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Application Of Derivatives

Question
CBSEENMA12035534
Wired Faculty App

A wire of length 25 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What would be the lengths of the two pieces, so,that combined area of the square and the circle is minimum?

Solution
Total length of wire = 25 metre. Let x metres be made into a circle and (25 – x) metres into a square.
space therefore space space space space space radius space of space circle space equals space fraction numerator straight x over denominator 2 straight pi end fraction metres space and space side space of space square space space equals space fraction numerator 25 minus straight x over denominator 4 end fraction metres space Let space straight S space be space the space sum space of space areas space of space two space figures. space therefore space space space space space space space straight S space equals space straight pi open parentheses fraction numerator straight x over denominator 2 straight pi end fraction close parentheses squared plus open parentheses fraction numerator 25 minus straight x over denominator 4 end fraction close parentheses squared space equals space fraction numerator straight x squared over denominator 4 straight pi end fraction plus fraction numerator left parenthesis 25 minus straight x right parenthesis squared over denominator 16 end fraction space space space space space space space space space dS over dx equals fraction numerator 2 straight x over denominator 4 straight pi end fraction plus fraction numerator 2 left parenthesis 25 minus straight x right parenthesis thin space left parenthesis negative 1 right parenthesis over denominator 16 end fraction space equals space fraction numerator straight x over denominator 2 straight pi end fraction minus fraction numerator 25 minus straight x over denominator 8 end fraction space space space space space space space space space space space space dS over dx equals 0 space space gives space us space space fraction numerator straight x over denominator 2 straight pi end fraction minus fraction numerator 25 minus straight x over denominator 8 end fraction equals 0 space
therefore space space space space 4 straight x minus 25 straight pi space plus space xπ space equals space 0 space space space space space space rightwards double arrow space space space space left parenthesis straight pi plus 4 right parenthesis space straight x space equals space 25 straight pi
therefore space space space space space space straight x space equals space fraction numerator 25 space straight pi over denominator straight pi plus 4 end fraction
          fraction numerator straight d squared straight S over denominator dx squared end fraction space equals fraction numerator 1 over denominator 2 straight pi end fraction plus 1 over 8
When straight x space equals fraction numerator 25 space straight pi over denominator straight pi space plus 4 end fraction comma space space space space fraction numerator straight d squared straight S over denominator dx squared end fraction space equals space fraction numerator 1 over denominator 2 straight pi end fraction plus 1 over 8 greater than 0
therefore space space space space space straight S space is space minimum space when space straight x space equals fraction numerator 25 space straight pi over denominator straight pi plus 4 end fraction therefore space space space the space wire space is space to space be space cut space at space straight a space distance space of space fraction numerator 25 space straight pi over denominator straight pi plus 4 end fraction metres space from space one space end. space

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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