| http://www.w3.org/ns/prov#value | - To find when the surface area is a minimum, we need to find dS/dr .dS = 4pr - 4000pdr r2When dS/dr = 0:4pr - (4000p)/r2 = 0Therefore 4pr = 4000p r2So 4pr3 = 4000pSo r3 = 1000So r = 10You should then check that this is indeed a minimum using the technique above.So the minimum area occurs when r = 10.
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