Optimization of $2x+3y+z$ under the constraint $x^2+ y^2+ z^2= 1$

Question

Let $x, y$ and $z$ be real numbers such that $x^2+ y^2+ z^2= 1$. Find the maximum and minimum values of $2x + 3y + z$ .

How can I able to solve the problem? Thanks for your help.

Answer 1

Using the Cauchy-Schwarz inequality, $2x+3y+z\leq \sqrt{2^2+3^2+1^2}\sqrt{x^2+y^2+z^2}=\sqrt{14}$ which is the maximum value.Since $x$,$y$ and $z$ are reals, the minimum value occurs when $x$,$y$ and $z$ are negative which can be used to infer that the minimum value is $-\sqrt{14}$.

Answer 2

Using Lagrange Multiplier, $$f(x,y,z,\lambda)=2x+3y+z-\lambda(x^2+y^2+z^2-1)$$

$$\frac{\delta f}{\delta x}=2-2\lambda x$$ etc,

For the extreme values of $f, \frac{\delta f}{\delta x}=0$ etc,

$\implies x=\frac1{\lambda},y=\frac3{2\lambda},z=\frac1{2\lambda}\implies 2x+3y+z=\frac7{\lambda}$

But $x^2+y^2+z^2=1\implies \lambda^2=\frac72$

If $\lambda=\sqrt{\frac72},2x+3y+z=\frac7{\lambda}=\sqrt {14}$

If $\lambda=-\sqrt{\frac72},2x+3y+z=\frac7{\lambda}=-\sqrt {14}$

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Alternatively, as $x,y,z$ are real, we can set $x=\cos A\cos B,y=\cos A\sin B,z=\sin A$

So, $$2x+3y+z=2\cos A\cos B+3\cos A\sin B+\sin A=\cos A(2\cos B+3\sin B)+\sin A$$

Let us find the extreme values of $F(C)=a\cos C+b\sin C$ (This can derived at least in two more ways without using calculus, see below)

So, $F'(C)=-a\sin C+b\cos C$

For the extreme values of $F, F'(C)=0\implies a\sin C-b\cos C=0\iff \frac a{\cos C}=\frac b{\sin C}=\pm \sqrt{a^2+b^2}$

$F''(C)=-a\cos C-b\sin C$

If $\frac a{\cos C}=\frac b{\sin C}=\sqrt{a^2+b^2}, F''(C)=-\sqrt{a^2+b^2}<0$

So, $F_{max}=\sqrt{a^2+b^2}$

Similarly, $F_{min}=-\sqrt{a^2+b^2}$

So, $-\sqrt{2^2+3^2}\le 2\cos B+3\sin B\le \sqrt {2^2+3^2}$

$\implies -\sqrt {13}\cos A+\sin A\le \cos A(2\cos B+3\sin B)+\sin A\le \sqrt {13}\cos A+\sin A$

Again, $\sqrt {13}\cos A+\sin A\le \sqrt{13+1}=\sqrt{14}$

and $-\sqrt {13}\cos A+\sin A\ge -\sqrt{13+1}=-\sqrt{14}$

$\implies -\sqrt {14} \le \cos A(2\cos B+3\sin B)+\sin A\le \sqrt {14}$

$\implies -\sqrt {14} \le 2x+3y+z\le \sqrt {14}$

[

(i)Let $y=a\cos C+b\sin C\implies (y-a\cos C)^2=(b\sin C)^2=b^2-b^2\cos^2C$

$\implies (a^2+b^2)\cos^2C-2ya\cos C+y^2-b^2=0$ which is a Quadratic equation in $\cos C$. As $\cos C$ is real, $$(-2ya)^2\ge4(a^2+b^2)(y^2-b^2)\implies y^2\le a^2+b^2\implies -\sqrt{a^2+b^2}\le y\le \sqrt{a^2+b^2}$$

(ii) Putting $a=R\cos D,b=R\sin D$ where $R>0$ so that $R^2=a^2+b^2$

So, $a\cos C+b\sin C=R\cos D\cos C+R\sin D\sin C=\sqrt{a^2+b^2}\cos(C-D)$

As $-1\le \cos(C-D)\le 1,$ $$-\sqrt{a^2+b^2}\le a\cos C+b\sin C\le \sqrt{a^2+b^2}$$

]

Answer 3

This is a simple geometry problem.

The set $x^2 + y^2 + z^2 = 1$ is a sphere. The set $2x + 3y + z = k$ is a plane perpendicular to the vector $(2,3,1)$ at a distance $k/\sqrt{14}$ from the origin. Optimal values occur for values of $k$ that cause the plane to be tangent to the sphere. This will happen when the directed distance from the plane to the origin is $\pm1$. This means $k/\sqrt{14} = \pm1$, so $k = \pm \sqrt{14}$.