Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ vertices












0












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Can't find absolute minimums and maximums!



Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ on the set D where D is the closed triangular region with vertices $(8,0), (0,6)$, and $(0,-6)$.



To find critical points:



$F_x = 0$



$F_y = 0$



$F_x$ (first order derivative with respect to x) $ = 2x - 4$



$F_y = 2y$



Setting them equal to $0$, to find critical points:



$2x - 4 = 0$



$2y = 0$



critical point: $(2,0)$



...



Now, the boundary of the triangle can be expressed in 3 lines:



Left side of triangle, $x = 0$:



Function can be expressed by the one variable function



$f(0, y) = y^2 + 9$



Absolute maximums:



You evaluate those at the given points, and I think also supposed to evaluate at the critical point? (derivative equal to $0$ etc)



$(0,6), (0,-6)$



Absolute minimums:



??? (8,0) <- why not at these points?
If y = 0, then that's when it has the absolute minimum, no?



...



Upper right side of the triangle:



between points: $(0,6)$ and $(8,0)$



$y= frac{-3}{4} x + 6 $



Function can be expressed by the one variable function



$f(x, frac{-3}{4} x + 6 ) = (frac{-3}{4} x + 6)^2 + x^2 - 4x + 9$



Absolute maximums:



$(0, 6)$



Absolute minimums:



(8,0) INCORRECT, WHY???



...



Lower right side of the triangle:



between points: $(0,-6)$ and $(8,0)$



$y = frac{3}{4} x - 6$



Function can be expressed by the one variable function



(3/4 x - 6)^2 + x^2 - 4x + 9



Absolute maximums:



????



Absolute minimums:



????



p.s. here's the screenshot










share|cite|improve this question









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    0












    $begingroup$


    Can't find absolute minimums and maximums!



    Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ on the set D where D is the closed triangular region with vertices $(8,0), (0,6)$, and $(0,-6)$.



    To find critical points:



    $F_x = 0$



    $F_y = 0$



    $F_x$ (first order derivative with respect to x) $ = 2x - 4$



    $F_y = 2y$



    Setting them equal to $0$, to find critical points:



    $2x - 4 = 0$



    $2y = 0$



    critical point: $(2,0)$



    ...



    Now, the boundary of the triangle can be expressed in 3 lines:



    Left side of triangle, $x = 0$:



    Function can be expressed by the one variable function



    $f(0, y) = y^2 + 9$



    Absolute maximums:



    You evaluate those at the given points, and I think also supposed to evaluate at the critical point? (derivative equal to $0$ etc)



    $(0,6), (0,-6)$



    Absolute minimums:



    ??? (8,0) <- why not at these points?
    If y = 0, then that's when it has the absolute minimum, no?



    ...



    Upper right side of the triangle:



    between points: $(0,6)$ and $(8,0)$



    $y= frac{-3}{4} x + 6 $



    Function can be expressed by the one variable function



    $f(x, frac{-3}{4} x + 6 ) = (frac{-3}{4} x + 6)^2 + x^2 - 4x + 9$



    Absolute maximums:



    $(0, 6)$



    Absolute minimums:



    (8,0) INCORRECT, WHY???



    ...



    Lower right side of the triangle:



    between points: $(0,-6)$ and $(8,0)$



    $y = frac{3}{4} x - 6$



    Function can be expressed by the one variable function



    (3/4 x - 6)^2 + x^2 - 4x + 9



    Absolute maximums:



    ????



    Absolute minimums:



    ????



    p.s. here's the screenshot










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Can't find absolute minimums and maximums!



      Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ on the set D where D is the closed triangular region with vertices $(8,0), (0,6)$, and $(0,-6)$.



      To find critical points:



      $F_x = 0$



      $F_y = 0$



      $F_x$ (first order derivative with respect to x) $ = 2x - 4$



      $F_y = 2y$



      Setting them equal to $0$, to find critical points:



      $2x - 4 = 0$



      $2y = 0$



      critical point: $(2,0)$



      ...



      Now, the boundary of the triangle can be expressed in 3 lines:



      Left side of triangle, $x = 0$:



      Function can be expressed by the one variable function



      $f(0, y) = y^2 + 9$



      Absolute maximums:



      You evaluate those at the given points, and I think also supposed to evaluate at the critical point? (derivative equal to $0$ etc)



      $(0,6), (0,-6)$



      Absolute minimums:



      ??? (8,0) <- why not at these points?
      If y = 0, then that's when it has the absolute minimum, no?



      ...



      Upper right side of the triangle:



      between points: $(0,6)$ and $(8,0)$



      $y= frac{-3}{4} x + 6 $



      Function can be expressed by the one variable function



      $f(x, frac{-3}{4} x + 6 ) = (frac{-3}{4} x + 6)^2 + x^2 - 4x + 9$



      Absolute maximums:



      $(0, 6)$



      Absolute minimums:



      (8,0) INCORRECT, WHY???



      ...



      Lower right side of the triangle:



      between points: $(0,-6)$ and $(8,0)$



      $y = frac{3}{4} x - 6$



      Function can be expressed by the one variable function



      (3/4 x - 6)^2 + x^2 - 4x + 9



      Absolute maximums:



      ????



      Absolute minimums:



      ????



      p.s. here's the screenshot










      share|cite|improve this question









      $endgroup$




      Can't find absolute minimums and maximums!



