1729 Problems solved!

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.

lagranges multipliers(probably making it too complicated but sharing all approaches). so, first $f=axy+byz-\lambda(x^2+y^2+z^2-1)$ hence $\begin{array}{c}ff_x=-2\lambda x+ay+0z\\ f_y=ax-2\lambda y+bz\\f_z=0x+by-2\lambda z\end{array}$ so $det(\begin{bmatrix}-2\lambda&a&0\\a&-2\lambda&b\\0&b&-2\lambda\end{bmatrix})=0$ or $-8\lambda^3+2(a+b)\lambda=0$ $\lambda_{max}=\dfrac{\sqrt{a^2+b^2}}{2}$ put a=17,b=29 for desired answer. $\boxed{16.808}$

Here is my version of the solution: $axy+byz \leq \sqrt{(a^2+b^2)(y^2(x^2+z^2))}\\ =\sqrt{(a^2+b^2)(y^2(1-y^2))} \leq \sqrt{(a^2+b^2)} \times \dfrac{1}{2}$ .Now just substitute the values of $a$ and $b$ and get the answer!

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 :

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