Application of Differentiation (20pts)

a)( h1 – R )2 + r12 = R2
h12 – 2h1R + R2 + r12 = R2
r12 = 2h1R – h12
r1 = ( 2h1R – h12 )0.5
V1 = πR2 h1 / 3
= π( 2h1R – h12 ) h1 / 3
= π( 2R – h1 ) h12 / 3
b) dV1/ dh1 = ( d / dh1 )[π( 2h12R – h13 ) / 3]
=(π/ 3 )( 4Rh1 – 3h12 )
d2V1 / dh12 = (π/ 3 )( 4R – 6h1 )
When dV / dh1 = 0,
(π/ 3 )( 4Rh1 – 3h12 ) = 0
R = 3h1 / 4
When R = 3h1 / 4,
d2V1 / dh12 = (π/ 3 )( 3h1 – 6h1 )
= -πh1 < 0
So V1 is a max. when R = 3h1 / 4, then h1 = 4R / 3
ci)By similar triangles,
R / r2 = [( h2 – R )2 – R2 ]0.5 / h2
R / r2 = [ h2 ( h2 – 2R ) ]0.5 / h2
r2 = Rh2 / [ h2 ( h2 – 2R ) ]0.5
V2 =π[(Rh2)2 / h2 ( h2 – 2R )](h2)( 1 / 3 )
=πR2 h22 / 3 ( h2 – 2R )
ii)dV2 / dh2 = (πR2 / 3 )( d / dh )[ h22 / ( h2 – 2R )]
= (πR2 / 3 )[ ( h22 – 4Rh2 ) / ( h – 2R )2 ]
When dV2 / dh2 > 0
(πR2 / 3 )[ ( h22 – 4Rh2 ) / ( h – 2R )2 ] > 0
h22 – 4Rh2 > 0
h2 < 0 or h2 > 4R for dV2 / dh2 is increasing
When dV2 / dh2 < 0,
h22 – 4Rh2 < 0
0 < h2 < 4R for dV2 / dh2 is decreasing
So h2 = 4R for V2 is a min.
d)V1 / V2 = [π( 2R – h1 ) h12 / 3] /[πR2 h22 / 3 ( h2 – 2R )]
= ( 2R – h1 )( h2 – 2R )( h12 ) / R2h22
= ( 2R – 4R / 3 )( 4R – 2R )( 4R / 3 )2 / R2 ( 4R )2
= 4 / 27
r1 / r2 = ( 2h1R – h12 )0.5 / [Rh2 / [ h2 ( h2 – 2R ) ]0.5]
= ( 8R2 / 3 – 16R2 / 9 )0.5 / [ 4R2 / [ 4R ( 4R – 2R )]0.5 ]
= ( 8 / 9 )0.5 R / ( 4R2 / 80.5 R )
= 2 / 3
( r1 / r2 )3 = ( 2 / 3 )3
= 4 / 27
So the two cones are similar.

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