      Find the absolute maximum and minimum values of $f(x,y) = y^{2}+x^{2}-4x+9$ on the set D where D is the closed triangular region with vertices $(8,0), (0,6)$, and $(0,-6)$.



      To find critical points:



      $F_x = 0$



      $F_y = 0$



      $F_x$ (first order derivative with respect to x) $ = 2x - 4$



      $F_y = 2y$



      Setting them equal to $0$, to find critical points:



      $2x - 4 = 0$



      $2y = 0$



      critical point: $(2,0)$



      ...



      Now, the boundary of the triangle can be expressed in 3 lines:



      Left side of triangle, $x = 0$:



      Function can be expressed by the one variable function



      $f(0, y) = y^2 + 9$



      Absolute maximums:



      You evaluate those at the given points, and I think also supposed to evaluate at the critical point? (derivative equal to $0$ etc)



      $(0,6), (0,-6)$



      Absolute minimums:



      ??? (8,0) <- why not at these points?
      If y = 0, then that's when it has the absolute minimum, no?



      ...



      Upper right side of the triangle:



      between points: $(0,6)$ and $(8,0)$



      $y= frac{-3}{4} x + 6 $



      Function can be expressed by the one variable function



      $f(x, frac{-3}{4} x + 6 ) = (frac{-3}{4} x + 6)^2 + x^2 - 4x + 9$



      Absolute maximums:



      $(0, 6)$



      Absolute minimums:



      (8,0) INCORRECT, WHY???



      ...



      Lower right side of the triangle:



      between points: $(0,-6)$ and $(8,0)$



      $y = frac{3}{4} x - 6$



      Function can be expressed by the one variable function



      (3/4 x - 6)^2 + x^2 - 4x + 9



      Absolute maximums:



      ????



      Absolute minimums:



      ????



      p.s. here's the screenshot







      calculus multivariable-calculus






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      share|cite|improve this question











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      asked Mar 2 '16 at 0:28









      JackJack

      311615




      311615






















          1 Answer
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          active

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          0












          $begingroup$

          enter image description here



          I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.



          edit 1:(add more details)



          $f(x,y)=(x-2)^2+y^2+5 implies (x-2)^2+y^2=f(x,y)-5=R^2$



          you want to find max and min of $f(x,y)$ is same to find max and min of $R$.



          note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$



          if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
            $endgroup$
            – Jack
            Mar 2 '16 at 3:22












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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          enter image description here



          I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.



          edit 1:(add more details)



          $f(x,y)=(x-2)^2+y^2+5 implies (x-2)^2+y^2=f(x,y)-5=R^2$



          you want to find max and min of $f(x,y)$ is same to find max and min of $R$.



          note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$



          if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
            $endgroup$
            – Jack
            Mar 2 '16 at 3:22
















          0












          $begingroup$

          enter image description here



          I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.



          edit 1:(add more details)



          $f(x,y)=(x-2)^2+y^2+5 implies (x-2)^2+y^2=f(x,y)-5=R^2$



          you want to find max and min of $f(x,y)$ is same to find max and min of $R$.



          note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$



          if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
            $endgroup$
            – Jack
            Mar 2 '16 at 3:22














          0












          0








          0





          $begingroup$

          enter image description here



          I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.



          edit 1:(add more details)



          $f(x,y)=(x-2)^2+y^2+5 implies (x-2)^2+y^2=f(x,y)-5=R^2$



          you want to find max and min of $f(x,y)$ is same to find max and min of $R$.



          note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$



          if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.






          share|cite|improve this answer











          $endgroup$



          enter image description here



          I suggest to take geo method directly which is easy to understand. check above picture, you need to find the circle(center is fixed) max and min R in a triangle region that is very straightforward.



          edit 1:(add more details)



          $f(x,y)=(x-2)^2+y^2+5 implies (x-2)^2+y^2=f(x,y)-5=R^2$



          you want to find max and min of $f(x,y)$ is same to find max and min of $R$.



          note the circle have to be inside the triangle so max of $R$ is $2$ ,min of R is $0$ which give $f_{max}=4+5=9, f_{min}=0+5=5$



          if op insist on pure Lagrange multiplier method, it has much more things to write and you have three boundary equations . Anyway, if you don't like geo method, simply ignore it. it is true that we can't use this simple method if $f(x,y)$ is not the circle.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 2 '16 at 6:15

























          answered Mar 2 '16 at 0:57









          chenbaichenbai

          6,60211024




          6,60211024












          • $begingroup$
            what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
            $endgroup$
            – Jack
            Mar 2 '16 at 3:22


















          • $begingroup$
            what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
            $endgroup$
            – Jack
            Mar 2 '16 at 3:22
















          $begingroup$
          what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
          $endgroup$
          – Jack
          Mar 2 '16 at 3:22




          $begingroup$
          what is that circle? Where does it come from? You're lucky you can draw this one, but how about instances when you can't? Can't always rely on visuals. Give me the mathematical method. For the left side of triangle, when x= 0. You have y^2+9. How do you find min/max? Do you plug the points into the equation y^2+9. and see which one yields minimum and maximum values or...?
          $endgroup$
          – Jack
          Mar 2 '16 at 3:22


















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