Minimize $3\sqrt{5-2x}+\sqrt{13-6y}$ subject to $x^2+y^2=4$

Question

If $x, y \in \mathbb{R}$ such that $x^2+y^2=4$, find the minimum value of $3\sqrt{5-2x}+\sqrt{13-6y}$.

I could observe that we can write $$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{x^2+y^2+1-2x}+\sqrt{x^2+y^2+9-6y}$$ $\implies$ $$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-3)^2}=3PA+PB$$ Where $P$ is a generic point on $x^2+y^2=4$ and $A(1,0),B(0,3)$. So the problem essentially means which point on the circle $x^2+y^2=4$ minimizes $3PA+PB$. I am really struggling to find the geometrical notion of this and hence unable to solve.

Answer 1

Comment: you may use this modification:

We rewrite final relation as:

$$A=\sqrt{[3(x-1)=a]^2+(3y=b)^2}+\sqrt{(x=a')^2+(y-3=b')^2}$$

and use this inequality:

$\sqrt{a^2+b^2}+\sqrt{a'^2+b'^2}\geq\sqrt{(a+a')^2+(b+b'^2)}$

we get:

$A\geq\sqrt{16(x^2+y^2)-24(x+y)+18}$

$x^2+y^2=4$

$\Rightarrow$

$A\geq \sqrt{82-24(x+y)}$

If $x=y=\sqrt 2$ then $A\geq\sqrt{84-67.7}\approx 4$

Update:Wolfram says minimum is $2\sqrt {10}=6.32$ at $(x,y)=(\frac25+\frac{3\sqrt6}5=1.87, \frac 65-\frac{\sqrt6}5)=0.71$. If we put this in $\sqrt{82-24(x+y)}$ we get 4.47. mind you x and y must suffice $x^2+y^2=4$, and what Wolfram gives does; $1.87^2+0.71^2=4$.Hence 40 can not be minimum .

Answer 2

We can parametrize the constraint condition as $$(x,y) = (2 \cos t, 2 \sin t), \quad t \in [0,2\pi),$$ hence we are interested in the extrema of $$f(t) = 3\sqrt{5-4\cos t} + \sqrt{13-12\sin t}.$$ Differentiating to locate critical points, we find $$\sin t \sqrt{13 - 12 \sin t} = \cos t \sqrt{5 - 4 \cos t},$$ or $$\tan^2 t = \frac{5 \sec t - 4}{13 \sec t - 12 \tan t}.$$ Collecting like terms yields $$(12 \tan^3 t - 4)^2 = (13 \tan^2 t - 5)^2 \sec^2 t,$$ hence with $z = \tan t$, $$\begin{align} 0 &= (13 \tan^2 t - 5)^2 (1 + \tan^2 t) - (12 \tan^3 t - 4)^2 \\ &= 25 z^6 + 39 z^4 + 96 z^3 - 105 z^2 + 9 \\ &= (5z^2 + 6z - 3)(5z^4 - 6z^3 + 18z^2 - 6z - 3). \end{align}$$ Consequently, the critical values correspond to$$\tan t = z = \frac{-3 \pm 2 \sqrt{6}}{5}.$$ The roots of the quartic factor are all complex-valued. It follows that $$(\cos t, \sin t) = \left\{ \left( \frac{3 \sqrt{6} - 2}{10}, \frac{3 \sqrt{6} + 2}{10} \right), \left( - \frac{6 + \sqrt{6}}{10}, \frac{6 - \sqrt{6}}{10} \right) \right\},$$ and $$f(t) = \left\{ \frac{\sqrt{101 + 6 \sqrt{6}} + 9 \sqrt{3} - 3 \sqrt{2}}{\sqrt{5}}, 2 \sqrt{10} \right\}.$$ The first one is the maximum and the second is the minimum. Attached is a plot of $f$:

