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20 meteers of wire is available for fencing of a flower bed in the form of a circular sector .then maximum areaof flower bed is

sri harshitha honey , 5 Years ago

Grade 12

1 Answers

Arun

Last Activity: 5 Years ago

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

hope it helps in case of any query please feel free to ask

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