215k views
5 votes
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?

1 Answer

4 votes

Answer:

A(t) = 41,47 in²

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

User Zemzela
by
4.2k points