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SHOW THAT THE SEMI-VERTICAL ANGLE OF RIGHT CIRCULAR CONE OF GIVEN SURFACE AREA AND MAX VOLUME IS SIN INVERSE(1/3)..??
Dear Studentlet r be the radius,l the slant height and h the height of the cone.let S denote the surface area and V the volume of the cone .then S = (πr² + πrl) = constantl = (S/πr - r)-------( 1 )now,V = 1/3πr²h = 1/3πr²√(l² - r²).V² = 1/9π²r⁴(l² - r²)V² = (1/9)π²r⁴[(S/πr - r)² - r²] ----- from ( 1 )V² = (1/9)S(Sr² - 2πr⁴).thus V² = (S²r²/9 - 2πSr⁴/9)2V•dV/dr = (2S²r/9 - 8πS*r³/9) = 2rS/9(S - 4πr²)----( 2 )now, dV/dr = 0r = 0 or (S - 4πr²) = 0r² = S/4π [neglecting r = 0].on differentiating (2) we get,2(dV/dr)² + 2V*d²V/dr² = 1/9S(2S - 24πr²).putting dV/dr = 0 and r² = S/4π, we get2V*d²V/dr² = 1/9*S(2S - 6S) = -4/9S² < 0.when the volume is maximum, we haver² = S/4π = (πr² + πrl)/4πl = 3r.now, if x is semivertical angle of the cone then,r/l = sin xr/3r = sin xsin x = 1/3x = sin^-1(1/3)hence,the semi Vertical angle of a right cone of a given surface and maximum value issin^-1(1/3) .Thanks
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