The minimum value of 6-x11-x2-x- x-0 is

Let nbsp;y=(6+x)(11+x)2+x=x^2+17x+662+x y'=(2+x)(2x+17) (x^2+17x+66)(2+x)^2 ⇒ y'=2x^2+21x+34 x^2 17x 66(2+x)^2 ⇒ y'=x^2+4x 32(2+x)^2 ⇒ y'=(x+8)(x 4)(2+x)^2 Cr ...

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Standard XII

Mathematics

Definition of a Determinant

The minimum v...

Question

The minimum value of $\frac{(6 + x) (11 + x)}{(2 + x)}, x \geq 0$ is

45

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25

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5

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15

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Solution

The correct option is B $25$
Let $y = \frac{(6 + x) (11 + x)}{2 + x} = \frac{x^{2} + 17 x + 66}{2 + x}$
$y^{'} = \frac{(2 + x) (2 x + 17) - (x^{2} + 17 x + 66)}{(2 + x)^{2}}$
$\Rightarrow y^{'} = \frac{2 x^{2} + 21 x + 34 - x^{2} - 17 x - 66}{(2 + x)^{2}}$
$\Rightarrow y^{'} = \frac{x^{2} + 4 x - 32}{(2 + x)^{2}}$
$\Rightarrow y^{'} = \frac{(x + 8) (x - 4)}{(2 + x)^{2}}$
Critical points are $x = - 8, - 2, 4$

But given that $x \geq 0$
Our critical point is $x = 4$ only
$f (x)$ changes sign from negative to positive as $x$ crosses $4$ from left to right. So $x = 4$ is the point of local minima.
Now compare value of function at $x = 4$ with value of function at the boundary points. So,
$f (4) = \frac{10 \times 15}{6} = 25$
$f (0) = 33$
when $x \to \infty, f \to \infty$
Hence, the minimum value of $f (x)$ is $25$

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The minimum value of (6+x)(11+x)(2+x), x≥0 is

The minimum value of $\frac{(6 + x) (11 + x)}{(2 + x)}, x \geq 0$ is