Find the maximum value of $17xy + 29yz$ if $x^2 + y^2 + z^2 = 1$ where $x, y, z$ are real numbers.

Give answer to 3 decimal places.

The answer is 16.807.

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This is a nice generalization...

Dev Sharma
- 5 years, 6 months ago

This is a wonderful solution! Please can you tell me how you thought of this? Was it from another problem? Thank you.

Michael Ng
- 5 years, 6 months ago

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Thanks! I have used this approach before. I thought of it then because when the equation of a sphere presented itself it seemed "natural" to convert to spherical coordinates, at which point the key was to assign the coordinates to $x,y,z$ in the most convenient manner. Accordingly, since in this case we have terms $xy$ and $yz$ , I assigned the "simplest" coordinate $\sin(\theta)$ to the shared variable $y.$

Brian Charlesworth
- 5 years, 6 months ago

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Ah ingenious! I hadn't thought of the condition in that way before; thanks for helping!

Michael Ng
- 5 years, 6 months ago

Nice! I think I have seen this problem and these solutions before ;)

Otto Bretscher
- 5 years, 6 months ago

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Haha Yes, I remembered my method but not the specific problem I used it on. I should have guessed it was one of yours. :)

Brian Charlesworth
- 5 years, 6 months ago

The convention of $y = \sin (\theta)$ suits $a x y + b y z$ very much for $y$ . The maximum occurs at all positive of x, y, z and also all negative of x, y, z.

Lu Chee Ket
- 5 years, 6 months ago

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My method is a bit longer than Brian sir's.

Consider the Cauchy-Schwarz Inequality,

$17xy+29yz ≤ \sqrt{17² + 29²} \times \sqrt{y² \times (1-y²) }$

$17xy+29yz ≤ \sqrt{17² + 29²} \times y\times \sqrt{(1-y²) }$ Let $y = \sin\theta$

$17xy+29yz ≤ \sqrt{17² + 29²} \times \sin\theta \times \cos\theta$

$17xy+29yz ≤ \frac{\sqrt{1130}}{2} \times \sin2\theta$

For max value, $\theta = \dfrac{\pi}{4}$

A Former Brilliant Member
- 5 years, 6 months ago

Aareyan, in which class you are?

Dev Sharma
- 5 years, 6 months ago

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standart 8.

Aareyan Manzoor
- 5 years, 6 months ago

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I didn't even know what Lagrange multipliers were till a few weeks ago, and this kid. :D Keep it up.

A Former Brilliant Member
- 5 years, 6 months ago

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Nice! This looks like Harsh' solution here , but you write it more succinctly.

Otto Bretscher
- 5 years, 6 months ago

I started with the inequality -

$(x - \frac{17}{\sqrt{(17)^2 + (29) ^2}}y)^2 + (z - \frac{29}{\sqrt{(17)^2 + (29) ^2}}y)^2 \geq 0$

expanding out gives -

$17xy + 29yz \leq \frac{\sqrt{(17)^2 + (29)^2}}{2} = \boxed{16.807}$

Also the equality occurs at :

$x = \frac{17}{\sqrt{2}.\sqrt{(17)^2 + (29)^2}} , y = \frac{1}{\sqrt{2}}, z = \frac{29}{\sqrt{2}.\sqrt{(17)^2 + (29)^2}}$

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Interesting approach. How did you know that was the inequality to start off with?

"This Person Is Awesome"

Prakhar Bindal
- 5 years, 5 months ago

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Yeah,Not many people who appreciate my work(nothing much to appreciate either) Or were you talking about yourself ?

Rohit Kumar
- 5 years, 5 months ago

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Nahh! i was talking about you only!

Prakhar Bindal
- 5 years, 5 months ago

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@Prakhar Bindal – Well thanks then ! How did you solve it by the way ?

Rohit Kumar
- 5 years, 5 months ago

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@Rohit Kumar – i considered a sphere in 3d space and then used parametric coordinates of a sphere!

Prakhar Bindal
- 5 years, 5 months ago

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We'll first look in general at maximizing $axy + byz$ given $x^{2} + y^{2} + z^{2} = 1.$

Let $x = \cos(\theta)\sin(\phi), y = \sin(\theta)$ and $z = \cos(\theta)\cos(\phi)$ . Then $x^{2} + y^{2} + z^{2} =1$ and

$axy + byz = \sin(\theta)\cos(\theta)(a\sin(\phi) + b\cos(\phi)) =$

$\dfrac{\sin(2\theta)}{2}*\sqrt{a^{2} + b^{2}}*\left(\dfrac{a}{\sqrt{a^{2} + b^{2}}}\sin(\phi) + \dfrac{b}{\sqrt{a^{2} + b^{2}}}\cos(\phi)\right) =$

$\dfrac{\sin(2\theta)}{2}*\sqrt{a^{2} + b^{2}}*\sin(\alpha + \phi),$ where $\alpha = \cos^{-1}\left(\dfrac{a}{\sqrt{a^{2} + b^{2}}}\right).$

This expression achieves a maximum value of $\dfrac{\sqrt{a^{2} + b^{2}}}{2}$ when $\theta = 45^{\circ}$ and $\alpha = 90^{\circ} - \phi.$

In this case $a = 17$ and $b = 29,$ and thus the desired maximum is $\dfrac{\sqrt{17^{2} + 29^{2}}}{2} = \boxed{16.808}$ to 3 decimal places.