What is the method to solve $min(5-15a-18b+15a^2+45ab+45b^2)$












1












$begingroup$



Let $(a,b)in mathbb{R}$



$A:=5-15a-18b+15a^2+45ab+45b^2$




My question is how to find $min(A)$



I know the method is to rearrange this expression in



$A=(A_1 )^2+ (A_2)^2+alpha$



My question is :is there a method?






algebra-precalculus arithmetic







share|cite|improve this question









$endgroup$












  • $begingroup$
    I guess its a problem of Relative minima
    $endgroup$
    – Yadati Kiran
    Dec 11 '18 at 22:16










  • $begingroup$
    I haven't studied yet this topics
    $endgroup$
    – Stu
    Dec 11 '18 at 22:22
















1












$begingroup$



Let $(a,b)in mathbb{R}$



$A:=5-15a-18b+15a^2+45ab+45b^2$




My question is how to find $min(A)$



I know the method is to rearrange this expression in



$A=(A_1 )^2+ (A_2)^2+alpha$



My question is :is there a method?






algebra-precalculus arithmetic







share|cite|improve this question









$endgroup$












  • $begingroup$
    I guess its a problem of Relative minima
    $endgroup$
    – Yadati Kiran
    Dec 11 '18 at 22:16










  • $begingroup$
    I haven't studied yet this topics
    $endgroup$
    – Stu
    Dec 11 '18 at 22:22














1












1








1





$begingroup$



Let $(a,b)in mathbb{R}$



$A:=5-15a-18b+15a^2+45ab+45b^2$




My question is how to find $min(A)$



I know the method is to rearrange this expression in



$A=(A_1 )^2+ (A_2)^2+alpha$



My question is :is there a method?






algebra-precalculus arithmetic







share|cite|improve this question









$endgroup$





Let $(a,b)in mathbb{R}$



$A:=5-15a-18b+15a^2+45ab+45b^2$




My question is how to find $min(A)$



I know the method is to rearrange this expression in



$A=(A_1 )^2+ (A_2)^2+alpha$



My question is :is there a method?






algebra-precalculus arithmetic




algebra-precalculus arithmetic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 11 '18 at 22:14









StuStu

1,1951414




1,1951414












  • $begingroup$
    I guess its a problem of Relative minima
    $endgroup$
    – Yadati Kiran
    Dec 11 '18 at 22:16










  • $begingroup$
    I haven't studied yet this topics
    $endgroup$
    – Stu
    Dec 11 '18 at 22:22


















  • $begingroup$
    I guess its a problem of Relative minima
    $endgroup$
    – Yadati Kiran
    Dec 11 '18 at 22:16










  • $begingroup$
    I haven't studied yet this topics
    $endgroup$
    – Stu
    Dec 11 '18 at 22:22
















$begingroup$
I guess its a problem of Relative minima
$endgroup$
– Yadati Kiran
Dec 11 '18 at 22:16




$begingroup$
I guess its a problem of Relative minima
$endgroup$
– Yadati Kiran
Dec 11 '18 at 22:16












$begingroup$
I haven't studied yet this topics
$endgroup$
– Stu
Dec 11 '18 at 22:22




$begingroup$
I haven't studied yet this topics
$endgroup$
– Stu
Dec 11 '18 at 22:22










