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

Question 23

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

Solution

Let $$x+4=y$$
So, the expression $$\frac{3\ \left(x+6\right)\left(x+12\right)}{2\left(x+4\right)}$$ equals $$\frac{3*(y+2)(y+8)}{2*y}$$
This equals $$\frac{3}{2} \times (y+16/y+10)$$
The minimum value of $$y+16/y$$ is 8 when $$y$$ equals 4
So, the minimum value of the expression is 3/2*18 = 27

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

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