Optimization AP Calculus problem












0












$begingroup$


Determine the point on the function $Y=(x-3)^2-1$ that is closest to the point $(-4,-3)$.




I used distance formula and got:





  • $d=x^4-12x^3+59x^2-124x+137$

  • $d'=4x^3-36x^2+118x-124$




I don't know how to factor from here if I have done the problem correctly so far.






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Ben W
    Jan 6 at 20:01






  • 1




    $begingroup$
    so, hi, what's the specific thing you need help with?
    $endgroup$
    – Marcus Müller
    Jan 6 at 20:01






  • 1




    $begingroup$
    I used distance formula and got d=x^4-12x^3+59x^2-124x+137. then I took the derivative and got 4x^3-36x^2+118x-124. I don't know how to factor from here if I have done the problem correctly so far.
    $endgroup$
    – cynthia
    Jan 6 at 20:09






  • 1




    $begingroup$
    Hint: the point $(a,b)$ satisfies the relation that the line through $(-4,-3)$ and $(a,b)$ is orthogonal to the graph at $(a,b)$.
    $endgroup$
    – A.Γ.
    Jan 6 at 20:27








  • 3




    $begingroup$
    @cynthia you should have included this in the original post so others can see where you encountered trouble and how to guide you back on track.
    $endgroup$
    – LoveTooNap29
    Jan 6 at 20:41
















0












$begingroup$


Determine the point on the function $Y=(x-3)^2-1$ that is closest to the point $(-4,-3)$.




I used distance formula and got:





  • $d=x^4-12x^3+59x^2-124x+137$

  • $d'=4x^3-36x^2+118x-124$




I don't know how to factor from here if I have done the problem correctly so far.






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Ben W
    Jan 6 at 20:01






  • 1




    $begingroup$
    so, hi, what's the specific thing you need help with?
    $endgroup$
    – Marcus Müller
    Jan 6 at 20:01






  • 1




    $begingroup$
    I used distance formula and got d=x^4-12x^3+59x^2-124x+137. then I took the derivative and got 4x^3-36x^2+118x-124. I don't know how to factor from here if I have done the problem correctly so far.
    $endgroup$
    – cynthia
    Jan 6 at 20:09






  • 1




    $begingroup$
    Hint: the point $(a,b)$ satisfies the relation that the line through $(-4,-3)$ and $(a,b)$ is orthogonal to the graph at $(a,b)$.
    $endgroup$
    – A.Γ.
    Jan 6 at 20:27








  • 3




    $begingroup$
    @cynthia you should have included this in the original post so others can see where you encountered trouble and how to guide you back on track.
    $endgroup$
    – LoveTooNap29
    Jan 6 at 20:41














0












0








0





$begingroup$


Determine the point on the function $Y=(x-3)^2-1$ that is closest to the point $(-4,-3)$.




I used distance formula and got:





  • $d=x^4-12x^3+59x^2-124x+137$

  • $d'=4x^3-36x^2+118x-124$




I don't know how to factor from here if I have done the problem correctly so far.






calculus







share|cite|improve this question











$endgroup$




Determine the point on the function $Y=(x-3)^2-1$ that is closest to the point $(-4,-3)$.




I used distance formula and got:





  • $d=x^4-12x^3+59x^2-124x+137$

  • $d'=4x^3-36x^2+118x-124$




I don't know how to factor from here if I have done the problem correctly so far.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 21:53









String

13.8k32756




13.8k32756










asked Jan 6 at 19:59









cynthiacynthia

1




1








  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Ben W
    Jan 6 at 20:01






  • 1




    $begingroup$
    so, hi, what's the specific thing you need help with?
    $endgroup$
    – Marcus Müller
    Jan 6 at 20:01






  • 1




    $begingroup$
    I used distance formula and got d=x^4-12x^3+59x^2-124x+137. then I took the derivative and got 4x^3-36x^2+118x-124. I don't know how to factor from here if I have done the problem correctly so far.
    $endgroup$
    – cynthia
    Jan 6 at 20:09






