Question

The perimeter of an isosceles triangle is 28cm. What is the maximum area possible? 

Answer 1

Let b be the length of the base, and a the length of each of the two equal legs.

Triangle area:

$S=\frac{b\sqrt{a^2-b^2/4}}{2}$

$b=28-2a$

$S=(14-a)\sqrt{28a-196}$

$S'(a)=-\sqrt{28a-196}+\frac{14(14-a)}{\sqrt{28a-196}}=0$

$-28a+196+196-14a=0$

$a=28/3$ cm

$b=28-56/3=28/3$

$S_{max}=\frac{28\sqrt{784/9-784/36}}{6}=\frac{196\sqrt{3}}{9}$