Let height of the cylindrical container be h cm
So πr² + 2πrh = 200π
=> r² + 2rh = 200
=> h = 200 − r 2 2 r \frac{200-r²}{2r} 2 r 200 − r 2
So volume of the cylindrical container is , V =
π r 2 h = ( 200 − r 2 2 r ) = π r ( 100 − r 2 2 ) πr²h=(\frac{200-r²}{2r}) = πr(100-\frac{r²}{2}) π r 2 h = ( 2 r 200 − r 2 ) = π r ( 100 − 2 r 2 )
= ( 100 π r − π r 3 2 ) (100πr-\frac{πr³}{2}) ( 100 π r − 2 π r 3 ) cm³
Now we find maximum volume....
d V d r = 100 π − 3 π r 2 2 \frac{dV}{dr} = 100π - \frac{3πr²}{2} d r d V = 100 π − 2 3 π r 2
d V d r = 0 = > \frac{dV}{dr} = 0 => d r d V = 0 => 100 π − 3 π r 2 2 = 0 100π - \frac{3πr²}{2}=0 100 π − 2 3 π r 2 = 0
=> r² = 200 3 \frac{200}{3} 3 200
=> r = 10 2 3 \frac{10\sqrt2}{\sqrt3} 3 10 2
Now d 2 V d r 2 = − 3 π r \frac{d²V}{dr²} = - 3πr d r 2 d 2 V = − 3 π r
[ d 2 V d r 2 ] r = 10 2 3 = − 3 π ∗ 10 2 3 < 0 [\frac{d²V}{dr²} ]_{r=\frac{10\sqrt2}{\sqrt3}}= - 3π*\frac{10\sqrt2}{\sqrt3} <0 [ d r 2 d 2 V ] r = 3 10 2 = − 3 π ∗ 3 10 2 < 0
So volume is maximum when r= 10 2 3 \frac{10\sqrt2}{\sqrt3} 3 10 2
So maximum volume is ( 100 π r − π r 3 2 ) (100πr-\frac{πr³}{2}) ( 100 π r − 2 π r 3 )
= π 2 [ 200 − r 2 ] r \frac{π}{2}[200-r²]r 2 π [ 200 − r 2 ] r
= π 2 [ 200 − 200 3 ] 10 2 3 \frac{π}{2}[200-\frac{200}{3}]\frac{10\sqrt2}{\sqrt3} 2 π [ 200 − 3 200 ] 3 10 2
= π 2 [ 400 3 ] 10 2 3 \frac{π}{2}[\frac{400}{3}]\frac{10\sqrt2}{\sqrt3} 2 π [ 3 400 ] 3 10 2
= 1710.06644 cm³
= 1710.07 cm³ ( Correct up to two decimal places)
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