Optimization exercise regarding a circle and a function
$begingroup$
A circle $omega$ centered at $K$, $(0.5,0.5)$ is tangent to the $x$- and $y$-axes. Consider the function $$f(x):=frac{8}{4+x^2} ; forall xinmathbb{R}$$ and a point $Pin f$ such that the distance between $omega$ and $P$ is minimal.
Determinate the coordinates of $P$.
My attempt so far:
It's trivial that the radius of $omega$ is $frac{1}{2}$ which might be proven for instance through reductio ad absurdum and the Pythagorean theorem.
Now, it is a well-known fact, that the Euclidean distance $d$ between $omega$ and some point $Q(x_q |f(x_q))$ is
$$d=mid sqrt{big(x_q-frac{1}{2}bigr)^2+big(f(x_q)-frac{1}{2}bigr)^2}-frac{1}{2}mid$$
This looks something like this
I've tried to simplify the expression and tried to determinate de zeros of the derivative without success since the expression I came up with was to ugly to work with. So, if you've achieved to solve the problem this way 'nicer', I would appreciate if you could show me how.
Anyhow, I was wondering if there is a nicer way to approach this problem, rather than the analytical one, which might be the case since this exercise is though for 15-years-old students...
Thanks in advanced
functions optimization circles
edited Jan 14 at 22:43
David G. Stork
12.2k41836
asked Jan 14 at 22:20
Dr. MathvaDr. Mathva
3,5581630
$endgroup$
add a comment |
$begingroup$
A circle $omega$ centered at $K$, $(0.5,0.5)$ is tangent to the $x$- and $y$-axes. Consider the function $$f(x):=frac{8}{4+x^2} ; forall xinmathbb{R}$$ and a point $Pin f$ such that the distance between $omega$ and $P$ is minimal.
Determinate the coordinates of $P$.
My attempt so far:
It's trivial that the radius of $omega$ is $frac{1}{2}$ which might be proven for instance through reductio ad absurdum and the Pythagorean theorem.
Now, it is a well-known fact, that the Euclidean distance $d$ between $omega$ and some point $Q(x_q |f(x_q))$ is
$$d=mid sqrt{big(x_q-frac{1}{2}bigr)^2+big(f(x_q)-frac{1}{2}bigr)^2}-frac{1}{2}mid$$
This looks something like this
I've tried to simplify the expression and tried to determinate de zeros of the derivative without success since the expression I came up with was to ugly to work with. So, if you've achieved to solve the problem this way 'nicer', I would appreciate if you could show me how.
Anyhow, I was wondering if there is a nicer way to approach this problem, rather than the analytical one, which might be the case since this exercise is though for 15-years-old students...
Thanks in advanced
functions optimization circles
edited Jan 14 at 22:43
David G. Stork
12.2k41836
asked Jan 14 at 22:20
Dr. MathvaDr. Mathva
3,5581630
$endgroup$
1
$begingroup$
Take the derivative of your distance function with respect to $x$, set it to zero and find the minimum distance.
$endgroup$
– David G. Stork
Jan 14 at 22:26
$begingroup$
@DavidG.Stork That's what I did, and I came up with a very ugly calculation. I'll rephrase my question in order to make clear what I'm looking for
$endgroup$
– Dr. Mathva
Jan 14 at 22:28
$begingroup$
As $d$ is larger than zero, you could minimize $d^2$. Another possible simplification is to transform point $K$ to $(0;0)$.
$endgroup$
– Axel Kemper
Jan 14 at 22:59
add a comment |
$begingroup$
A circle $omega$ centered at $K$, $(0.5,0.5)$ is tangent to the $x$- and $y$-axes. Consider the function $$f(x):=frac{8}{4+x^2} ; forall xinmathbb{R}$$ and a point $Pin f$ such that the distance between $omega$ and $P$ is minimal.
Determinate the coordinates of $P$.
My attempt so far:
It's trivial that the radius of $omega$ is $frac{1}{2}$ which might be proven for instance through reductio ad absurdum and the Pythagorean theorem.
Now, it is a well-known fact, that the Euclidean distance $d$ between $omega$ and some point $Q(x_q |f(x_q))$ is
$$d=mid sqrt{big(x_q-frac{1}{2}bigr)^2+big(f(x_q)-frac{1}{2}bigr)^2}-frac{1}{2}mid$$
This looks something like this
I've tried to simplify the expression and tried to determinate de zeros of the derivative without success since the expression I came up with was to ugly to work with. So, if you've achieved to solve the problem this way 'nicer', I would appreciate if you could show me how.
Anyhow, I was wondering if there is a nicer way to approach this problem, rather than the analytical one, which might be the case since this exercise is though for 15-years-old students...
Thanks in advanced
functions optimization circles
edited Jan 14 at 22:43
David G. Stork
12.2k41836
asked Jan 14 at 22:20
Dr. MathvaDr. Mathva
3,5581630
$endgroup$
A circle $omega$ centered at $K$, $(0.5,0.5)$ is tangent to the $x$- and $y$-axes. Consider the function $$f(x):=frac{8}{4+x^2} ; forall xinmathbb{R}$$ and a point $Pin f$ such that the distance between $omega$ and $P$ is minimal.
