Minima !

Geometry Level 3

Find the minimum positive value of the function

$f(x) = \dfrac{9x^2 \sin^2 x + 4}{x \sin x}.$

The answer is 12.

2 solutions

Department 8
Feb 7, 2016

Method 1: By using AM-GM ,

$\large{\frac { 9{ x }^{ 2 }\sin ^{ 2 }{ x+4 } }{ x\sin { x } } =9x\sin { x } +\frac { 4 }{ x\sin { x } } \ge 2\sqrt { 9x\sin { x } \times \frac { 4 }{ x\sin { x } } } =\boxed { 12 } }$

Method 2: Graphing

Nice solution, I was thinking that there has to be another way of doing it besides calculus. That was why it had a whole number as a solution.

Akshay Yadav - 5 years, 4 months ago

Nice solution! But I have Method 3 also: Now,as the minimum value of a square can be 0 .Hence , Minimum value of the function will be

Jaideep Khare - 5 years, 4 months ago

Ah! Completing the square method that's nice.

Department 8 - 5 years, 4 months ago

Akshay Yadav
Feb 7, 2016

$y=\frac{9x^2\sin^2x+4}{x\sin x}$

Let $z=x \sin x$ ,

$y=\frac{9z^2+4}{z}$

Now,

$\frac{dy}{dz}=\frac{9z^2-4}{z^2}$

Also,

$\frac{dz}{dx}=x \cos x+\sin x$

So,

$\frac{dy}{dx}=\frac{9x^2 \sin^2 x-4}{x^2 \sin^2}(x \cos x+\sin x)$

For minima (or maxima) $\frac{dy}{dx}=0$ ,

$0=\frac{9x^2 \sin^2 x-4}{x^2 \sin^2}(x \cos x+\sin x)$

Solving for $x$ we get many solution,

For one of them $x=\pm0.87$ , $y\approx 12$ .

Minima !

The answer is 12.

2 solutions

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