Question

Perimeter of a triangle is $10cm$ and one of its sides is $4cm.$ If $A$ is its maximum area, then the value of $A^2$ is equal to

Answer 1

Let $a,b,c=4$ be the sides and perimeter $2s=10$
$\therefore a+b=6$
Now, $A^2=s(s-a)(s-b)(s-c)=5(5-a)(a-1)\left(1\right)=f\left(a\right)$
So, $f\left(a\right)=5(6a-a^2-5)$
$f^{\prime }\left(a\right)=5(6-2a)\text{and}{f}^{\prime \prime }\left(a\right)=-10<0$
$\therefore f\left(a\right)$ will be maximum, when $f^{\prime }\left(a\right)=0$
$\Rightarrow a=3$
Hence, Maximum of $A^2=5(18-9-5)=20$