Determinate the coordinates of $P$.
My attempt so far:
It's trivial that the radius of $omega$ is $frac{1}{2}$ which might be proven for instance through reductio ad absurdum and the Pythagorean theorem.
Now, it is a well-known fact, that the Euclidean distance $d$ between $omega$ and some point $Q(x_q |f(x_q))$ is
$$d=mid sqrt{big(x_q-frac{1}{2}bigr)^2+big(f(x_q)-frac{1}{2}bigr)^2}-frac{1}{2}mid$$
This looks something like this
I've tried to simplify the expression and tried to determinate de zeros of the derivative without success since the expression I came up with was to ugly to work with. So, if you've achieved to solve the problem this way 'nicer', I would appreciate if you could show me how.
Anyhow, I was wondering if there is a nicer way to approach this problem, rather than the analytical one, which might be the case since this exercise is though for 15-years-old students...
Thanks in advanced
functions optimization circles
functions optimization circles
edited Jan 14 at 22:43
David G. Stork
12.2k41836
asked Jan 14 at 22:20
Dr. MathvaDr. Mathva
3,5581630
edited Jan 14 at 22:43
David G. Stork
12.2k41836
asked Jan 14 at 22:20
Dr. MathvaDr. Mathva
3,5581630
edited Jan 14 at 22:43
David G. Stork
12.2k41836
edited Jan 14 at 22:43
David G. Stork
12.2k41836
edited Jan 14 at 22:43
David G. Stork
12.2k41836
12.2k41836
asked Jan 14 at 22:20
Dr. MathvaDr. Mathva
3,5581630
asked Jan 14 at 22:20
Dr. MathvaDr. Mathva
3,5581630
asked Jan 14 at 22:20
Dr. MathvaDr. Mathva
3,5581630
3,5581630
1
$begingroup$
Take the derivative of your distance function with respect to $x$, set it to zero and find the minimum distance.
$endgroup$
– David G. Stork
Jan 14 at 22:26
$begingroup$
@DavidG.Stork That's what I did, and I came up with a very ugly calculation. I'll rephrase my question in order to make clear what I'm looking for
$endgroup$
– Dr. Mathva
Jan 14 at 22:28
$begingroup$
As $d$ is larger than zero, you could minimize $d^2$. Another possible simplification is to transform point $K$ to $(0;0)$.
$endgroup$
– Axel Kemper
Jan 14 at 22:59
add a comment |
1
$begingroup$
Take the derivative of your distance function with respect to $x$, set it to zero and find the minimum distance.
$endgroup$
– David G. Stork
Jan 14 at 22:26
$begingroup$
@DavidG.Stork That's what I did, and I came up with a very ugly calculation. I'll rephrase my question in order to make clear what I'm looking for
$endgroup$
– Dr. Mathva
Jan 14 at 22:28
$begingroup$
As $d$ is larger than zero, you could minimize $d^2$. Another possible simplification is to transform point $K$ to $(0;0)$.
$endgroup$
– Axel Kemper
Jan 14 at 22:59
1
1
$begingroup$
Take the derivative of your distance function with respect to $x$, set it to zero and find the minimum distance.
$endgroup$
– David G. Stork
Jan 14 at 22:26
$begingroup$
Take the derivative of your distance function with respect to $x$, set it to zero and find the minimum distance.
$endgroup$
– David G. Stork
Jan 14 at 22:26
$begingroup$
@DavidG.Stork That's what I did, and I came up with a very ugly calculation. I'll rephrase my question in order to make clear what I'm looking for
$endgroup$
– Dr. Mathva
Jan 14 at 22:28
$begingroup$
@DavidG.Stork That's what I did, and I came up with a very ugly calculation. I'll rephrase my question in order to make clear what I'm looking for
$endgroup$
– Dr. Mathva
Jan 14 at 22:28
$begingroup$
As $d$ is larger than zero, you could minimize $d^2$. Another possible simplification is to transform point $K$ to $(0;0)$.
$endgroup$
– Axel Kemper
Jan 14 at 22:59
$begingroup$
As $d$ is larger than zero, you could minimize $d^2$. Another possible simplification is to transform point $K$ to $(0;0)$.
$endgroup$
– Axel Kemper
Jan 14 at 22:59
add a comment |
1 Answer
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$begingroup$
I feel a bit blocked because I don't know what your students can use... Even though the following ideas could help.
Since the circle and the curve do not intersect:
The distance $d$ is minimal $Longleftrightarrow d+frac{1}{2} = sqrt{left(x-frac{1}{2}right)^2+left(f(x)-frac{1}{2}right)^2}$ is minimal,
and it is $quadquadquadquadquadquadquadLongleftrightarrow left(x-frac{1}{2}right)^2+left(f(x)-frac{1}{2}right)^2$ is minimal.
A point $P$ located on the curve $mathcal{C}_f$ at a minimal distance to $omega$ lies on a circle $rho,$ centered at $K$ and tangent to $mathcal{C}_f$ at $P.$ Thus the line $KP$ is normal to the curve at $P.$
Putting together: $P$ lies at a circle centered at $K,$ which has only one common point with $mathcal{C}_f.$
answered Feb 5 at 22:20
user376343user376343
3,9834829
$endgroup$
add a comment |
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$begingroup$
I feel a bit blocked because I don't know what your students can use... Even though the following ideas could help.
Since the circle and the curve do not intersect:
The distance $d$ is minimal $Longleftrightarrow d+frac{1}{2} = sqrt{left(x-frac{1}{2}right)^2+left(f(x)-frac{1}{2}right)^2}$ is minimal,
and it is $quadquadquadquadquadquadquadLongleftrightarrow left(x-frac{1}{2}right)^2+left(f(x)-frac{1}{2}right)^2$ is minimal.
A point $P$ located on the curve $mathcal{C}_f$ at a minimal distance to $omega$ lies on a circle $rho,$ centered at $K$ and tangent to $mathcal{C}_f$ at $P.$ Thus the line $KP$ is normal to the curve at $P.$
Putting together: $P$ lies at a circle centered at $K,$ which has only one common point with $mathcal{C}_f.$
answered Feb 5 at 22:20
user376343user376343
3,9834829
$endgroup$
add a comment |
$begingroup$
I feel a bit blocked because I don't know what your students can use... Even though the following ideas could help.
Since the circle and the curve do not intersect:
The distance $d$ is minimal $Longleftrightarrow d+frac{1}{2} = sqrt{left(x-frac{1}{2}right)^2+left(f(x)-frac{1}{2}right)^2}$ is minimal,
and it is $quadquadquadquadquadquadquadLongleftrightarrow left(x-frac{1}{2}right)^2+left(f(x)-frac{1}{2}right)^2$ is minimal.
A point $P$ located on the curve $mathcal{C}_f$ at a minimal distance to $omega$ lies on a circle $rho,$ centered at $K$ and tangent to $mathcal{C}_f$ at $P.$ Thus the line $KP$ is normal to the curve at $P.$
Putting together: $P$ lies at a circle centered at $K,$ which has only one common point with $mathcal{C}_f.$
answered Feb 5 at 22:20
user376343user376343
3,9834829
$endgroup$
add a comment |
$begingroup$
I feel a bit blocked because I don't know what your students can use... Even though the following ideas could help.
Since the circle and the curve do not intersect:
The distance $d$ is minimal $Longleftrightarrow d+frac{1}{2} = sqrt{left(x-frac{1}{2}right)^2+left(f(x)-frac{1}{2}right)^2}$ is minimal,
and it is $quadquadquadquadquadquadquadLongleftrightarrow left(x-frac{1}{2}right)^2+left(f(x)-frac{1}{2}right)^2$ is minimal.
A point $P$ located on the curve $mathcal{C}_f$ at a minimal distance to $omega$ lies on a circle $rho,$ centered at $K$ and tangent to $mathcal{C}_f$ at $P.$ Thus the line $KP$ is normal to the curve at $P.$
Putting together: $P$ lies at a circle centered at $K,$ which has only one common point with $mathcal{C}_f.$
answered Feb 5 at 22:20
user376343user376343
3,9834829
$endgroup$
I feel a bit blocked because I don't know what your students can use... Even though the following ideas could help.
Since the circle and the curve do not intersect:
The distance $d$ is minimal $Longleftrightarrow d+frac{1}{2} = sqrt{left(x-frac{1}{2}right)^2+left(f(x)-frac{1}{2}right)^2}$ is minimal,
and it is $quadquadquadquadquadquadquadLongleftrightarrow left(x-frac{1}{2}right)^2+left(f(x)-frac{1}{2}right)^2$ is minimal.
A point $P$ located on the curve $mathcal{C}_f$ at a minimal distance to $omega$ lies on a circle $rho,$ centered at $K$ and tangent to $mathcal{C}_f$ at $P.$ Thus the line $KP$ is normal to the curve at $P.$
Putting together: $P$ lies at a circle centered at $K,$ which has only one common point with $mathcal{C}_f.$
answered Feb 5 at 22:20
user376343user376343
3,9834829
answered Feb 5 at 22:20
user376343user376343
3,9834829
answered Feb 5 at 22:20
user376343user376343
3,9834829
answered Feb 5 at 22:20
user376343user376343
3,9834829
3,9834829
add a comment |
add a comment |
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$begingroup$
Take the derivative of your distance function with respect to $x$, set it to zero and find the minimum distance.
$endgroup$
– David G. Stork
Jan 14 at 22:26
$begingroup$
@DavidG.Stork That's what I did, and I came up with a very ugly calculation. I'll rephrase my question in order to make clear what I'm looking for
$endgroup$
– Dr. Mathva
Jan 14 at 22:28
$begingroup$
As $d$ is larger than zero, you could minimize $d^2$. Another possible simplification is to transform point $K$ to $(0;0)$.
$endgroup$
– Axel Kemper
Jan 14 at 22:59