Yes!malawi_glenn said:Is 13 correct?
Ok, will try to type my solution. It is not short...anemone said:Yes!
Oh thanks for the reminder, I forgot to post my solution :)bob012345 said: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!
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]
Ok, so your solution is a different strategy but still lengthy?malawi_glenn said:Oh thanks for the reminder, I forgot to post my solution :)
I did not use partial derivatives
To optimize a square root expression for positive x and y, you must first simplify the expression by factoring out any perfect square factors. Then, you can use the property of square roots to rewrite the expression as the square root of the product of the perfect square factors. Finally, evaluate the expression by taking the square root of each perfect square factor and multiplying them together.
Optimizing a square root expression for positive x and y can help simplify the expression and make it easier to work with. It can also help identify any potential errors or mistakes in the original expression.
Sure, let's say we have the expression √(16x^2y^4). We can first factor out the perfect square 4 from both x and y, giving us √(4^2x^2y^4). Then, we can rewrite this as √(4^2)√(x^2)√(y^4), which simplifies to 4xy^2. This is the optimized form of the original expression.
Yes, there are a few limitations to keep in mind. First, this method only works for expressions with positive x and y values. Additionally, the expression must be a perfect square in order for it to be optimized in this way.
Optimizing square root expressions can be useful in various fields such as engineering, physics, and finance. It can help simplify complex equations and make them easier to solve, making it a valuable tool in problem-solving and decision-making processes.