2 Answers
2






active

oldest

votes


















2












$begingroup$

You can complete the squares. First write
$$5-15a-18b+15a^2+45ab+45b^2=5-15a-18b+frac {45}4(a+2b)^2+frac {15}4a^2$$
where I chose $(a+2b)^2$ because the $ab$ and $b^2$ terms will have the same coefficient. That got rid of the cross term. Now we can write
$$5-15a-18b+frac {45}4(a+2b)^2+frac {15}4a^2=frac {45}4(a+2b)^2-9(a+2b)+frac {15}4a^2-6a+5$$
and you can make terms $(a+2b+c)^2$ and $(a+d)^2$ and sort out what the left over constant is. Then set $a=-d, b=frac 12(-c-a)$ to zero out the squares and you are left with the constant.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Ok here are two methods.
    The first and imo easiest is to consider this as two single variable problems. Consider
    $$z= Ax^2+Bxy+Cy^2+Dx+Ey+F$$ you could rewrite this as $$z=Ax^2+(By+D)x+(Cy^2+Ey+F)$$ which is a parabola with a vertex at $x=frac{-(By+D)}{2A}$ if you consider $y$ a constant and $x$ a variable. You can then consider $x$ a constant and $y$ a variable to get another parabola with a vertex at $y=frac{-(Bx+E)}{2C}$. Then you solve the simultaneous equation $x=frac{-(By+D)}{2A}$,$y=frac{-(Bx+E)}{2C}$. If both parabolas are concave down you get a maximum, and if both are concave up you'll get a minimum. You can plug in the $x$ and $y$ values to solve for $z$



    The second method is rewriting $z$ in terms of new variables so you don't need to deal with the $xy$ term. This makes the problem much easier as you can then just complete the square in each individual variable to get a minimum. To do this you rotate the $x$ and $y$ axis around by some angle $theta$ to get a new $X$ and $Y$ axis.



    pic i stole to illustrate this



    Now it turns out that $x=Xcostheta-Ysintheta$ and $y=Xsintheta + Ycostheta$. And it also turns out that there will be no $XY$ term if $cot 2theta =frac{A-C}{B}$. For explanations of this I will refer you to pdf that explains it better than I would. Once you do that you'll have something like this. $$z=aX^2+cY^2+dX+eY+f$$
    and you can complete the square to get $$z=a(X-u)^2+b(X-v)^2+w$$ and the minimum (if there is one) will just be $w$. It would be a good exercise to give both of these a go to see if you get the same answer.






    share|cite|improve this answer











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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      2












      $begingroup$

      You can complete the squares. First write
      $$5-15a-18b+15a^2+45ab+45b^2=5-15a-18b+frac {45}4(a+2b)^2+frac {15}4a^2$$
      where I chose $(a+2b)^2$ because the $ab$ and $b^2$ terms will have the same coefficient. That got rid of the cross term. Now we can write
      $$5-15a-18b+frac {45}4(a+2b)^2+frac {15}4a^2=frac {45}4(a+2b)^2-9(a+2b)+frac {15}4a^2-6a+5$$
      and you can make terms $(a+2b+c)^2$ and $(a+d)^2$ and sort out what the left over constant is. Then set $a=-d, b=frac 12(-c-a)$ to zero out the squares and you are left with the constant.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        You can complete the squares. First write
        $$5-15a-18b+15a^2+45ab+45b^2=5-15a-18b+frac {45}4(a+2b)^2+frac {15}4a^2$$
        where I chose $(a+2b)^2$ because the $ab$ and $b^2$ terms will have the same coefficient. That got rid of the cross term. Now we can write
        $$5-15a-18b+frac {45}4(a+2b)^2+frac {15}4a^2=frac {45}4(a+2b)^2-9(a+2b)+frac {15}4a^2-6a+5$$
        and you can make terms $(a+2b+c)^2$ and $(a+d)^2$ and sort out what the left over constant is. Then set $a=-d, b=frac 12(-c-a)$ to zero out the squares and you are left with the constant.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          You can complete the squares. First write
          $$5-15a-18b+15a^2+45ab+45b^2=5-15a-18b+frac {45}4(a+2b)^2+frac {15}4a^2$$
          where I chose $(a+2b)^2$ because the $ab$ and $b^2$ terms will have the same coefficient. That got rid of the cross term. Now we can write
          $$5-15a-18b+frac {45}4(a+2b)^2+frac {15}4a^2=frac {45}4(a+2b)^2-9(a+2b)+frac {15}4a^2-6a+5$$
          and you can make terms $(a+2b+c)^2$ and $(a+d)^2$ and sort out what the left over constant is. Then set $a=-d, b=frac 12(-c-a)$ to zero out the squares and you are left with the constant.






