The minimum value of \frac{3(6+x)(x+12)}{2(4+x)}, where x > -4 is

CMAT 2021 Solved Question Paper

Question : 23 of 100

Marks: +1, -0

The minimum value of

{3(6+x)(x+12)}/{2(4+x)}

, where

x > -4

Solution:

Since,

x > -4

∴ At

x=0

, the minimum value of

{3(6+x)(x+12)}/{2(4+x)}

={3(6+0)(0+12)}/{2(4+0)}

={3 ×6 ×12}{8}

=27

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