MHB How Can You Minimize This Complex Square Root Expression for Positive x and y?

[anemone]

Here is this week's POTW:

-----
Minimize $\sqrt{x^2-7\sqrt{2}x+49}+\sqrt{x^2-\sqrt{2}xy+y^2}+\sqrt{y^2-10y+50}$ for positive $x$ and $y$.

-----

 

[malawi_glenn]

Spoiler

Is 13 correct?

 

[anemone]

Spoiler

Is 13 correct?

Yes!

 

[malawi_glenn]

Yes!

Ok, will try to type my solution. It is not short...

 

[bob012345]

I got to the same answer as @malawi_glenn going through the process directly. I had hoped to find some clever shortcut but in the end I had to just do all the work and I agree, it was definitely not short!

Spoiler

Given, $$f(x,y) = \sqrt{x^2-7\sqrt{2}x+49}+\sqrt{x^2-\sqrt{2}xy+y^2}+\sqrt{y^2-10y+50}$$then

$$\frac{ \partial f(x,y)}{\partial x}= \frac{1}{2} \left( \frac{2x-\sqrt{2}y}{\sqrt{x^2 -\sqrt{2}xy +y^2}}+ \frac{2x-7\sqrt{2}}{\sqrt{x^2 -7\sqrt{2}x+49}}\right)$$and

$$\frac{ \partial f(x,y)}{\partial y}= \frac{1}{2} \left( \frac{2y-\sqrt{2}x}{\sqrt{x^2 -\sqrt{2}xy +y^2}}+ \frac{2y-10}{\sqrt{y^2 -10y+50}}\right)$$setting these equal to zero and solving (see appendix) gives us;

$y= \large \frac{-7x}{x-7\sqrt{2}}$ and $y =\large \frac{10x}{x+5\sqrt{2}}$

The intersection of these two functions is the minimum. In the positive quadrant these are both monotonically increasing towards asymptotes and thus intersect only once besides at $(0,0)$. Setting them equal finds the intersection point;

$$ \frac{-7x}{x-7\sqrt{2}} = \frac{10x}{x+5\sqrt{2}}$$ or

$$ -7x (x+5\sqrt{2})= 10x(x-7\sqrt{2})$$ or

$$x(17x -35\sqrt{2})$$ giving $x=0$ or $x=\large\frac{35\sqrt{2}}{17}$ and thus $y= \large\frac{35}{12}$ putting these into the original function gives $13$ as shown in the following appendix.

Appendix:

Set $$\frac{ \partial f(x,y)}{\partial x}= \frac{1}{2} \left( \frac{2x-\sqrt{2}y}{\sqrt{x^2 -\sqrt{2}xy +y^2}}+ \frac{2x-7\sqrt{2}}{\sqrt{x^2 -7\sqrt{2}x+49}}\right) = 0$$ then

$$(2x-\sqrt{2}y)\sqrt{x^2 -7\sqrt{2}x+49} + (2x-7\sqrt{2})\sqrt{x^2 -\sqrt{2}xy +y^2}=0$$or
$$(2x-\sqrt{2}y)\sqrt{x^2 -7\sqrt{2}x+49} = (-2x+7\sqrt{2})\sqrt{x^2 -\sqrt{2}xy +y^2}$$ square both sides;

$$(4x^2 -4\sqrt{2}xy+2y^2)(x^2-7\sqrt{2}x+49)=(4x^2 -28\sqrt{2}x +98)(x^2-\sqrt{2}xy+y^2)$$ the LHS is
$$4x^4 -4\sqrt{2}x^3y +2x^2y^2 -28\sqrt{2}x^3 +49x^2y-14\sqrt{2}xy^2+196x^2-196\sqrt{2}xy+98y^2$$
the RHS is
$$4x^4 -4\sqrt{2}x^3y +4x^2y^2 -28\sqrt{2}x^3 +49x^2y-28\sqrt{2}xy^2+98x^2-98\sqrt{2}xy+98y^2$$ some terms cancel out leaving

$$2x^2y^2-14\sqrt{2}xy^2 +196x^2 -196\sqrt{2} xy=4x^2y^2 -28\sqrt{2}xy^2 +98x^2-98\sqrt{2}xy$$ or
$$2x^2y^2-14\sqrt{2}xy^2-98x^2-98\sqrt{2}xy=0$$ or
$$y^2(x-7\sqrt{2}) -49\sqrt{2}y -49x=0$$ giving a solution $y= \large \frac{-7x}{x-7\sqrt{2}}$

