Question

The perimeter of a triangle is $8cm$. If one of the side is 3 cm, what are the other two sides for maximum area of the triangle?

Answer 1

Perimeter of triangle $=x+y+3\Rightarrow 8$

$x+y=5,y=5-x$

To find $x,y$ so that area is maximum

Now

Area $=\sqrt{s(s-a)(s-b)(s-c)}$

$\frac{dA}{dx}=\sqrt{8(8-x)(8-y)(8-3)}$

To find maximum we diff and put $\frac{dA}{dx}=0$

$\frac{dA}{dx}=\frac{d}{dx}\sqrt{8(8-x)\times 5(8-5+x)}$

$=\frac{1}{\sqrt{8(8-x)\times 5(3+x)}}$

$-3-x+8-x=0$

$2x\Rightarrow 5$

$x=2.5$

$y=2.5$

hence for maximum are $x=y=2.5$