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
7.9k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories