Question Number 56037 by ajfour last updated on 08/Mar/19
Commented by ajfour last updated on 08/Mar/19
Commented by mr W last updated on 08/Mar/19
Commented by 121194 last updated on 08/Mar/19
Commented by ajfour last updated on 08/Mar/19
Commented by mr W last updated on 09/Mar/19
Answered by ajfour last updated on 11/Mar/19
![let side of equilateral △ be s. ⇒ 3s=a+b+c let bottom segments be x and s−x. The right side segments are c−s+x and 2s−x−c. The left side segments are s−b+x, and b−x. For the inner △ area to be maximum, The remaining area has to be minimum. The blue area A=(1/2)sin 60°[x(b−x)+(s−x)(c−s+x) +(2s−c−x)(s−b+x) (dA/dx)=((√3)/4)[(b−x−x)+(−c+s−x+s −x)+(−s+b−x+2s−c−x) ]= 0 ⇒ 3s−6x+2b−2c = 0 (a+b+c)+2b−2c = 6x ⇒ x_0 =((a+3b−c)/6) (d^2 A/dx^2 )= −6 ⇒ A is maximum for x_0 =((a+3b−c)/6) . ⇒ b−x = ((−a+3b+c)/6) Minimum inner △ area A is =((√3)/(36))(a+b+c)^2 −((3(√3))/4)(((9b^2 −(c−a)^2 )/(36))) =((√3)/(36))(a+b+c)^2 − (((√3)[9b^2 −(c−a)^2 ])/(48)) .](https://www.tinkutara.com/question/Q56122.png)
Commented by mr W last updated on 10/Mar/19
Commented by ajfour last updated on 11/Mar/19
