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I think you are supposed to use quadratics, but I don't know how.

 

A piece of wire 10 m long is to be cut into two pieces. One piece is bent into a square, and the other into a circle. How can the wire be cut so that the area enclosed by the two shapes is minimized? Give your answer to the nearest tenth.

 Feb 23, 2020
 #1
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Let  x  be  the part  of the wire reserved for the  circle

This  must  be  the perimeter of the  circle

Perimeter =  2pi*r

x  =2pi* r

[ x / (2pi) ]   = r

So   the  area  of the  circle =  pi* r^2 =  pi  [ x / (2pi)]^2 = [x^2] / [ 4pi ]

 

The part  reserved for the squre is just the perimeter of the square  =   [ 10- x ]

The side of the square  =  perimeter /4 =   [ 10 - x ] /4

Area of the square =  side^2  =  [ 10 - x ]^2  / 16

 

 

So....the  total area of the circle and square , A, can be expressed  as

 

A =  x^2 / [ 4pi ]  +  [ 10 - x ]^2 / 16         simplify

 

A = x^2 / [ 4pi ]  +  [ x^2 - 20x + 100] /16

 

A =  x^2 / [4pi]  + (1/16)x^2  - (5/4)x  + 25/4

 

This  is  most easily done with Calculus.....I'll  graph the  curve to show you that  the  value for x obtained is the  correct one

 

Take the  derivative  of  A  and set  it to  0 

 

2x /(4pi) + (2/16)x  - 5/4  = 0

 

x / [ 2pi]  +  x /8  = 5/4

 

x  [   1/[ 2pi] +  1/8]  =  5/4 

 

x [ 1/(2pi) + 1/8 ]  = 1.25   

 

x  =  1.25  / [  1/(2pi)  + 1/8]  ≈  4.399  ≈  4.4  

 

Cut the wire  at  about  4.4 inches.....this is the part reserved  for the  circle

So  the part reserved for the square  =10 - 4.4  = 5.6  in

 

Here is a graph  showing that the  value for  x  is  correct  :

 

https://www.desmos.com/calculator/d8t5qos9c3

 

 

cool cool cool

 Feb 23, 2020

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