          share|cite|improve this answer









          $endgroup$



          You can complete the squares. First write
          $$5-15a-18b+15a^2+45ab+45b^2=5-15a-18b+frac {45}4(a+2b)^2+frac {15}4a^2$$
          where I chose $(a+2b)^2$ because the $ab$ and $b^2$ terms will have the same coefficient. That got rid of the cross term. Now we can write
          $$5-15a-18b+frac {45}4(a+2b)^2+frac {15}4a^2=frac {45}4(a+2b)^2-9(a+2b)+frac {15}4a^2-6a+5$$
          and you can make terms $(a+2b+c)^2$ and $(a+d)^2$ and sort out what the left over constant is. Then set $a=-d, b=frac 12(-c-a)$ to zero out the squares and you are left with the constant.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 12 '18 at 6:35









          Ross MillikanRoss Millikan

          297k23198371




          297k23198371























              1












              $begingroup$

              Ok here are two methods.
              The first and imo easiest is to consider this as two single variable problems. Consider
              $$z= Ax^2+Bxy+Cy^2+Dx+Ey+F$$ you could rewrite this as $$z=Ax^2+(By+D)x+(Cy^2+Ey+F)$$ which is a parabola with a vertex at $x=frac{-(By+D)}{2A}$ if you consider $y$ a constant and $x$ a variable. You can then consider $x$ a constant and $y$ a variable to get another parabola with a vertex at $y=frac{-(Bx+E)}{2C}$. Then you solve the simultaneous equation $x=frac{-(By+D)}{2A}$,$y=frac{-(Bx+E)}{2C}$. If both parabolas are concave down you get a maximum, and if both are concave up you'll get a minimum. You can plug in the $x$ and $y$ values to solve for $z$



              The second method is rewriting $z$ in terms of new variables so you don't need to deal with the $xy$ term. This makes the problem much easier as you can then just complete the square in each individual variable to get a minimum. To do this you rotate the $x$ and $y$ axis around by some angle $theta$ to get a new $X$ and $Y$ axis.



              pic i stole to illustrate this



              Now it turns out that $x=Xcostheta-Ysintheta$ and $y=Xsintheta + Ycostheta$. And it also turns out that there will be no $XY$ term if $cot 2theta =frac{A-C}{B}$. For explanations of this I will refer you to pdf that explains it better than I would. Once you do that you'll have something like this. $$z=aX^2+cY^2+dX+eY+f$$
              and you can complete the square to get $$z=a(X-u)^2+b(X-v)^2+w$$ and the minimum (if there is one) will just be $w$. It would be a good exercise to give both of these a go to see if you get the same answer.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Ok here are two methods.
                The first and imo easiest is to consider this as two single variable problems. Consider
                $$z= Ax^2+Bxy+Cy^2+Dx+Ey+F$$ you could rewrite this as $$z=Ax^2+(By+D)x+(Cy^2+Ey+F)$$ which is a parabola with a vertex at $x=frac{-(By+D)}{2A}$ if you consider $y$ a constant and $x$ a variable. You can then consider $x$ a constant and $y$ a variable to get another parabola with a vertex at $y=frac{-(Bx+E)}{2C}$. Then you solve the simultaneous equation $x=frac{-(By+D)}{2A}$,$y=frac{-(Bx+E)}{2C}$. If both parabolas are concave down you get a maximum, and if both are concave up you'll get a minimum. You can plug in the $x$ and $y$ values to solve for $z$



                The second method is rewriting $z$ in terms of new variables so you don't need to deal with the $xy$ term. This makes the problem much easier as you can then just complete the square in each individual variable to get a minimum. To do this you rotate the $x$ and $y$ axis around by some angle $theta$ to get a new $X$ and $Y$ axis.