  • 1




    $begingroup$
    Hint: the point $(a,b)$ satisfies the relation that the line through $(-4,-3)$ and $(a,b)$ is orthogonal to the graph at $(a,b)$.
    $endgroup$
    – A.Γ.
    Jan 6 at 20:27








  • 3




    $begingroup$
    @cynthia you should have included this in the original post so others can see where you encountered trouble and how to guide you back on track.
    $endgroup$
    – LoveTooNap29
    Jan 6 at 20:41














  • 1




    $begingroup$
    What have you tried?
    $endgroup$
    – Ben W
    Jan 6 at 20:01






  • 1




    $begingroup$
    so, hi, what's the specific thing you need help with?
    $endgroup$
    – Marcus Müller
    Jan 6 at 20:01






  • 1




    $begingroup$
    I used distance formula and got d=x^4-12x^3+59x^2-124x+137. then I took the derivative and got 4x^3-36x^2+118x-124. I don't know how to factor from here if I have done the problem correctly so far.
    $endgroup$
    – cynthia
    Jan 6 at 20:09






  • 1




    $begingroup$
    Hint: the point $(a,b)$ satisfies the relation that the line through $(-4,-3)$ and $(a,b)$ is orthogonal to the graph at $(a,b)$.
    $endgroup$
    – A.Γ.
    Jan 6 at 20:27








  • 3




    $begingroup$
    @cynthia you should have included this in the original post so others can see where you encountered trouble and how to guide you back on track.
    $endgroup$
    – LoveTooNap29
    Jan 6 at 20:41








1




1




$begingroup$
What have you tried?
$endgroup$
– Ben W
Jan 6 at 20:01




$begingroup$
What have you tried?
$endgroup$
– Ben W
Jan 6 at 20:01




1




1




$begingroup$
so, hi, what's the specific thing you need help with?
$endgroup$
– Marcus Müller
Jan 6 at 20:01




$begingroup$
so, hi, what's the specific thing you need help with?
$endgroup$
– Marcus Müller
Jan 6 at 20:01




1




1




$begingroup$
I used distance formula and got d=x^4-12x^3+59x^2-124x+137. then I took the derivative and got 4x^3-36x^2+118x-124. I don't know how to factor from here if I have done the problem correctly so far.
$endgroup$
– cynthia
Jan 6 at 20:09




$begingroup$
I used distance formula and got d=x^4-12x^3+59x^2-124x+137. then I took the derivative and got 4x^3-36x^2+118x-124. I don't know how to factor from here if I have done the problem correctly so far.
$endgroup$
– cynthia
Jan 6 at 20:09




1




1




$begingroup$
Hint: the point $(a,b)$ satisfies the relation that the line through $(-4,-3)$ and $(a,b)$ is orthogonal to the graph at $(a,b)$.
$endgroup$
– A.Γ.
Jan 6 at 20:27






$begingroup$
Hint: the point $(a,b)$ satisfies the relation that the line through $(-4,-3)$ and $(a,b)$ is orthogonal to the graph at $(a,b)$.
$endgroup$
– A.Γ.
Jan 6 at 20:27






3




3




$begingroup$
@cynthia you should have included this in the original post so others can see where you encountered trouble and how to guide you back on track.
$endgroup$
– LoveTooNap29
Jan 6 at 20:41




$begingroup$
@cynthia you should have included this in the original post so others can see where you encountered trouble and how to guide you back on track.
$endgroup$
– LoveTooNap29
Jan 6 at 20:41










2 Answers
2






active

oldest

votes


















1












$begingroup$

DISCLAIMER: I originally made an error in my calculations. The expressions you gave are in fact correct. By using the squared distance formula we have:
$$
begin{align}
d(x) &=(-4-x)^2+(-3-f(x))^2 \
&=x^4-12x^3+59x^2-124x+137
end{align}
$$
This has derivative:
$$
d'(x)=4x^3-36x^2+118x-124
$$
and second order derivative:
$$
d''(x)=12x^2-72x+118
$$
This quadratic polynomial has its vertex at $(x,y)=(3,10)$ and since the leading coefficient is positive this implies that $d''(x)geq10$ and $d'$ has only a single zero.