Now set $$ \frac{ \partial f(x,y)}{\partial y}=\frac{1}{2} \left( \frac{2y-\sqrt{2}x}{\sqrt{x^2 -\sqrt{2}xy +y^2}}+ \frac{2y-10}{\sqrt{y^2 -10y+50}}\right)=0$$ giving

$$(2y-\sqrt{2}x)\sqrt{y^2 -10y+50}+(2y-10)\sqrt{x^2 -\sqrt{2}xy +y^2}=0$$or

$$(2y-\sqrt{2}x)\sqrt{y^2 -10y+50}=(-2y+10)\sqrt{x^2 -\sqrt{2}xy +y^2}$$again, squaring both sides

$$(4y^2-4\sqrt{2}xy+2x^2)(y^2 -10y+50)=(4y^2-40y+100)(x^2 -\sqrt{2}xy +y^2)$$ the LHS is
$$4y^4 -4\sqrt{2}xy^3+2x^2y^2-40y^3+40\sqrt{2}xy^2-20x^2y+200y^2-200\sqrt{2}xy+100x^2$$ the RHS is
$$4y^4 -4\sqrt{2}xy^3+4x^2y^2-40y^3+40\sqrt{2}xy^2-40x^2 y+100y^2-100\sqrt{2}xy+100x^2$$ giving
$$2x^2y^2-20x^2y-100y^2-100\sqrt{2}xy=0$$ or
$$y(2x^2-100)=20x^2+100\sqrt{2}x$$ or
$$y= \frac{20x(x+5\sqrt{2})}{2(x^2+50)}= \frac{10x}{x-5\sqrt{2}}$$

Now put the values for $(x,y)$ into the original equation with $x=\large\frac{35\sqrt{2}}{17}$, $y=\large\frac{35}{12}$, $x^2=\large\frac{2450}{289}$ and $y^2=\large\frac{1225}{144}$

$$\sqrt{x^2-7\sqrt{2}x+49}+\sqrt{x^2-\sqrt{2}xy+y^2}+\sqrt{y^2-10y+50}$$is
$$\sqrt{ \frac{2450}{289}-7\sqrt{2}\frac{35\sqrt{2}}{17} +49}+\sqrt{\frac{2450}{289}-\sqrt{2}\frac{35\sqrt{2}}{17}\frac{35}{12}+\frac{1225}{144}}+\sqrt{\frac{1225}{144}-10\frac{35}{12}+50}$$
with
$$\sqrt{ \frac{2450}{289}-7\sqrt{2}\frac{35\sqrt{2}}{17} +49}=\sqrt{\frac{2450}{289} -\frac{8330}{289} +\frac{14161}{289}}= \sqrt{\frac{8281}{289}}=\frac{91}{17}$$and

$$\sqrt{\frac{2450}{289}-\sqrt{2}\frac{35\sqrt{2}}{17}\frac{35}{12}+\frac{1225}{144}}=\sqrt{\frac{2450}{289}-\frac{2450}{204} + \frac{1225}{144}} = \sqrt{\frac{207025}{41616}} = \frac{455}{204}$$ and
$$\sqrt{\frac{1225}{144}-10\frac{35}{12}+50}=\sqrt{\frac{1225}{144}-\frac{4200}{144}+\frac{7200}{144}}=\sqrt{\frac{12625}{144}}= \frac{65}{12}$$thus
$$f_{min}=\frac{91}{17} +\frac{455}{204}+ \frac{65}{12}=\frac{1092}{204}+\frac{455}{204}+\frac{1105}{204}=\frac{2652}{204}=13$$[\spoiler]

 

[malawi_glenn]

I got to the same answer as @malawi_glenn going through the process directly. I had hoped to find some clever shortcut but in the end I had to just do all the work and I agree, it was definitely not short!

