# 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:

A. 12.5

B.10

C.25

D.30

Last updated date: 27th Mar 2023

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Answer

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Hint: Use the concept of maxima- minima. Single derivative or double derivative rule. It’s up to you. Get the extreme points and then find the maximum value.

We know that the area of the circular sector is $A = \dfrac{1}{2} \times l \times r$, where l is length and r is the radius of circular sector. Also, $20 = 2r + l$ which implies $l = 20 - 2r$.

Now, using the value of l in the area formula we get, $A = \dfrac{1}{2}(20 - 2r)r = 10r - {r^2}$. On differentiating it, $\dfrac{{dA}}{{dr}} = 10 - 2r$. Notice that, we need the r where $\dfrac{{dA}}{{dr}}$ is maximum. So, we need to consider $\dfrac{{dA}}{{dr}} = 0$ to get the extreme points of r. So, $\dfrac{{dA}}{{dr}} = 0 \Rightarrow 2r = 10 \Rightarrow r = 5$. Double differentiation of A will give us, $\dfrac{{{d^2}A}}{{d{r^2}}} = - 2$. Which is negative.

It means whatever point we have got on considering $\dfrac{{dA}}{{dr}} = 0$ will be the point of maximum. Hence, $r = 5$ is the maximum point. The maximum value will be $A = 10 \times 5 - {5^2} = 50 - 25 = 25{\text{ }}{m^2}$.

Hence, option C is the correct option.

Note: Here, we have used the double differentiation rule of maxima-minima. One can also use other rules. For example: - single differentiation rule.

We know that the area of the circular sector is $A = \dfrac{1}{2} \times l \times r$, where l is length and r is the radius of circular sector. Also, $20 = 2r + l$ which implies $l = 20 - 2r$.

Now, using the value of l in the area formula we get, $A = \dfrac{1}{2}(20 - 2r)r = 10r - {r^2}$. On differentiating it, $\dfrac{{dA}}{{dr}} = 10 - 2r$. Notice that, we need the r where $\dfrac{{dA}}{{dr}}$ is maximum. So, we need to consider $\dfrac{{dA}}{{dr}} = 0$ to get the extreme points of r. So, $\dfrac{{dA}}{{dr}} = 0 \Rightarrow 2r = 10 \Rightarrow r = 5$. Double differentiation of A will give us, $\dfrac{{{d^2}A}}{{d{r^2}}} = - 2$. Which is negative.

It means whatever point we have got on considering $\dfrac{{dA}}{{dr}} = 0$ will be the point of maximum. Hence, $r = 5$ is the maximum point. The maximum value will be $A = 10 \times 5 - {5^2} = 50 - 25 = 25{\text{ }}{m^2}$.

Hence, option C is the correct option.

Note: Here, we have used the double differentiation rule of maxima-minima. One can also use other rules. For example: - single differentiation rule.

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