Question

show that it the perimeter of a triangle is constant its area is maximum when the is triangle equilateral.​

Answer 1

Answer:

A=S−B2

Step-by-step explanation:

This one is pretty fun.

One formula that is going to be really useful here is Heron’s formula. If  S  is the perimeter of a triangle with sides  A,B,C , then the area is equal to  S(S−A)(S−B)(S−C)−−−−−−−−−−−−−−−−−−−−√ .

We have that  S  is constant, so we want to find the values of  A,B  and  C(=S−A−B)  that maximize  (S−A)(S−B)(A+B)−−−−−−−−−−−−−−−−−−−√ .

This still has two variables,  A  and  B . Let’s see what we can do if we keep one of them fixed. Given a value of  B , what value of  A  will make the above formula maximal?

Using some calculus, we get that the derivative of the formula with respect to  A  is equal to  (S−B)(S−A)−(S−B)(A+B)Z  where  Z  is twice the above formula (but we don’t care about that part — we’re only really interested in the numerator). We can rewrite the numerator as  (S−B)(S−2A−B) . For a given value of  B  somewhere between  0  and  S , this can only be  0  if  2A=S−B , so  A=S−B2 .

This means that, if  B  is fixed, the best we can do is to make  A  and  C  equal. Applying the same reasoning to the other sides, we can conclude that the largest area is found when all three sides are equal.