Inspired by Sandeep Bhardwaj

Algebra Level 3

If 0 < θ < π 0 < \theta < \pi , what is the minimum value of

4 sin ⁡ 2 θ + 9 sin ⁡ θ ? \frac{ 4 \sin ^ 2 \theta + 9 } { \sin \theta } ?

This problem is inspired by Sandeep Bhardwaj's problem .

12 14 13 11

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Calvin Lin by Calvin Lin

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5 solutions

Tijmen Veltman
Dec 3, 2014

Basically we're looking for the minimum of f ( x ) = 4 x + 9 x f(x)=4x+\frac9x with the constraint that x ∈ ( 0 , 1 ] x\in (0,1] . To find this minimum we can set the derivative 4 − 9 x 2 4-\frac9{x^2} to zero and we will find x = ± 3 2 x=\pm\frac32 , which is not in our domain. It follows that f ( x ) f(x) is monotonous on ( 0 , 1 ] (0,1] , therefore our minimum must lie at the edge. Hence we find f ( 1 ) = 13 f(1)=\boxed{13} .

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More accurately, because f ′ ( x ) < 0 f'(x) < 0 in the domain ( 0 , 1 ] (0, 1] , hence the minimum occurs at x = 1 x=1 .

Calvin Lin Staff - 6 years, 6 months ago

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When I took the derivative of f ( θ ) f(\theta) directly using the quotient rule, (instead of converting to the variable x x first), I found that an option to have f ′ ( θ ) = 0 f'(\theta) = 0 was cos ⁡ ( θ ) = 0 ⟹ θ = π 2 \cos(\theta) = 0 \Longrightarrow \theta = \frac{\pi}{2} . (The others were sin ⁡ ( θ ) = ± 3 2 \sin(\theta) = \pm \frac{3}{2} ).

So this direct approach does result in a valid value for θ \theta .

It seems like this type of question always generates the most spirited discussions. Is that what you have found too?

Brian Charlesworth - 6 years, 6 months ago
Akshay Bodhare
Dec 2, 2014

we can also use weighted AM-GM

sin ⁡ θ + sin ⁡ θ + sin ⁡ θ + sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ 13 ≥ sin ⁡ 4 θ sin ⁡ 9 θ 13 ≥ 1 \frac { \sin \theta + \sin \theta + \sin \theta + \sin \theta + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } } {13} \\ \geq \sqrt[13]{ \frac {\sin^4 \theta} { \sin^9 \theta }} \\ \geq 1

First Equality will hold when all terms are equal, IE sin ⁡ θ = 1 sin ⁡ θ \sin \theta = \frac{ 1 } { \sin \theta } .
Second Equality will hold when sin ⁡ θ \sin \theta = 1 ).

Clearly, this both conditions are satisfied by sin ⁡ θ = 1 \sin \theta = 1 .

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How would you use the weighted form ? Note my comment to Mardokay that the 'standard' AM-GM inequality would lead to an erroneous answer.

Brian Charlesworth - 6 years, 6 months ago

The point of this question is that weighted AM-GM will not work. Can you show how you arrived at the answer of 13, as opposed to 12?

To me, weighted AM-GM will give

4 sin ⁡ 2 θ + 9 sin ⁡ θ ≥ 2 × 2 sin ⁡ θ × 3 sin ⁡ θ = 12. \frac{ 4 \sin^2 \theta + 9 } { \sin \theta } \geq \frac{ 2 \times 2 \sin \theta \times 3 } { \sin \theta } = 12.

The equality condition will occur when 4 sin ⁡ 2 θ = 9 4 \sin ^2 \theta = 9 , or that sin ⁡ θ = 3 2 \sin \theta = \frac{3}{2} , which is not allowed.

Of course, there may be another way to approach this problem using AM-GM, where the equality condition happens at sin ⁡ θ = 1 \sin \theta = 1 . If so, I would like to see it.

Calvin Lin Staff - 6 years, 6 months ago

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we can write the AM-GM inequality as

sin ⁡ θ + sin ⁡ θ + sin ⁡ θ + sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ + 1 sin ⁡ θ 13 ≥ sin ⁡ 4 θ sin ⁡ 9 θ 13 \frac { \sin \theta + \sin \theta + \sin \theta + \sin \theta + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } + \frac {1} { \sin \theta } } {13} \geq \sqrt[13]{ \frac {\sin^4 \theta} { \sin^9 \theta }}

Equality will hold when all terms are equal i.e.

sin ⁡ θ = 1 sin ⁡ θ \sin \theta = \frac {1} { \sin \theta }

Akshay Bodhare - 6 years, 6 months ago

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Oh wow! Indeed. I forgot that we could do that!!! This is now my favourite solution!

P.S. I've edited your comment into the solution, so that it is easier for others to understand what you did.

Calvin Lin Staff - 6 years, 6 months ago
Pranav Dev
Dec 15, 2014

maximum value of sin is at 90 sin=1 so putting value=1 answer comes 13

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Michael Stevenson
Dec 13, 2014

Let s i n θ = 1 − x , 1 > x > = 0 sin \theta = 1 - x, 1>x>=0 . Then we get 8 + 5 + x 2 1 − x > = 13 8+\frac{5+x^2}{1-x} >= 13

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Humberto Bento
Dec 3, 2014

Why not to find the derivative and equal it to zero? Easy and not subject to errors!

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That approach doesn't work in this case, which is one of the points of my question.

As pointed out by Tijmen, trying to solve for the derivative equals to 0 results in an invalid value of θ \theta .

Calvin Lin Staff - 6 years, 6 months ago

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