Spoiler

Given, $$f(x,y) = \sqrt{x^2-7\sqrt{2}x+49}+\sqrt{x^2-\sqrt{2}xy+y^2}+\sqrt{y^2-10y+50}$$then

$$\frac{ \partial f(x,y)}{\partial x}= \frac{1}{2} \left( \frac{2x-\sqrt{2}y}{\sqrt{x^2 -\sqrt{2}xy +y^2}}+ \frac{2x-7\sqrt{2}}{\sqrt{x^2 -7\sqrt{2}x+49}}\right)$$and

$$\frac{ \partial f(x,y)}{\partial y}= \frac{1}{2} \left( \frac{2y-\sqrt{2}x}{\sqrt{x^2 -\sqrt{2}xy +y^2}}+ \frac{2y-10}{\sqrt{y^2 -10y+50}}\right)$$setting these equal to zero and solving (see appendix) gives us;

$y= \large \frac{-7x}{x-7\sqrt{2}}$ and $y =\large \frac{10x}{x+5\sqrt{2}}$

The intersection of these two functions is the minimum. In the positive quadrant these are both monotonically increasing towards asymptotes and thus intersect only once besides at $(0,0)$. Setting them equal finds the intersection point;

$$ \frac{-7x}{x-7\sqrt{2}} = \frac{10x}{x+5\sqrt{2}}$$ or

$$ -7x (x+5\sqrt{2})= 10x(x-7\sqrt{2})$$ or

$$x(17x -35\sqrt{2})$$ giving $x=0$ or $x=\large\frac{35\sqrt{2}}{17}$ and thus $y= \large\frac{35}{12}$ putting these into the original function gives $13$ as shown in the following appendix.

Appendix:

Set $$\frac{ \partial f(x,y)}{\partial x}= \frac{1}{2} \left( \frac{2x-\sqrt{2}y}{\sqrt{x^2 -\sqrt{2}xy +y^2}}+ \frac{2x-7\sqrt{2}}{\sqrt{x^2 -7\sqrt{2}x+49}}\right) = 0$$ then

$$(2x-\sqrt{2}y)\sqrt{x^2 -7\sqrt{2}x+49} + (2x-7\sqrt{2})\sqrt{x^2 -\sqrt{2}xy +y^2}=0$$or
$$(2x-\sqrt{2}y)\sqrt{x^2 -7\sqrt{2}x+49} = (-2x+7\sqrt{2})\sqrt{x^2 -\sqrt{2}xy +y^2}$$ square both sides;

$$(4x^2 -4\sqrt{2}xy+2y^2)(x^2-7\sqrt{2}x+49)=(4x^2 -28\sqrt{2}x +98)(x^2-\sqrt{2}xy+y^2)$$ the LHS is
$$4x^4 -4\sqrt{2}x^3y +2x^2y^2 -28\sqrt{2}x^3 +49x^2y-14\sqrt{2}xy^2+196x^2-196\sqrt{2}xy+98y^2$$
the RHS is
$$4x^4 -4\sqrt{2}x^3y +4x^2y^2 -28\sqrt{2}x^3 +49x^2y-28\sqrt{2}xy^2+98x^2-98\sqrt{2}xy+98y^2$$ some terms cancel out leaving

$$2x^2y^2-14\sqrt{2}xy^2 +196x^2 -196\sqrt{2} xy=4x^2y^2 -28\sqrt{2}xy^2 +98x^2-98\sqrt{2}xy$$ or
$$2x^2y^2-14\sqrt{2}xy^2-98x^2-98\sqrt{2}xy=0$$ or
$$y^2(x-7\sqrt{2}) -49\sqrt{2}y -49x=0$$ giving a solution $y= \large \frac{-7x}{x-7\sqrt{2}}$

