Calculus Question?

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 cm^2#, in terms of r, is #A=(72theta)/(theta+2)^2#. Hence determine the maximum value of A as r varies.

1038 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T13:58:45

Please see the explanation below

The area of the sector is

#A=1/2r^2theta#

The perimeter of the sector is

#P=2r+rtheta#

As #P=12#

#2r+rtheta=12#

#=#, #r(2+theta)=12#

#r=12/(2+theta)#

Therefore,

#A=1/2*(12/(2+theta))^2theta#

#=144/2*theta/(2+theta)^2#

#A=(72theta)/(2+theta)^2#

#A=f(theta)#

#(dA)/(d theta)=(72(2+theta)^2-2(2+theta)72theta)/(2+theta)^4#

#=(144+72theta-144theta)/(2+theta)^3#

#=(144-72theta)/(2+theta)^3=(72(2-theta))/(2+theta)^3#

The maximum is when #(dA)/(d theta)=0#

That is #theta=2#, #=>#, #A=9#

Also,

#theta=(12-2r)/r#

#A=(72(12-2r))/(r(144/r^2))=r/2(12-2r)=6r-r^2#

#A=f(r)#

Then,

#(dA)/(dr)=6-2r#

The maximum is when #(dA)/(dr)=0#

That is

#6-2r=0#, #=>#, #r=3#

And #A=9#