Application of differentiation

知識就是力量

2011-01-20 5:41 am

A wire of length 10 m is cut into two parts. Once part is bent to form a circle of radius r m and the other part is bent to form a square of side s m. Let A m^2 be the sum of the areas of the circle and square formed.
(a) Find r and s in terms of π when A is a minimum.
(b) Hence, find the minimum areas of the circle and square formed.

回答 (1)

良師益友

2011-01-20 6:06 am

✔ 最佳答案

2 π r + 4s = 10

s = ( 10 - 2 π r ) / 4

A = π r^2 + s^2

A = π r^2 + ( 10 - 2 π r ) ^2 / 16

A = [ 16 π r^2 + ( 100 - 40 π r + 4 π^2 r^2 ) ] / 16

A = [ ( 16π + 4 π^2 ) r^2 - 40 π r + 100 ] / 16

A' = 1/16 [ 2( 16π + 4 π^2 ) r - 40π ]

A' = π/16 [ 8( 4 + π ) r - 40 ]

A' = 0 when 8( 4 + π ) r = 40

A' = 0 when r = 5 / ( 4 + π )

A" = π( 4 + π )/2 > 0

Hence, A is min. when r = 5 / ( 4 + π )
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Corresponding s = ( 10 - 2 π r ) / 4

s = 1/4[ 10 - 10π / ( 4 + π ) ]

s = 10 / ( 4 + π )
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A = π r^2 + s^2

Min A = 25π / ( 4 + π )^2 + 100 / ( 4 + π )^2

Min A = 25( 4 + π ) / ( 4 + π )^2

Min A = 25 / ( 4 + π )

Hence, the min. area is [ 25 / ( 4 + π ) ] m^2
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Hope I have answered the question correctly.

參考: Mathematics Teacher Mr. Ip

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