Now set $$ \frac{ \partial f(x,y)}{\partial y}=\frac{1}{2} \left( \frac{2y-\sqrt{2}x}{\sqrt{x^2 -\sqrt{2}xy +y^2}}+ \frac{2y-10}{\sqrt{y^2 -10y+50}}\right)=0$$ giving

$$(2y-\sqrt{2}x)\sqrt{y^2 -10y+50}+(2y-10)\sqrt{x^2 -\sqrt{2}xy +y^2}=0$$or

$$(2y-\sqrt{2}x)\sqrt{y^2 -10y+50}=(-2y+10)\sqrt{x^2 -\sqrt{2}xy +y^2}$$again, squaring both sides

$$(4y^2-4\sqrt{2}xy+2x^2)(y^2 -10y+50)=(4y^2-40y+100)(x^2 -\sqrt{2}xy +y^2)$$ the LHS is
$$4y^4 -4\sqrt{2}xy^3+2x^2y^2-40y^3+40\sqrt{2}xy^2-20x^2y+200y^2-200\sqrt{2}xy+100x^2$$ the RHS is
$$4y^4 -4\sqrt{2}xy^3+4x^2y^2-40y^3+40\sqrt{2}xy^2-40x^2 y+100y^2-100\sqrt{2}xy+100x^2$$ giving
$$2x^2y^2-20x^2y-100y^2-100\sqrt{2}xy=0$$ or
$$y(2x^2-100)=20x^2+100\sqrt{2}x$$ or
$$y= \frac{20x(x+5\sqrt{2})}{2(x^2+50)}= \frac{10x}{x-5\sqrt{2}}$$

Now put the values for $(x,y)$ into the original equation with $x=\large\frac{35\sqrt{2}}{17}$, $y=\large\frac{35}{12}$, $x^2=\large\frac{2450}{289}$ and $y^2=\large\frac{1225}{144}$

$$\sqrt{x^2-7\sqrt{2}x+49}+\sqrt{x^2-\sqrt{2}xy+y^2}+\sqrt{y^2-10y+50}$$is
$$\sqrt{ \frac{2450}{289}-7\sqrt{2}\frac{35\sqrt{2}}{17} +49}+\sqrt{\frac{2450}{289}-\sqrt{2}\frac{35\sqrt{2}}{17}\frac{35}{12}+\frac{1225}{144}}+\sqrt{\frac{1225}{144}-10\frac{35}{12}+50}$$
with
$$\sqrt{ \frac{2450}{289}-7\sqrt{2}\frac{35\sqrt{2}}{17} +49}=\sqrt{\frac{2450}{289} -\frac{8330}{289} +\frac{14161}{289}}= \sqrt{\frac{8281}{289}}=\frac{91}{17}$$and

$$\sqrt{\frac{2450}{289}-\sqrt{2}\frac{35\sqrt{2}}{17}\frac{35}{12}+\frac{1225}{144}}=\sqrt{\frac{2450}{289}-\frac{2450}{204} + \frac{1225}{144}} = \sqrt{\frac{207025}{41616}} = \frac{455}{204}$$ and
$$\sqrt{\frac{1225}{144}-10\frac{35}{12}+50}=\sqrt{\frac{1225}{144}-\frac{4200}{144}+\frac{7200}{144}}=\sqrt{\frac{12625}{144}}= \frac{65}{12}$$thus
$$f_{min}=\frac{91}{17} +\frac{455}{204}+ \frac{65}{12}=\frac{1092}{204}+\frac{455}{204}+\frac{1105}{204}=\frac{2652}{204}=13$$[\spoiler]

Oh thanks for the reminder, I forgot to post my solution :)

Spoiler

I did not use partial derivatives

 

[bob012345]

Oh thanks for the reminder, I forgot to post my solution :)

Spoiler

I did not use partial derivatives

Ok, so your solution is a different strategy but still lengthy?