# Calculus

A piece of wire 40cm long is cut into two pieces. One piece is bent into the shape of a square and the other is bent into the shape of a circle. How should the wire be cut so that the total area enclosed is
a) a maximum? /b) a minimum?

Perimeter of square = x
Circumference of circle = 40-x

Circumference of circle = 2pir
r=(40-x)/2pi

Area total = (x/4)²+pi((40-x)2pi)²
Area total'=2(x/4)(1/4)+2pi(40-x)(-1/2pi)
=(pix-160+4x)/8pi
square=x= 160/(pi+4)
circle=4pi/(pi+4)

restriction:0≤x≤40

How do I find the maximum and minimum?

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1. line used for square = (40 - x)
line used for circle = x

Area = (area of square) + (area of circle)

Area = (40-x)^2 + (pi*x^2)/4

Area' = (2)(-1)(40-x) + (pi*x)/2 = 0

x = 22.4

Is this max or min?

Using x=20 for before the critical point, and x=30 for after the critical point.

Area' = (2)(-1)(40-20) + (pi*20)/2 = a negative number
So at x=20, slope is negative

Area' = (2)(-1)(40-30) + (pi*30)/2 = a positive number
So at x = 30, slove is positive

So it is a minimum

What is max?
x=0 --> x^2 = 1600
x = 40 --> (pi*x^2)/4 --> 1256.6

so max is when x = 0

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2. af

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