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JEE MainMathematicsEasyApplication of derivative

The maximum volume (in cu. m ) of the right circular cone having
slant height 3m is

Show answer and solution
V=13πr2h,r2+h2=9V = \frac{1}{3}\pi r^{2}h,r^{2} + h^{2} = 9
V=13πh(9−h2)V = \frac{1}{3}\pi h\left( 9 - h^{2} \right)
dvdh=13π(9−3h2)=0\frac{dv}{dh} = \frac{1}{3}\pi\left( 9 - 3h^{2} \right) = 0
9−3h2=09 - 3h^{2} = 0
h2=3,h=3h^{2} = 3,h = \sqrt{3}
v=13(π)(6)3=23πv = \frac{1}{3}(\pi)(6)\sqrt{3} = 2\sqrt{3}\pi

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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