Application of differentiation 2

Suppose x m of wire is bent into a square and ( 6 – x ) m of wire is bent into a circle.

Each side of square = x / 4 m and the radius of circle = ( 6 – x ) / 2π m



A = ( x / 4 )2 + {( 6 – x ) / 2π}2 π

= x2 / 16 + ( 36 – 12x + x2 ) / 4π



dA / dx = 2x / 16 + ( 1 / 4π)( - 12 + 2x )

= x / 8 – 3 /π + x / 2π



When dA / dx = 0,

x/8 + x / 2π – 3 /π = 0

2 πx + 8x = 48

x = 24 / ( π + 4 )



d2A / dx2 = 1 / 8 + 1 / 2π > 0



But when the area is a maximum, d2A / dx2 < 0, so the above will be the minimum area only. As the above is a continuous function, so the area can increase indefinitely and hence no maximum area can be found that is bounded by the circle and square. So either the circle or square will make up the maximum area.



When x = 6, A = ( 6 / 4 )2 = 2.25,

When x = 0, A = ( 6 / 2π)2 π= 2.86 > 2.25

So to obtain a maximum area, x = 0 and hence the length of the square is 0.



The above method always works but there are errors in the differentiation part.

dA / dd = 2 ( 3 – 2d )( - 2 ) / π + 2d

= (- 12 + 8d ) / π + 2d

d2A / dd2 = 8 / π + 2 > 0

let x be the length of piece is bent into a cirule, and r be the radius of circule
y be the length of the other piece into a square, and d be length of the side of the square
x = 2兀r and y = 4d and x + y = 6--&gt;x=6 - 4d so r = (6-4d)/2兀=(3-2d)/兀
A = 兀r^2 + d^2
= 兀 ((3-2d)/兀)^2 + d^2
= (3-2d)^2/兀 + d^2
dA/dd = -2x(3-2d)/兀 + 2d
because d^2A/dd^2=4/兀+d&gt;0 so that this is no max. turning point.
when d=0, A = 3^2/兀=2.86478
and when if y=6, d=6/4=1.5, A=1.5^2=2.25
so the max. in region 0=&lt;1.5 = 2.86478 when d=0

2007-07-15 20:24:36 補充：
so the max. area in d region [0,1.5] = 2.86478 when d=0