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Grade 12 math
A string of length k cm can be cut into 2 pieces,one piece to form an equilateral triangle with side length x (x in between 0 and k/3) and the other to form a circle of radius r
Determine the most appropriate way of using the string if the total area is to be maximised
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- By SreedeviS |
- Category: calculus |
- Grade level: 12th grade (18)
- Aug/26/2008 |
- Answers to this question: 1
|
Answer by: Ganesha
First Answer
The lengt h of the String = k
Le x be the sides .of the equilateral triangle . Therefore. each side of the triangle =x/3 and the area = (sqrt3/4)side2 = (sqrt3/4)(x/3)2=(sqrt3/36)x2-------------------(1)
Rest of the sring =k-x which is equal to the cicumference of a circle whose radius is = (k-x)/2Pi.
The area of this cicle =Pi.Radius^2 = Pi *[(k-x)/2Pi]2 = (k-x)2/(4Pi)------------(2)
The total area A formed by the wire = (1)+(2)= A = (sqrt3/36)x2 + (k-x)2/(4Pi)--------(3)
dA/dx=0 Gives, 0=(sqrt3/36)2x+2(k-x)(-1)/(4Pi)]==>
(2sqrt3/36 + 2/4Pi]x = 2k/(4Pi)
x =(k/2Pi)/[sqrt3/18 +1/2Pi] = 9k*Pi/{(sqrt3)Pi +9} = 0.623208358k is the piece length used for the equilateral triangle and 1-x =0.376791642k is the piece used for the circle to get the highest area.
d2A/dx2 =sqrt3/36 - 1/2Pi= -0.11104 which is -ve.
Therefore,
Le x be the sides .of the equilateral triangle . Therefore. each side of the triangle =x/3 and the area = (sqrt3/4)side2 = (sqrt3/4)(x/3)2=(sqrt3/36)x2-------------------(1)
Rest of the sring =k-x which is equal to the cicumference of a circle whose radius is = (k-x)/2Pi.
The area of this cicle =Pi.Radius^2 = Pi *[(k-x)/2Pi]2 = (k-x)2/(4Pi)------------(2)
The total area A formed by the wire = (1)+(2)= A = (sqrt3/36)x2 + (k-x)2/(4Pi)--------(3)
dA/dx=0 Gives, 0=(sqrt3/36)2x+2(k-x)(-1)/(4Pi)]==>
(2sqrt3/36 + 2/4Pi]x = 2k/(4Pi)
x =(k/2Pi)/[sqrt3/18 +1/2Pi] = 9k*Pi/{(sqrt3)Pi +9} = 0.623208358k is the piece length used for the equilateral triangle and 1-x =0.376791642k is the piece used for the circle to get the highest area.
d2A/dx2 =sqrt3/36 - 1/2Pi= -0.11104 which is -ve.
Therefore,
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- Aug/28/2008 |
- Answers by Ganesha: 1111 |
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