Answer 3

Making

$$ \cases{ u^2 = 5-2x\\ v^2 = 13-6y } $$

we have equivalently

$$ \min_{u,v} 3u+v\ \ \ \text{s. t.}\ \ \ \frac{1}{4} \left(5-u^2\right)^2+\frac{1}{36} \left(13-v^2\right)^2=4 $$

now, as the objective function is linear, and the orbits defined by $\frac 14\left(5-u^2\right)^2+\frac{1}{36} \left(13-v^2\right)^2=4$ are regular and closed, the solution should be equivalent to:

find $\lambda$ such that $3u+v = \lambda$ is tangent to $g(u,v)=\left(5-u^2\right)^2+\frac{1}{36} \left(13-v^2\right)^2=4$. To accomplish that, after substituting $v = \lambda-3u$ into $g(u,v)$ we obtain a polynomial

$$ p(v,\lambda) = \frac{\lambda ^4}{90}-\frac{2 \lambda ^3 u}{15}+\frac{3 \lambda ^2 u^2}{5}-\frac{13 \lambda ^2}{45}-\frac{6 \lambda u^3}{5}+\frac{26 \lambda u}{15}+u^4-\frac{18 u^2}{5}+\frac{25}{9} $$

At tangency, this polynomial should have a double real root or

$$ p(v,\lambda) = (v-r_1(\lambda))^2(v^2+c_1(\lambda)v+c_2(\lambda))) $$

and identifying coefficients we have

$$ \left\{ \begin{array}{rcl} -c_2 r_1^2+\frac{\lambda ^4}{90}-\frac{13 \lambda ^2}{45}+\frac{25}{9} & = & 0\\ -c_1 r_1^2+2 c_2 r_1-\frac{2 \lambda ^3}{15}+\frac{26 \lambda }{15} & = & 0\\ 2 c_1 r_1-c_2+\frac{3 \lambda ^2}{5}-r_1^2-\frac{18}{5} & = & 0\\ -c_1-\frac{6 \lambda }{5}+2 r_1 & = & 0 \\ \end{array} \right. $$

solving now for $r_1,c_1,c_2,\lambda$ we obtain the feasible results

$$ \left[ \begin{array}{cccc} r_1 & c_1 & c_2 & \lambda\\ 3 \sqrt{\frac{2}{5}}-\sqrt{\frac{3}{5}} & \frac{1}{5} \left(10 \left(3 \sqrt{\frac{2}{5}}-\sqrt{\frac{3}{5}}\right)-12 \sqrt{10}\right) & \frac{27480384-2892672 \sqrt{10} \left(3 \sqrt{\frac{2}{5}}-\sqrt{\frac{3}{5}}\right)}{2410560} & 2 \sqrt{10} \\ 3 \sqrt{\frac{2}{5}}+\sqrt{\frac{3}{5}} & \frac{1}{5} \left(10 \left(3 \sqrt{\frac{2}{5}}+\sqrt{\frac{3}{5}}\right)-12 \sqrt{10}\right) & \frac{27480384-2892672 \sqrt{10} \left(3 \sqrt{\frac{2}{5}}+\sqrt{\frac{3}{5}}\right)}{2410560} & 2 \sqrt{10} \\ \end{array} \right] $$

so the minimum is at $\lambda = 2\sqrt{10}$

Answer 4

Nice problem! Here is a solution using geometry.

For the moment, lets forget $P$ is lying on the circle $x^2+y^2=4$. Call the origin $O=(0,0)$. $A=(1,0), B=(0,3)$. So $OA=1, OB=3$ and $\angle BOA = 90^\circ$. Draw the quadrilateral $OAPB$.

Note that we have to minimize $$3\cdot PA+1\cdot PB=OB\cdot PA + OA\cdot PB$$

We know from Ptolemy's inequality that $$OB\cdot PA + OA\cdot PB \ge AB \cdot OP$$

with equality only when $O,A,P,B$ lie on a circle.

Thus the desired minimum is attained when $P$ is the intersection of circle with diameter $AB$, call it $S_2$ and $S_1:x^2+y^2=4$.

The value of this minimum is $AB\cdot OP $. Since $AB=\sqrt{10}$ and $OP$ is radius of $S_1$, this minimum value is $2\sqrt{10}$.

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Though coordinates of $P$ are not needed, they can be quickly found as intersection of common chord of the two circles and the circle $S_1$. Equation for common chord is $x+3y=4$ hence ordinate of $P$ satisfies $5y^2-12y+6=0$ whose only geometrically valid solution is $$y_P=\frac{6-\sqrt{6}}{5} \Rightarrow x_P=\frac{2+3\sqrt{6}}{5}$$

which matches other answers and WA.