Now since $d'(0)=-124$ and $d''(x)geq10$ we only have to look for positive values $xin(0,12.4]$. Furthermore $d'(1)=-38$ and $d''(1)=58$ and decreasing so the zero appears to be nearby.




BTW the perpendicular slope approach suggested by a user in the comments yields (essentially) the same cubic equation:
$$
frac{-4-f(x)}{-3-x}cdot f'(x)=-1iff d'(x)=0
$$






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Like other users have said, the distance formula is the way you want to go.
    Since we want to find the shortest distance between a point on the function and the point $(-4,-3)$ we have to somehow incorporate the distance formula. So we begin by first taking what we know and putting that into the formula. Namely,



    $$f(x) = (x-3)^2 - 1$$



    $$distance = sqrt{((x-(-4))^2 + (f(x)-(-3))^2}$$



    $$or$$



    $$distance = sqrt{(x-(-4))^2 + (((x-3)^2 - 1)-(-3))^2}$$



    We now have a function to optimize!



    So that is what we do (i.e take the derivative and set equal to zero).
    We note that the derivative of any square root function ($f(x)$) of another function ($g(x)$) will yield



    $$frac {g'(x)}{sqrt{f'(g(x)}}$$



    And when we set that equal to zero and solve only the numerator will matter,
    $g'(x)$. So, we have



    $$g(x) = (x-(-4))^2 + ((x-3)^2 - 1)-(-3))^2$$



    $$or$$



    $$g(x) = (x+4)^2 + (x-3)^4 - 4(x-3)^2 +4$$



    So the derivative is:



    $$g'(x) = 2(x+4) + 4(x-3)^3 - 8(x-3)$$



    or after some work:



    $$g'(x) = 4x^3-36x^2+118x-124$$



    And we can factor this (using the rational root theorem) to be:



    $$g'(x) = (x-2)(4x^2-28x+62)$$



    Since we are optimizing:



    $$(x-2)(4x^2-28x+62) = 0$$



    Which only has the real solution of $x = 2$.
    We plug this into our $f(x)$ and we get that the closest point is $(2,0)$.
    Again we can check this answer to see if we are even close to right by using any graphing utility.



    https://www.desmos.com/calculator/fd8pyqknhw






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Wow! Thank you so much for answering this question!! This is a huge help!
      $endgroup$
      – cynthia
      Jan 7 at 5:32











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    DISCLAIMER: I originally made an error in my calculations. The expressions you gave are in fact correct. By using the squared distance formula we have:
    $$
    begin{align}
    d(x) &=(-4-x)^2+(-3-f(x))^2 \
    &=x^4-12x^3+59x^2-124x+137
    end{align}
    $$
    This has derivative:
    $$
    d'(x)=4x^3-36x^2+118x-124
    $$
    and second order derivative:
    $$
    d''(x)=12x^2-72x+118
    $$
    This quadratic polynomial has its vertex at $(x,y)=(3,10)$ and since the leading coefficient is positive this implies that $d''(x)geq10$ and $d'$ has only a single zero.



    Now since $d'(0)=-124$ and $d''(x)geq10$ we only have to look for positive values $xin(0,12.4]$. Furthermore $d'(1)=-38$ and $d''(1)=58$ and decreasing so the zero appears to be nearby.