                pic i stole to illustrate this



                Now it turns out that $x=Xcostheta-Ysintheta$ and $y=Xsintheta + Ycostheta$. And it also turns out that there will be no $XY$ term if $cot 2theta =frac{A-C}{B}$. For explanations of this I will refer you to pdf that explains it better than I would. Once you do that you'll have something like this. $$z=aX^2+cY^2+dX+eY+f$$
                and you can complete the square to get $$z=a(X-u)^2+b(X-v)^2+w$$ and the minimum (if there is one) will just be $w$. It would be a good exercise to give both of these a go to see if you get the same answer.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Ok here are two methods.
                  The first and imo easiest is to consider this as two single variable problems. Consider
                  $$z= Ax^2+Bxy+Cy^2+Dx+Ey+F$$ you could rewrite this as $$z=Ax^2+(By+D)x+(Cy^2+Ey+F)$$ which is a parabola with a vertex at $x=frac{-(By+D)}{2A}$ if you consider $y$ a constant and $x$ a variable. You can then consider $x$ a constant and $y$ a variable to get another parabola with a vertex at $y=frac{-(Bx+E)}{2C}$. Then you solve the simultaneous equation $x=frac{-(By+D)}{2A}$,$y=frac{-(Bx+E)}{2C}$. If both parabolas are concave down you get a maximum, and if both are concave up you'll get a minimum. You can plug in the $x$ and $y$ values to solve for $z$



                  The second method is rewriting $z$ in terms of new variables so you don't need to deal with the $xy$ term. This makes the problem much easier as you can then just complete the square in each individual variable to get a minimum. To do this you rotate the $x$ and $y$ axis around by some angle $theta$ to get a new $X$ and $Y$ axis.



                  pic i stole to illustrate this



                  Now it turns out that $x=Xcostheta-Ysintheta$ and $y=Xsintheta + Ycostheta$. And it also turns out that there will be no $XY$ term if $cot 2theta =frac{A-C}{B}$. For explanations of this I will refer you to pdf that explains it better than I would. Once you do that you'll have something like this. $$z=aX^2+cY^2+dX+eY+f$$
                  and you can complete the square to get $$z=a(X-u)^2+b(X-v)^2+w$$ and the minimum (if there is one) will just be $w$. It would be a good exercise to give both of these a go to see if you get the same answer.






                  share|cite|improve this answer











                  $endgroup$



                  Ok here are two methods.
                  The first and imo easiest is to consider this as two single variable problems. Consider
                  $$z= Ax^2+Bxy+Cy^2+Dx+Ey+F$$ you could rewrite this as $$z=Ax^2+(By+D)x+(Cy^2+Ey+F)$$ which is a parabola with a vertex at $x=frac{-(By+D)}{2A}$ if you consider $y$ a constant and $x$ a variable. You can then consider $x$ a constant and $y$ a variable to get another parabola with a vertex at $y=frac{-(Bx+E)}{2C}$. Then you solve the simultaneous equation $x=frac{-(By+D)}{2A}$,$y=frac{-(Bx+E)}{2C}$. If both parabolas are concave down you get a maximum, and if both are concave up you'll get a minimum. You can plug in the $x$ and $y$ values to solve for $z$



                  The second method is rewriting $z$ in terms of new variables so you don't need to deal with the $xy$ term. This makes the problem much easier as you can then just complete the square in each individual variable to get a minimum. To do this you rotate the $x$ and $y$ axis around by some angle $theta$ to get a new $X$ and $Y$ axis.



                  pic i stole to illustrate this



                  Now it turns out that $x=Xcostheta-Ysintheta$ and $y=Xsintheta + Ycostheta$. And it also turns out that there will be no $XY$ term if $cot 2theta =frac{A-C}{B}$. For explanations of this I will refer you to pdf that explains it better than I would. Once you do that you'll have something like this. $$z=aX^2+cY^2+dX+eY+f$$
                  and you can complete the square to get $$z=a(X-u)^2+b(X-v)^2+w$$ and the minimum (if there is one) will just be $w$. It would be a good exercise to give both of these a go to see if you get the same answer.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 12 '18 at 6:32

























                  answered Dec 12 '18 at 6:13









                  B.MartinB.Martin

                  1341110




                  1341110






























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