Rightarrow \Delta^{2}=5(\mathrm{c}-1)(5-\mathrm{c})\)
\(=5\left(5 c-c^{2}-5+c\right)=5\left(-c^{2}+6 c-5\right)\)
Let \(f(c)=5\left(-c^{2}+6 c-5\right) \Rightarrow f^{\prime}(c)=5(-2 c+6) \Rightarrow\) \(f^{\prime \prime}(c)=5(-2)=-10\)
For extreme value, \(f^{\prime}(c)=0\) i.e. \(5(-2 c+6)=0 \Rightarrow c=3\)
Thus \(f^{\prime \prime}(3)=-10 < 0 \Rightarrow f(c)\) has maximum value at \(c=3\)
\(\therefore \mathrm{b}=6-3=3\) i.e. the lengths of the remaining sides are \(3 \mathrm{~cm}\) and \(3 \mathrm{~cm}\).