SpletNow by pythagorus theorem ((h-r)^2)+(R^2)=(r^2). Now express R in terms of h and r. We get (R^2)=(2hr-hh). Substitute in the volume of cone equation V=(pi/3)(R^2)h. Now V=(pi/3)(2hr-hh)h. Differentiate wrt h with r as … http://sofia.nmsu.edu/~breakingaway/ebookofcalculus/HandsOnUnitsInvolvingDerivatives/ConeWithMaximalVolume/ConeWithMaximalVolume.html

Maximizing the volume of a cone - Math Central

http://jwilson.coe.uga.edu/emt725/Class/Lanier/Cone/cone.html Splet09. okt. 2024 · Max volume occurs when d V d x = 0 Then 2 3 π ⋅ r 2 x ⋅ r 2 − ( r x) 2 = 1 3 π ( r x) 2 ⋅ − r 2 x r 2 − ( r x) 2 Which reduces to 2 r 2 = 3 ( r x) 2 And x = 2 3 Therefore the ratio … birth preparation

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Splet03. jul. 2024 · From Sector to Cone - Volume in terms of central angle Carmel Schettino 17 subscribers Subscribe 2.5K views 4 years ago This video helps students to see how a sector of circle can be … Splet24. sep. 2015 · Now, the volume of the cylinder V = π r 2 h = π r 2 ( A 2 π r − r) = A 2 r − π r 3 differentiating V w.r.t. r, we get d V d r = A 2 − 3 π r 2 d 2 V d r 2 = − 6 π r < 0 ( ∀ r > 0) Hence, the volume is maximum, now, setting d V d r = 0 for maxima A 2 − 3 π r 2 = 0 r = A 6 π Setting value of r in (1), we get SpletTo calculate the area of a sector of a circle we have to multiply the central angle by the radius squared, and divide it by 2. Area of a sector of a circle = (θ × r 2 )/2 where θ is measured in radians. The formula can also be represented as Sector Area = (θ/360°) × πr 2, where θ is measured in degrees. darden smith youtube