    BTW the perpendicular slope approach suggested by a user in the comments yields (essentially) the same cubic equation:
    $$
    frac{-4-f(x)}{-3-x}cdot f'(x)=-1iff d'(x)=0
    $$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      DISCLAIMER: I originally made an error in my calculations. The expressions you gave are in fact correct. By using the squared distance formula we have:
      $$
      begin{align}
      d(x) &=(-4-x)^2+(-3-f(x))^2 \
      &=x^4-12x^3+59x^2-124x+137
      end{align}
      $$
      This has derivative:
      $$
      d'(x)=4x^3-36x^2+118x-124
      $$
      and second order derivative:
      $$
      d''(x)=12x^2-72x+118
      $$
      This quadratic polynomial has its vertex at $(x,y)=(3,10)$ and since the leading coefficient is positive this implies that $d''(x)geq10$ and $d'$ has only a single zero.



      Now since $d'(0)=-124$ and $d''(x)geq10$ we only have to look for positive values $xin(0,12.4]$. Furthermore $d'(1)=-38$ and $d''(1)=58$ and decreasing so the zero appears to be nearby.




      BTW the perpendicular slope approach suggested by a user in the comments yields (essentially) the same cubic equation:
      $$
      frac{-4-f(x)}{-3-x}cdot f'(x)=-1iff d'(x)=0
      $$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        DISCLAIMER: I originally made an error in my calculations. The expressions you gave are in fact correct. By using the squared distance formula we have:
        $$
        begin{align}
        d(x) &=(-4-x)^2+(-3-f(x))^2 \
        &=x^4-12x^3+59x^2-124x+137
        end{align}
        $$
        This has derivative:
        $$
        d'(x)=4x^3-36x^2+118x-124
        $$
        and second order derivative:
        $$
        d''(x)=12x^2-72x+118
        $$
        This quadratic polynomial has its vertex at $(x,y)=(3,10)$ and since the leading coefficient is positive this implies that $d''(x)geq10$ and $d'$ has only a single zero.



        Now since $d'(0)=-124$ and $d''(x)geq10$ we only have to look for positive values $xin(0,12.4]$. Furthermore $d'(1)=-38$ and $d''(1)=58$ and decreasing so the zero appears to be nearby.




        BTW the perpendicular slope approach suggested by a user in the comments yields (essentially) the same cubic equation:
        $$
        frac{-4-f(x)}{-3-x}cdot f'(x)=-1iff d'(x)=0
        $$






        share|cite|improve this answer











        $endgroup$



        DISCLAIMER: I originally made an error in my calculations. The expressions you gave are in fact correct. By using the squared distance formula we have:
        $$
        begin{align}
        d(x) &=(-4-x)^2+(-3-f(x))^2 \
        &=x^4-12x^3+59x^2-124x+137
        end{align}
        $$
        This has derivative:
        $$
        d'(x)=4x^3-36x^2+118x-124
        $$
        and second order derivative:
        $$
        d''(x)=12x^2-72x+118
        $$
        This quadratic polynomial has its vertex at $(x,y)=(3,10)$ and since the leading coefficient is positive this implies that $d''(x)geq10$ and $d'$ has only a single zero.



        Now since $d'(0)=-124$ and $d''(x)geq10$ we only have to look for positive values $xin(0,12.4]$. Furthermore $d'(1)=-38$ and $d''(1)=58$ and decreasing so the zero appears to be nearby.




        BTW the perpendicular slope approach suggested by a user in the comments yields (essentially) the same cubic equation:
        $$
        frac{-4-f(x)}{-3-x}cdot f'(x)=-1iff d'(x)=0
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 6 at 21:49









        John Hughes

        64.8k24292




        64.8k24292










        answered Jan 6 at 20:38









        StringString

        13.8k32756




        13.8k32756























            1












            $begingroup$

            Like other users have said, the distance formula is the way you want to go.
            Since we want to find the shortest distance between a point on the function and the point $(-4,-3)$ we have to somehow incorporate the distance formula. So we begin by first taking what we know and putting that into the formula. Namely,



            $$f(x) = (x-3)^2 - 1$$



            $$distance = sqrt{((x-(-4))^2 + (f(x)-(-3))^2}$$



            $$or$$



            $$distance = sqrt{(x-(-4))^2 + (((x-3)^2 - 1)-(-3))^2}$$



            We now have a function to optimize!



