What is the method to solve $min(5-15a-18b+15a^2+45ab+45b^2)$
$begingroup$
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
asked Dec 11 '18 at 22:14
StuStu
1,1951414
$endgroup$
add a comment |
$begingroup$
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
asked Dec 11 '18 at 22:14
StuStu
1,1951414
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Dec 11 '18 at 22:14
StuStu
1,1951414
$endgroup$
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
asked Dec 11 '18 at 22:14
StuStu
1,1951414
asked Dec 11 '18 at 22:14
StuStu
1,1951414
asked Dec 11 '18 at 22:14
StuStu
1,1951414
asked Dec 11 '18 at 22:14
StuStu
1,1951414
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 12 '18 at 6:35
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
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.
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.
answered Dec 12 '18 at 6:13
B.MartinB.Martin
1341110
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
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.
answered Dec 12 '18 at 6:35
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 6:35
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 12 '18 at 6:35
Ross MillikanRoss Millikan
297k23198371
$endgroup$
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.
answered Dec 12 '18 at 6:35
Ross MillikanRoss Millikan
297k23198371
answered Dec 12 '18 at 6:35
Ross MillikanRoss Millikan
297k23198371
answered Dec 12 '18 at 6:35
Ross MillikanRoss Millikan
297k23198371
answered Dec 12 '18 at 6:35
Ross MillikanRoss Millikan
297k23198371
297k23198371
add a comment |
add a comment |
$begingroup$
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.
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.
answered Dec 12 '18 at 6:13
B.MartinB.Martin
1341110
$endgroup$
add a comment |
$begingroup$
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.
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.
answered Dec 12 '18 at 6:13
B.MartinB.Martin
1341110
$endgroup$
add a comment |
$begingroup$
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.
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.
answered Dec 12 '18 at 6:13
B.MartinB.Martin
1341110
$endgroup$
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.
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.
answered Dec 12 '18 at 6:13
B.MartinB.Martin
1341110
edited Dec 12 '18 at 6:32
answered Dec 12 '18 at 6:13
B.MartinB.Martin
1341110
answered Dec 12 '18 at 6:13
B.MartinB.Martin
1341110
answered Dec 12 '18 at 6:13
B.MartinB.Martin
1341110
1341110
add a comment |
add a comment |
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$begingroup$
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