1. Class 12
2. Important Question for exams Class 12
3. Chapter 6 Class 12 Application of Derivatives

Transcript

Ex 6.5,20 Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base. Let ð , â be the Radius & Height of Cylinder respectively & ð , ð be the Volume & Surface area of Cylinder respectively Given Surface Area of Cylinder = 2ð ðï·®2ï·¯+ 2ððâ S = 2ð ðï·®2ï·¯+ 2ððâ S â 2ð ðï·®2ï·¯= 2ððâ ð â 2ð ðï·®2ï·¯ï·®2ððï·¯=â â= ð â 2ð ðï·®2ï·¯ï·®2ððï·¯ Volume of Cylinder = ðð2â V = ðð2â We need to maximum volume Now, V = Ïr2h V = Ïr2 ð â 2ð ðï·®2ï·¯ï·®2ððï·¯ï·¯ V = ð ðï·®2ï·¯ï·®2ððï·¯ ð â2ð ðï·®2ï·¯ï·¯ V = ðï·®2ï·¯ ð â2ð ðï·®2ï·¯ï·¯ V = 1ï·®2ï·¯ ðð â2ð ðï·®3ï·¯ï·¯ Diff w.r.t ð ððï·®ððï·¯= 1ï·®2ï·¯ ð ð ðâ2ð ðï·®3ï·¯ï·¯ï·®ððï·¯ ððï·®ððï·¯= 1ï·®2ï·¯ ð â6ð ðï·®2ï·¯ï·¯ Putting ðð£ï·®ððï·¯=0 1ï·®2ï·¯ ð â6ð ðï·®2ï·¯ï·¯=0 ð â6ð ðï·®2ï·¯=0 Putting value of ð = 2ðð2+ 2ððâ 2ð ðï·®2ï·¯+2ððâï·¯â6ð ðï·®2ï·¯=0 â4ð ð2 + 2ððâ = 0 2ððâ â2ð+âï·¯=0 2ðð ââ2ðï·¯=0 ââ2ð=0 â=2ð Finding ðï·®2ï·¯ð£ï·®ð ðï·®2ï·¯ï·¯ ððï·®ððï·¯= 1ï·®2ï·¯ ð â6ð ðï·®2ï·¯ï·¯ ðï·®2ï·¯ðï·®ð ðï·®2ï·¯ï·¯= 1ï·®2ï·¯ ð ð â 6ð ðï·®2ï·¯ï·¯ï·®ððï·¯ ðï·®2ï·¯ð£ï·®ð ðï·®2ï·¯ï·¯= 1ï·®2ï·¯ 0â12ððï·¯ ðï·®2ï·¯ð£ï·®ð ðï·®2ï·¯ï·¯=â6ðð < 0 â´ ðï·®2ï·¯ð£ï·®ð ðï·®2ï·¯ï·¯<0 for â=2ð Hence volume of a cylinder is Maximum when ð=ðð

Chapter 6 Class 12 Application of Derivatives

Class 12
Important Question for exams Class 12