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  • Help me solve this Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is :

Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector.  Then the maximum area (in sq. m) of the flower-bed, is :

  • Option 1)

    10

  • Option 2)

     25

  • Option 3)

    30

  • Option 4)

    12.5

     

 

Answers (1)

best_answer

As we have learned

Method for maxima or minima -

By second derivative method :

Step\:1.\:\:find\:values\:of\:x\:for\:\frac{dy}{dx}=0

Step\:\:2.\:\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:maximum\:if  f''(x)<0\:\:and\:local\:minimum\:if\:f''(x)>0

- wherein

Where\:\:y=f(x)

\frac{dy}{dx}=f'(x)

 

 So,  P= 2r + r \theta = 20 .......(1)\\A = 1/2 r^2 \theta

So , 

\frac{dA }{d\theta }= \frac{d(1/2 r^2\theta )}{d\theta }= \frac{d(1/2 (\frac{20}{2+\theta })^2\theta )}{d\theta }

\left ( from (1) , r = \frac{20}{2+\theta } \right )

\frac{dA }{d\theta } = 200 \frac{d}{d\theta }\left ( \frac{\theta }{(2+\theta )^2} \right )= 200\times \frac{(2+\theta )^2-2\theta (2+\theta )}{(2+\theta )4}= 0

\therefore (2+ \theta )^2 - 2 \theta (2+\theta )= 0 \\\Rightarrow (2+\theta )(2-\theta )=0 \\\rightarrow \theta = \pm 2\\\Rightarrow r = 5

Hence A_{max} = \frac{1}{2}(5)^2(2)= 25

 

 

 

 

 


Option 1)

10

Option 2)

 25

Option 3)

30

Option 4)

12.5

 

Posted by

Himanshu

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