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Minimum value of $fracb+1a+b-2$
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If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $dfracb+1a+b-2$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = dfrac2-k2(k+1)$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
calculus algebra-precalculus trigonometry derivatives quadratics
asked Jul 15 at 11:02
Abcd
2,3831624
add a comment |Â
up vote
5
down vote
favorite
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $dfracb+1a+b-2$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = dfrac2-k2(k+1)$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
calculus algebra-precalculus trigonometry derivatives quadratics
asked Jul 15 at 11:02
Abcd
2,3831624
1
Just an idea: you can interpret the question as having the points $(a,b)$ on the unit circle. Perhaps you can try creating a bijection by expressing $a$ and $b$ in terms of the angle $theta$.
– user496634
Jul 15 at 11:27
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up vote
5
down vote
favorite
up vote
5
down vote
favorite
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $dfracb+1a+b-2$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = dfrac2-k2(k+1)$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
calculus algebra-precalculus trigonometry derivatives quadratics
asked Jul 15 at 11:02
Abcd
2,3831624
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $dfracb+1a+b-2$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = dfrac2-k2(k+1)$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
calculus algebra-precalculus trigonometry derivatives quadratics
asked Jul 15 at 11:02
Abcd
2,3831624
edited Jul 15 at 19:44
asked Jul 15 at 11:02
Abcd
2,3831624
asked Jul 15 at 11:02
Abcd
2,3831624
asked Jul 15 at 11:02
Abcd
2,3831624
2,3831624
1
Just an idea: you can interpret the question as having the points $(a,b)$ on the unit circle. Perhaps you can try creating a bijection by expressing $a$ and $b$ in terms of the angle $theta$.
– user496634
Jul 15 at 11:27
add a comment |Â
1
Just an idea: you can interpret the question as having the points $(a,b)$ on the unit circle. Perhaps you can try creating a bijection by expressing $a$ and $b$ in terms of the angle $theta$.
– user496634
Jul 15 at 11:27
1
1
Just an idea: you can interpret the question as having the points $(a,b)$ on the unit circle. Perhaps you can try creating a bijection by expressing $a$ and $b$ in terms of the angle $theta$.
– user496634
Jul 15 at 11:27
Just an idea: you can interpret the question as having the points $(a,b)$ on the unit circle. Perhaps you can try creating a bijection by expressing $a$ and $b$ in terms of the angle $theta$.
– user496634
Jul 15 at 11:27
add a comment |Â
7 Answers
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Note that $$u=fracb+1sqrt1-b^2+b-2$$ so $$fracdudb=frac1big(sqrt1-b^2+b-2big)-(b+1)left(-frac2bsqrt1-b^2+1right)big(sqrt1-b^2+b-2big)^2$$ and setting to zero gives $$-3sqrt1-b^2+b+1=0implies 1-b^2=fracb^2+2b+19implies 5b^2+b-4=0$$ and we see that $b=4/5,-1$ are roots.
Checking second derivatives, we have that $4/5$ is a minimum.
Hence $$u^2=left(fracfrac45+1frac35+frac45-2right)^2=9.$$
Note that the negative root ($-3/5$) is also possible, but that yields a lower value of $u^2$ since $$bigg|-frac35+frac45-2bigg|>bigg|frac35+frac45-2bigg|$$
answered Jul 15 at 11:19
TheSimpliFire
9,69261951
4
You may need to consider $v=fracb+1-sqrt1-b^2+b-2$ as well
– Henry
Jul 15 at 11:24
@Henry Of course! Edited.
