Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector.

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20 meters of wire is available for fencing off a flower-bed in the form of a circular sector.



Then what is the maximum area (in sq. m) of the flower-bed?







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    up vote
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    20 meters of wire is available for fencing off a flower-bed in the form of a circular sector.



    Then what is the maximum area (in sq. m) of the flower-bed?







    share|cite|improve this question






















      up vote
      2
      down vote

      favorite
      2









      up vote
      2
      down vote

      favorite
      2






      2





      20 meters of wire is available for fencing off a flower-bed in the form of a circular sector.



      Then what is the maximum area (in sq. m) of the flower-bed?







      share|cite|improve this question












      20 meters of wire is available for fencing off a flower-bed in the form of a circular sector.



      Then what is the maximum area (in sq. m) of the flower-bed?









      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Apr 3 '17 at 5:51









      starunique2016

      533326




      533326




















          1 Answer
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          This is a very simple question. Let $M$ be the angle, the arc of the sector $AOB$ makes with the center $O$. Let OA = OB = radius r.



          Given
          $$ OA + arc AB + OB = 20 $$
          $$2r + rM = 20 $$
          $$ M = 20-2rr$$



          Area $A$ of the sector =$ fracM2r^2 = 10r-r^2$



          $$fracdAdr= 10 - 2r = 0$$
          $$r=5$$



          Second derivative is negative for $r =5$



          We get maximum area at $r=5$



          $A = 25 $






          share|cite|improve this answer






















          • can you tidy up the maths typesetting? It's quite difficult to follow.
            – Harambe
            Apr 5 '17 at 9:48










          • an edit is suggested @Sanjay Bhatnagar
            – starunique2016
            Apr 10 '17 at 5:15










          Your Answer




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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote



          accepted










          This is a very simple question. Let $M$ be the angle, the arc of the sector $AOB$ makes with the center $O$. Let OA = OB = radius r.



          Given
          $$ OA + arc AB + OB = 20 $$
          $$2r + rM = 20 $$
          $$ M = 20-2rr$$



          Area $A$ of the sector =$ fracM2r^2 = 10r-r^2$



          $$fracdAdr= 10 - 2r = 0$$
          $$r=5$$



          Second derivative is negative for $r =5$



          We get maximum area at $r=5$



          $A = 25 $






          share|cite|improve this answer






















          • can you tidy up the maths typesetting? It's quite difficult to follow.
            – Harambe
            Apr 5 '17 at 9:48










          • an edit is suggested @Sanjay Bhatnagar
            – starunique2016
            Apr 10 '17 at 5:15














          up vote
          2
          down vote



          accepted










          This is a very simple question. Let $M$ be the angle, the arc of the sector $AOB$ makes with the center $O$. Let OA = OB = radius r.



          Given
          $$ OA + arc AB + OB = 20 $$
          $$2r + rM = 20 $$
          $$ M = 20-2rr$$



          Area $A$ of the sector =$ fracM2r^2 = 10r-r^2$



          $$fracdAdr= 10 - 2r = 0$$
          $$r=5$$



          Second derivative is negative for $r =5$



          We get maximum area at $r=5$



          $A = 25 $






          share|cite|improve this answer






















          • can you tidy up the maths typesetting? It's quite difficult to follow.
            – Harambe
            Apr 5 '17 at 9:48










          • an edit is suggested @Sanjay Bhatnagar
            – starunique2016
            Apr 10 '17 at 5:15












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          This is a very simple question. Let $M$ be the angle, the arc of the sector $AOB$ makes with the center $O$. Let OA = OB = radius r.



          Given
          $$ OA + arc AB + OB = 20 $$
          $$2r + rM = 20 $$
          $$ M = 20-2rr$$



          Area $A$ of the sector =$ fracM2r^2 = 10r-r^2$



          $$fracdAdr= 10 - 2r = 0$$
          $$r=5$$



          Second derivative is negative for $r =5$



          We get maximum area at $r=5$



          $A = 25 $






          share|cite|improve this answer














          This is a very simple question. Let $M$ be the angle, the arc of the sector $AOB$ makes with the center $O$. Let OA = OB = radius r.



          Given
          $$ OA + arc AB + OB = 20 $$
          $$2r + rM = 20 $$
          $$ M = 20-2rr$$



          Area $A$ of the sector =$ fracM2r^2 = 10r-r^2$



          $$fracdAdr= 10 - 2r = 0$$
          $$r=5$$



          Second derivative is negative for $r =5$



          We get maximum area at $r=5$



          $A = 25 $







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 14 at 12:23









          Nehal Samee

          1469




          1469










          answered Apr 5 '17 at 9:18









          Sanjay Mohan Bhatnagar

          693




          693











          • can you tidy up the maths typesetting? It's quite difficult to follow.
            – Harambe
            Apr 5 '17 at 9:48










          • an edit is suggested @Sanjay Bhatnagar
            – starunique2016
            Apr 10 '17 at 5:15
















          • can you tidy up the maths typesetting? It's quite difficult to follow.
            – Harambe
            Apr 5 '17 at 9:48










          • an edit is suggested @Sanjay Bhatnagar
            – starunique2016
            Apr 10 '17 at 5:15















          can you tidy up the maths typesetting? It's quite difficult to follow.
          – Harambe
          Apr 5 '17 at 9:48




          can you tidy up the maths typesetting? It's quite difficult to follow.
          – Harambe
          Apr 5 '17 at 9:48












          an edit is suggested @Sanjay Bhatnagar
          – starunique2016
          Apr 10 '17 at 5:15




          an edit is suggested @Sanjay Bhatnagar
          – starunique2016
          Apr 10 '17 at 5:15












           

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