The point at the curve y = 12x − x2 where the slope of the tangent is zero will be _____________ .

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None of these 	ext { Let }\left( x_1 , y_1 ight)	ext { be the required point.} 	ext { Since, the point lies on the curve }, 	ext { Hence }, y_1 = 12 x_1 - {x_1}^2 	ext { Now }, y = 12x - x^2 \Rightarrow dy/dx = 12 - 2x 	ext { Slope of the tangent } = \left( dy/dx ight)_\left( x_1 , y_1 ight) =12 - 2 x_1 	ext { Given }: 	ext { Slope of the tangent }=0 12 - 2 x_1 = 0 \Rightarrow x_1 = 6 	ext { Now }, y_1 = 12 x_1 - {x_1}^2 = 72 - 36 = 36 	herefore \left( x_1 , y_1 ight) = \left( 6, 36 ight)

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