Minima/ maxima (statement questions) - Geometry questions

Chapter 6 Class 12 Application of Derivatives
Concept wise

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Ex 6.5, 25 Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan β1 β2Let π be the slant height & ΞΈ be the semi vertical angle of the cone. Now, Height of cone = h = π cos ΞΈ Radius of cone = r = π sin ΞΈ We need to maximize volume of cone V = 1/3 ππ^2 β V = 1/3 ππ^2sin2π l cos π V= 1/3 ππ^3sin2π cos π Differentiating ππ£/ππ = 1/3 ππ^3 (2 sin π cos π . cos π + sin2 π (sin π)) ππ£/ππ = 1/3 ππ^3 [2 sinβ‘π cos^2β‘πβsin^3β‘π ] ππ£/ππ = 1/3 ππ^3 (2 sin π cos2 π sin3 π) ππ£/ππ = 1/3 ππ^3 sin π (2 cos2 π β sin2 π) ππ£/ππ = 1/3 ππ^3 sin π (β2 cosβ‘γπ+sinβ‘γπ)(γ γ β2 cos π β sin π) ππ£/ππ = 1/3 ππ^3 sin π cos π ((β2 cosβ‘γπ+sinβ‘π γ))/cosβ‘π Γ cos π ((β2 cosβ‘γπβsinβ‘π γ))/cosβ‘π ππ£/ππ = 1/3 ππ^3 sin π cos2 π (β2β‘γ+γtan πγβ‘γ )(γ γ β2 β tan π) Putting ππ£/ππ = 0 1/3 ππ^3 sin π cos2 π (β2β‘γ+γtan πγβ‘γ )(γ γ β2 β tan π) = 0 sin π cos2 π (β2β‘γ+γtan πγβ‘γ )(γ γ β2 β tan π) = 0 sin π cos2 π (β2β‘γ+γtan πγβ‘γ )(γ γ β2 β tan π) = 0 sin π½ = 0 π = 0Β° π cannot be 0Β° for cone. cos2 π½ = 0 π = 90Β° π cannot be 90Β° for cone. βπ + tan π½ = 0 tan π = ββ2 For cone, 0Β° < π < 90Β° tan π is (β) ve in II & IV quadrant so tan π = β β2 is not possible βπ β tan π½ = 0 tan π = β2 π = tanβ1 β2 β΄ tan π = β2 is the possible value for cone. The value is either maxima or minima. Finding maxima or minima by first derivative test ππ£/ππ = 1/3 ππ^3 sin π cos2 π (β2 + tanπ) (β2 β tan π) sin π½ 0Β° < π < 90Β° Since π is in 1st quadrant sin π > 0 cos π½ 0Β° < π < 90Β° Since π is in 1st quadrant β΄ cos π > 0 (βπ+πππ π½) 0Β° < ΞΈ < 90Β° Since ΞΈ is in 1st quadrant So, tan ΞΈ > 0 So (β2+π‘ππ ΞΈ) is also (+)ve. Since ππ£/πΞΈ = changes sign from (+) ve to (β) ve π½ = tanβ 1βπ is the maxima. Hence proved