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Minimum Value of Expression

CMAT 2021 · Quantitative Techniques & Data Interpretation · Medium · Algebra

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

  1. A.

    18

  2. B.

    27

  3. C.

    36

  4. D.

    45

Answer

B

Explanation

Let y=x+4>0  ⟹  x=y−4y = x + 4 > 0 \implies x = y - 4. Expression = 3(y+2)(y+8)2y=3(y2+10y+16)2y=32(y+16y+10)\frac{3(y + 2)(y + 8)}{2y} = \frac{3(y^2 + 10y + 16)}{2y} = \frac{3}{2}\left(y + \frac{16}{y} + 10\right). By AM-GM, minimum of y+16yy + \frac{16}{y} is 216=82\sqrt{16} = 8. Minimum value = 32(8+10)=32(18)=27\frac{3}{2}(8 + 10) = \frac{3}{2}(18) = 27.

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