A 24 inch wire is cut in two and shaped into a square and a regular octagon . What is the minimum possible sum of the two areas?

A(t) = 41,47 in²

Step-by-step explanation:

Let´s call "x" the cut of point to get to pieces of wire, we make a square from x and the regular octagon will be shaped with 24-x

Then Area of the square  A(s)  =  x²

Area of the octagon is  A(o)  = 1/2*p*length of apothem (d)

p = ( 24 - x )

length of apothem (d) :

The side of the octagon is equal to ( 24 - x ) / 8 half the side is  

( 24  -  x ) / 16

tan α  = ( 24- x ) 16 / d      since ∡s in octagon are 360 / 8  = 45°

α ( ∡ between apothem and one of the interiors ∡ of the octagon )half of 45 is  α = 22,5°  

tanα = 0,41

d = (24 - x ) / 16*0,41        d = ( 24 - x ) / 6,56

Then

A(t) = A(s) + A(o)

A(t) =  x²  + (1/2)* ( 24 - x ) ( 24 - x ) / 6,56

Note  A(t) = A(x)

A(x) =  x²  + (1/2) * (24 - x )²/ 6,56

A(x) = x² + ( 1/ 2*6,56) * ( (24)² -48*x + x² )

Taking derivatives on both sides of the equation

A´(x) = 2*x + ( 1/13,12)* ( - 48 + 2x )

A´(x) = 2*x - 48/ 13,12 + 2*x

A´(x) = 4*x - 3,66

A´(x) = 0     4x = 3,66        x  = 0,91 in    and  d =( 24 - x ) / 6,56

d = ( 24 - 0,91 ) / 6,56        d = 3,52

Then  A(s) = (0,91)²      A(s) = 0,83 in²

A(o) = 1/2 * ( 24 - 0,91 )* 3,52

A(o) = 40,63 in²

A(t) = 40,63 + 0,83

A(t) = 41,47 in²

i think it is x = 1.2