MBA
Exam
((a^2+a+1)+(b^2+b+1)+(c^2+c+1)+(d^2+d+1)+(e^2+e+1))/abcde,where a,b,c,d,e are positive integers then find the minimum value of the above expression.
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- lets consider (a^2+a+1)/a=y
so, dy/da= a^2-1/a^2
so, dy/da=0
a^2-1/a^2=0
a=1 or -1
now d2y/da2 = 2/a^3
so for min, d2y/da2 should be max which is at a=1
so similarly a=b=c=d=e=1
so the above expression yields 15 as the minimum value - 10 years agoHelpfull: Yes(1) No(0)
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