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Question

If a,b,c>0, ab2c3=64 and a+b+c=k, then

A
maximum value of k=33
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B
at maximum value of k, a:b:c=3:2:1
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C
at minimum value of k, a:b:c=1:2:3
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D
minimum value of k=6×(1627)1/6
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Solution

The correct option is D minimum value of k=6×(1627)1/6
a+2b2+3c366a(b2)2(c3)3
k6×6644×27=6×(1627)1/6
So, the minimum value of k=6×(1627)1/6

At this value a=b2=c3
a:b:c=1:2:3

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