HSC Science (Electronics) 12th Board ExamMaharashtra State Board
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# Solution for A wire of length l is cut into two parts. One part is bent into a circle and other into a square. Show that the sum of areas of the circle and square is the least, if the radius of circle is half the side of the square. - HSC Science (Electronics) 12th Board Exam - Mathematics and Statistics

ConceptMaxima and Minima - Introduction of Extrema and Extreme Values

#### Question

A wire of length l is cut into two parts. One part is bent into a circle and other into a square. Show that the sum of areas of the circle and square is the least, if the radius of circle is half the side of the square.

#### Solution

Length of the wire is 'l'.
Let the part bent to make circle is of length 'x',
and the part bent to make square is of length 'l - x'.
Circumference of the circle = 2πr = x

r=x/(2pi)

Area of the circle =pir^2=pi(x/(2pi))^2=x^2/(4pi)

Perimeter of the square = 4a=l-x rArra=(l-x)/4

Area of the square = ((1-x)/4)^2=(1-x)^2/16

Sum of the areas A(x) =  x^2/(4pi)+(l-x)^2/16

For extrema, (dAx)/(dx)=0

(2x)/(4pi)+(2(l-x)(-1))/16=0

(4(2x)+2pi(x-l))/16=0

4x+pix-pil=0

x=(pil)/(4+pi)

Since there is one point of extremum, it has to be the minimum in this case.

r=x/(2pi)=l/(2(4+pi)).........(1)

Side of the square a=(l-x)/4=(l-(pil)/(4+pi))/4=l/(4+pi)..................(2)

From (1) and (2), we get that the radius of the circle is half the side of the square, for least sum of areas. (Proved)

Is there an error in this question or solution?

#### APPEARS IN

2015-2016 (March) (with solutions)
Question 6.2.2 | 4.00 marks
Solution for question: A wire of length l is cut into two parts. One part is bent into a circle and other into a square. Show that the sum of areas of the circle and square is the least, if the radius of circle is half the side of the square. concept: Maxima and Minima - Introduction of Extrema and Extreme Values. For the courses HSC Science (Electronics), HSC Science (General) , HSC Arts, HSC Science (Computer Science)
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