Prove that if the perimeter of the triangle is constant its area is maximum. When it is equilateral triangle
Answers
Given : perimeter of the triangle is constant
To Find : its area is maximum When it is equilateral triangle
Solution:
a + b + c = 2s ( 2s is constant)
Area A = √s(s - a)(s - b)(s - c)
c = 2s - a - b
=> A = √s(s - a)(s - b)(s - (2s-a - b)))
=> A = √s(s - a)(s - b)(a + b - s))
A is max then A² is also max as A is +ve
Z= A² = s(s - a)(s - b)(a + b - s)
∂Z/∂a = s(-1)(s - b)(a + b - s) + s(s - a)(s - b)(1)
= - s(s - b)(a + b - s) + s(s - a)(s - b)
= s(s - b)( - a - b + s + s - a)
= s( s - b)( 2s - 2a - b)
s = 0 not possible
s = b
2s - 2a - b = 0
Similarly
∂Z/∂b = s( s - a)( 2s - 2b - a)
s = a
2s - 2b - a = 0
if we consider s = b and s= a then c = 0 which is not possible
2s - 2a - b = 0
2s - 2b - a = 0
solving a = b = 2s/3
=> c = 2s/3
a = b = c = 2s/3
∂²Z/∂a² = s( s - b)( - 2) < 0 and ∂²Z/∂b² = s( s - a)( - 2) < 0
Hence Maximum Z when a = b = c = 2s/3
Hence Area is maximum when a = b = c
Hence triangle is Equilateral triangle
QED
if the perimeter of the triangle is constant its area is maximum. When it is equilateral triangle
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