122 



PRACTICAL MATHEMATICS 



That is, when 



(R 2 - 



- 3x z ) = 



R 2 - 2Rr - 3x z = 



(R - 80) (R + or) = 

 R 



Height of cone 

 Radius of base 



Volume of cone 



4R 

 - 



2A/2 



II 



TT 4R 8R 

 339 

 32?cR 3 

 81 



76. Example 4. Find the dimensions of a cylindrical vessel 

 of greatest capacity which can be made from a given amount of 

 sheet metal (1) When the vessel has no lid; (2) When the 

 vessel has a lid. 



Let S be the area of sheet metal used (1) Without lid. 



S = surface area of vessel = 2rcxy + TO 2 , where x = radius of 

 base and y = height. 



Volume of vessel = TO 2 */ 



S 



S - TC 

 i 



I 27KB 



= -(S*- 



Then J = i(S - 

 dx 2 V 



Now v is a maximum when -7- = 



ax 



That is, when 



Height of vessel = y 



