Elegant way to maximizing linear function subject to being on the surface of a sphere

Question

Problem statement:$$\text{Maximize $x+2y+3z$}\\\text{given $x^2+y^2+z^2=1$.}$$

My solution:

Let $\vec u=(1,2,3),\vec v=(x,y,z)$, then the problem is equivalent to maximizing $\vec{u}\cdot\vec v$. By Cauchy Schwarz , $$(\vec u\cdot\vec v)^2\leq ||\vec u||^2\cdot||\vec v||^2\iff x+2y+3z\le\sqrt{14}$$And from here we can find for which $z,y,z$ the maximum is achieved.

While this proof could be described as concise and elegant, I am seeking a more "elementary" approach—ideally one that avoids the use of the Cauchy-Schwarz inequality, linear algebra, or any advanced mathematical concepts.

Answer 1

• Obviously, $x^2+y^2+z^2=1$ describes a sphere of radius 1.
• The equation $x+2y+3z=k$ for any given $k$ describes a plane passing through the points $(k, 0, 0)$, $(0, k/2, 0)$ and $(0, 0, k/3)$.

So, geometrically, we are trying to maximise the intersection of this plane with the $x$-axis. I claim it is geometrically obvious that a maximal such plane is tangent to the sphere: if it weren't, then you could increase the $x$-intercept by "pushing" the plane out away from the origin.

• A plane tangent to the unit sphere at the point $(x_0, y_0, z_0)$ has equation $x_0 x + y_0 y + z_0 z = 1$.
• Since $1^2+2^2+3^2=14$, the plane $\frac{1}{\sqrt{14}}x+\frac{2}{\sqrt{14}}y+\frac{3}{\sqrt{14}}z=1$ is such a tangent plane.
• Therefore $x+2y+3z=\pm\sqrt{14}$ are the only planes of the form $x+2y+3z=k$ which are tangent to the sphere.

So the maximal $k$ is $\sqrt{14}$.

Answer 2

Here is a solution using 'elementary manipulations' such as completing the squares and AM-GM. This trick works for some problems involving multiple constraints, e.g. this question.

First, we hope to find the constants $M, s$ such that $M - (x + 2y + 3z) + s(x^2 + y^2 + z^2 - 1)$ is a sum of squares (note that it is quadratic in $x, y, z$; we complete the squares in $x, y, z$, respectively). By completing the squares, we have \begin{align*} &M - (x + 2y + 3z) + s(x^2 + y^2 + z^2 - 1)\\ ={}& s\left(x - \frac{1}{2s}\right)^2 + s\left(y - \frac{1}{s}\right)^2 + s\left(z - \frac{3}{2s}\right)^2 + M - s - \frac{7}{2s} \end{align*} Note that $s + \frac{7}{2s} \ge 2\sqrt{s \cdot \frac{7}{2s}} = \sqrt{14}$ by AM-GM. Thus, if we choose $s = \frac12\sqrt{14}$ and $M = \sqrt{14}$, then $M - (x + 2y + 3z) + s(x^2 + y^2 + z^2 - 1)$ is a sum of squares, i.e. \begin{align*} &\sqrt{14} - (x + 2y + 3z) + \frac12\sqrt{14}(x^2 + y^2 + z^2 - 1)\\ ={}& \frac12\sqrt{14}\left(x - \frac{\sqrt{14}}{14}\right)^2 + \frac12\sqrt{14}\left(y - \frac{\sqrt{14}}{7}\right)^2 + \frac12\sqrt{14}\left(z - \frac{3\sqrt{14}}{14}\right)^2. \tag{1} \end{align*} Thus, we have $x + 2y + 3z \le \sqrt{14}$ if $x^2 + y^2 + z^2 = 1$.

Second, from (1), we have $x + 2y + 3z = \sqrt{14}$ if $x = \frac{\sqrt{14}}{14}, y = \frac{\sqrt{14}}{7}, z = \frac{3\sqrt{14}}{14}$ which satisfies $x^2 + y^2 + z^2 = 1$.

Thus, the maximum of $x + 2y + 3z$ subject to $x^2 + y^2 + z^2 = 1$ is $\sqrt{14}$ when $x = \frac{\sqrt{14}}{14}, y = \frac{\sqrt{14}}{7}, z = \frac{3\sqrt{14}}{14}$.

Answer 3

Solving for $z$

$$ \cases{ x + 2 y + 3z = u\\ x^2+y^2+z^2=1 } $$

we have

$$ u^2-2 u x-4 u y+10 x^2+4 x y+13 y^2-9=0 $$

now

$$ x=\frac{1}{10} \left(u-2y\pm 3 \sqrt{4 u y-14 y^2-u^2+10}\right) $$

but $x$ as a solution point on a maximum, should be unique so

$$ 4 u y-14 y^2-u^2+10 = 0\Rightarrow y=\frac{1}{14} \left(2 u\pm\sqrt{10} \sqrt{14-u^2}\right) = 0 $$

and $y$ should be as well, unique then

$$ 14-u^2 = 0 $$