            So that is what we do (i.e take the derivative and set equal to zero).
            We note that the derivative of any square root function ($f(x)$) of another function ($g(x)$) will yield



            $$frac {g'(x)}{sqrt{f'(g(x)}}$$



            And when we set that equal to zero and solve only the numerator will matter,
            $g'(x)$. So, we have



            $$g(x) = (x-(-4))^2 + ((x-3)^2 - 1)-(-3))^2$$



            $$or$$



            $$g(x) = (x+4)^2 + (x-3)^4 - 4(x-3)^2 +4$$



            So the derivative is:



            $$g'(x) = 2(x+4) + 4(x-3)^3 - 8(x-3)$$



            or after some work:



            $$g'(x) = 4x^3-36x^2+118x-124$$



            And we can factor this (using the rational root theorem) to be:



            $$g'(x) = (x-2)(4x^2-28x+62)$$



            Since we are optimizing:



            $$(x-2)(4x^2-28x+62) = 0$$



            Which only has the real solution of $x = 2$.
            We plug this into our $f(x)$ and we get that the closest point is $(2,0)$.
            Again we can check this answer to see if we are even close to right by using any graphing utility.



            https://www.desmos.com/calculator/fd8pyqknhw






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Wow! Thank you so much for answering this question!! This is a huge help!
              $endgroup$
              – cynthia
              Jan 7 at 5:32
















            1












            $begingroup$

            Like other users have said, the distance formula is the way you want to go.
            Since we want to find the shortest distance between a point on the function and the point $(-4,-3)$ we have to somehow incorporate the distance formula. So we begin by first taking what we know and putting that into the formula. Namely,



            $$f(x) = (x-3)^2 - 1$$



            $$distance = sqrt{((x-(-4))^2 + (f(x)-(-3))^2}$$



            $$or$$



            $$distance = sqrt{(x-(-4))^2 + (((x-3)^2 - 1)-(-3))^2}$$



            We now have a function to optimize!



            So that is what we do (i.e take the derivative and set equal to zero).
            We note that the derivative of any square root function ($f(x)$) of another function ($g(x)$) will yield



            $$frac {g'(x)}{sqrt{f'(g(x)}}$$



            And when we set that equal to zero and solve only the numerator will matter,
            $g'(x)$. So, we have



            $$g(x) = (x-(-4))^2 + ((x-3)^2 - 1)-(-3))^2$$



            $$or$$



            $$g(x) = (x+4)^2 + (x-3)^4 - 4(x-3)^2 +4$$



            So the derivative is:



            $$g'(x) = 2(x+4) + 4(x-3)^3 - 8(x-3)$$



            or after some work:



            $$g'(x) = 4x^3-36x^2+118x-124$$



            And we can factor this (using the rational root theorem) to be:



            $$g'(x) = (x-2)(4x^2-28x+62)$$



            Since we are optimizing:



            $$(x-2)(4x^2-28x+62) = 0$$



            Which only has the real solution of $x = 2$.
            We plug this into our $f(x)$ and we get that the closest point is $(2,0)$.
            Again we can check this answer to see if we are even close to right by using any graphing utility.



            https://www.desmos.com/calculator/fd8pyqknhw






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Wow! Thank you so much for answering this question!! This is a huge help!
              $endgroup$
              – cynthia
              Jan 7 at 5:32














            1












            1








            1





            $begingroup$

            Like other users have said, the distance formula is the way you want to go.
            Since we want to find the shortest distance between a point on the function and the point $(-4,-3)$ we have to somehow incorporate the distance formula. So we begin by first taking what we know and putting that into the formula. Namely,



            $$f(x) = (x-3)^2 - 1$$



            $$distance = sqrt{((x-(-4))^2 + (f(x)-(-3))^2}$$



            $$or$$



            $$distance = sqrt{(x-(-4))^2 + (((x-3)^2 - 1)-(-3))^2}$$



            We now have a function to optimize!



