Calculus Question?

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Calculus Question?

The diagram shows a sector of a circle with radius r cm. The perimeter of the sector is 12 cm. Show that the area of the sector, $A c m^{2}$ $A cm^2$ , in terms of r, is $A = \frac{72 θ}{{(θ + 2)}^{2}}$ $A=(72theta)/(theta+2)^2$ . Hence determine the maximum value of A as r varies.

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Answer 1 · 2018-07-27T13:58:45

Please see the explanation below

Explanation:

The area of the sector is

$A = \frac{1}{2} r^{2} θ$ $A=1/2r^2theta$

The perimeter of the sector is

$P = 2 r + r θ$ $P=2r+rtheta$

As $P = 12$ $P=12$

$2 r + r θ = 12$ $2r+rtheta=12$

$=$ $=$ , $r (2 + θ) = 12$ $r(2+theta)=12$

$r = \frac{12}{2 + θ}$ $r=12/(2+theta)$

Therefore,

$A = \frac{1}{2} \cdot {(\frac{12}{2 + θ})}^{2} θ$ $A=1/2*(12/(2+theta))^2theta$

$= \frac{144}{2} \cdot \frac{θ}{{(2 + θ)}^{2}}$ $=144/2*theta/(2+theta)^2$

$A = \frac{72 θ}{{(2 + θ)}^{2}}$ $A=(72theta)/(2+theta)^2$

$A = f (θ)$ $A=f(theta)$

$\frac{d A}{d θ} = \frac{72 {(2 + θ)}^{2} - 2 (2 + θ) 72 θ}{{(2 + θ)}^{4}}$ $(dA)/(d theta)=(72(2+theta)^2-2(2+theta)72theta)/(2+theta)^4$

$= \frac{144 + 72 θ - 144 θ}{{(2 + θ)}^{3}}$ $=(144+72theta-144theta)/(2+theta)^3$

$= \frac{144 - 72 θ}{{(2 + θ)}^{3}} = \frac{72 (2 - θ)}{{(2 + θ)}^{3}}$ $=(144-72theta)/(2+theta)^3=(72(2-theta))/(2+theta)^3$

The maximum is when $\frac{d A}{d θ} = 0$ $(dA)/(d theta)=0$

That is $θ = 2$ $theta=2$ , $\Rightarrow$ $=>$ , $A = 9$ $A=9$

Also,

$θ = \frac{12 - 2 r}{r}$ $theta=(12-2r)/r$

$A = \frac{72 (12 - 2 r)}{r (\frac{144}{r^{2}})} = \frac{r}{2} (12 - 2 r) = 6 r - r^{2}$ $A=(72(12-2r))/(r(144/r^2))=r/2(12-2r)=6r-r^2$

$A = f (r)$ $A=f(r)$

Then,

$\frac{d A}{d r} = 6 - 2 r$ $(dA)/(dr)=6-2r$

The maximum is when $\frac{d A}{d r} = 0$ $(dA)/(dr)=0$

That is

$6 - 2 r = 0$ $6-2r=0$ , $\Rightarrow$ $=>$ , $r = 3$ $r=3$

And $A = 9$ $A=9$

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Calculus Question?

The diagram shows a sector of a circle with radius r cm. The perimeter of the sector is 12 cm. Show that the area of the sector, $A c m^{2}$ $A cm^2$ , in terms of r, is $A = \frac{72 θ}{{(θ + 2)}^{2}}$ $A=(72theta)/(theta+2)^2$ . Hence determine the maximum value of A as r varies.

1 Answer

Explanation:

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Calculus Question?

The diagram shows a sector of a circle with radius r cm. The perimeter of the sector is 12 cm. Show that the area of the sector, Acm2A cm^2, in terms of r, is A=72θ(θ+2)2A=(72theta)/(theta+2)^2. Hence determine the maximum value of A as r varies.

1 Answer

Explanation:

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Impact of this question

The diagram shows a sector of a circle with radius r cm. The perimeter of the sector is 12 cm. Show that the area of the sector, $A c m^{2}$ $A cm^2$ , in terms of r, is $A = \frac{72 θ}{{(θ + 2)}^{2}}$ $A=(72theta)/(theta+2)^2$ . Hence determine the maximum value of A as r varies.