            So that is what we do (i.e take the derivative and set equal to zero).
            We note that the derivative of any square root function ($f(x)$) of another function ($g(x)$) will yield



            $$frac {g'(x)}{sqrt{f'(g(x)}}$$



            And when we set that equal to zero and solve only the numerator will matter,
            $g'(x)$. So, we have



            $$g(x) = (x-(-4))^2 + ((x-3)^2 - 1)-(-3))^2$$



            $$or$$



            $$g(x) = (x+4)^2 + (x-3)^4 - 4(x-3)^2 +4$$



            So the derivative is:



            $$g'(x) = 2(x+4) + 4(x-3)^3 - 8(x-3)$$



            or after some work:



            $$g'(x) = 4x^3-36x^2+118x-124$$



            And we can factor this (using the rational root theorem) to be:



            $$g'(x) = (x-2)(4x^2-28x+62)$$



            Since we are optimizing:



            $$(x-2)(4x^2-28x+62) = 0$$



            Which only has the real solution of $x = 2$.
            We plug this into our $f(x)$ and we get that the closest point is $(2,0)$.
            Again we can check this answer to see if we are even close to right by using any graphing utility.



            https://www.desmos.com/calculator/fd8pyqknhw






            share|cite|improve this answer











            $endgroup$



            Like other users have said, the distance formula is the way you want to go.
            Since we want to find the shortest distance between a point on the function and the point $(-4,-3)$ we have to somehow incorporate the distance formula. So we begin by first taking what we know and putting that into the formula. Namely,



            $$f(x) = (x-3)^2 - 1$$



            $$distance = sqrt{((x-(-4))^2 + (f(x)-(-3))^2}$$



            $$or$$



            $$distance = sqrt{(x-(-4))^2 + (((x-3)^2 - 1)-(-3))^2}$$



            We now have a function to optimize!



            So that is what we do (i.e take the derivative and set equal to zero).
            We note that the derivative of any square root function ($f(x)$) of another function ($g(x)$) will yield



            $$frac {g'(x)}{sqrt{f'(g(x)}}$$



            And when we set that equal to zero and solve only the numerator will matter,
            $g'(x)$. So, we have



            $$g(x) = (x-(-4))^2 + ((x-3)^2 - 1)-(-3))^2$$



            $$or$$



            $$g(x) = (x+4)^2 + (x-3)^4 - 4(x-3)^2 +4$$



            So the derivative is:



            $$g'(x) = 2(x+4) + 4(x-3)^3 - 8(x-3)$$



            or after some work:



            $$g'(x) = 4x^3-36x^2+118x-124$$



            And we can factor this (using the rational root theorem) to be:



            $$g'(x) = (x-2)(4x^2-28x+62)$$



            Since we are optimizing:



            $$(x-2)(4x^2-28x+62) = 0$$



            Which only has the real solution of $x = 2$.
            We plug this into our $f(x)$ and we get that the closest point is $(2,0)$.
            Again we can check this answer to see if we are even close to right by using any graphing utility.



            https://www.desmos.com/calculator/fd8pyqknhw







            share|cite|improve this answer














            share|cite|improve this answer



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            edited Jan 6 at 23:38

























            answered Jan 6 at 22:24









            Anthony MontemayorAnthony Montemayor

            113




            113












            • $begingroup$
              Wow! Thank you so much for answering this question!! This is a huge help!
              $endgroup$
              – cynthia
              Jan 7 at 5:32


















            • $begingroup$
              Wow! Thank you so much for answering this question!! This is a huge help!
              $endgroup$
              – cynthia
              Jan 7 at 5:32
















            $begingroup$
            Wow! Thank you so much for answering this question!! This is a huge help!
            $endgroup$
            – cynthia
            Jan 7 at 5:32




            $begingroup$
            Wow! Thank you so much for answering this question!! This is a huge help!
            $endgroup$
            – cynthia
            Jan 7 at 5:32


















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