– TheSimpliFire
Jul 15 at 18:39
add a comment |Â
up vote
13
down vote
Try with $b=cos 2x$ and $a= sin 2x$.
begineqnarrayb+1over a+b-2&=& 2cos^2 xover -cos^2x+2sin x cos x -3sin^2x\
&=& 2over -1+2tan x -3tan^2x\
&=& 2over -1+2t -3t^2
endeqnarray
where $t= tan x $. So the expression will take a minimum when quadratic function $g(t)=-3t^2+2t-1$ will take a maximum. Note that $g(t)<0$ for all $t in mathbbR$
So $$ u= 2over -2over 3 = -3implies ....$$
answered Jul 15 at 11:04
greedoid
26.6k93574
2
This was the first thing that came to my mind.
– prog_SAHIL
Jul 15 at 11:33
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4
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A bit geometry;
1)$x^2+y^2 = 1$, a circle, centre $(0,0)$, $r=1$.
2) Minimum of $C$:
$C:=dfracy+1x+y-2$
(Note: $x+y-2 not =0$).
$C(x+y-2) = y+1$, or
$Cx +y(C-1) -(2C+1)= 0$, a straight line.
The line touches or intersects the circle 1)
if the distance line-to-origin $le 1$ (radius).
Distance to $(0,0):$
$d =dfracsqrtC^2+(C-1)^2 le 1.$
$(2C+1)^2 le C^2 + (C-1)^2;$
$4C^2 +4C +1 le 2C^2 -2C+1;$
$2C(C+3) le 0.$
Hence: $-3 le C le 0$.
Minimum at $C =-3$.
Used: Line to point distance formula:
http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
answered Jul 15 at 13:44
Peter Szilas
7,9552617
What's the geometrical interpretation of minimum at C= -3 ?
– Abcd
Jul 15 at 14:17
C=-3 ; Line: -3x -4y +5=0 , y intercept 5/4 , slope: -3/4, distance from (0,0) : =1, hence a tangent to the circle. You can sketch it.
– Peter Szilas
Jul 15 at 16:10
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up vote
4
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Let $displaystyle u=fracb+1a+b-2Rightarrow ua+ub-2u=b+1$
So $ua+(u-1)b=1+2u$
Now Using Cauchy Schwarz Inequality
$displaystyle bigg[u^2+(u-1)^2bigg]cdot bigg[a^2+b^2bigg]geq bigg[ua+(u-1)bbigg]^2$
So $displaystyle 2u^2-2u+1geq (1+2u)^2Rightarrow 2u^2+6uleq 0$
So $displaystyle 2u(u+3)leq 0Rightarrow -3 leq uleq 0$
So $displaystyle u^2 geq 9.$
answered Jul 15 at 11:51
Durgesh Tiwari
4,7882426
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Just for a variation, using Lagrange’s method:
$$
f(a,b,t)=fracb+1a+b-2-t(a^2+b^2-1)
$$
Then
beginalign
fracpartial fpartial a&=-fracb+1(a+b-2)^2-2at \[6px]
fracpartial fpartial b&=fraca-3(a+b-2)^2-2bt
endalign
If these equal $0$, then
$$
-fracb+1a(a+b-2)^2=fraca-3b(a+b-2)^2
$$
so that $-b^2-b=a^2-3a$, that gives $3a-b=1$ and from $a^2+b^2=1$ we derive $a=0$ or $a=3/5$.
The critical points are thus $(0,-1)$ and $(3/5,4/5)$. We have
$$
f(0,-1,0)=frac23,quad
f(3/5,4/5,0)=-3,quad
$$
This also shows the maximum.
answered Jul 15 at 13:27
egreg
164k1180187
Why should the partial derivatives equal 0?
– Abcd
Jul 18 at 7:52
@Abcd Because we are looking for critical points.
– egreg
Jul 18 at 8:10
With respect to the maximum part, can't maximum be at the endpoints of the definition interval? How do we know if it is or not?
– Abcd
Jul 20 at 2:10
1
@Abcd There's no “definition intervalâ€Â: the function is restricted to the circle $a^2+b^2=1$, which is compact and has no “endpointsâ€Â.
– egreg
Jul 20 at 8:32
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up vote
2
down vote
From
$$
fracb+1a+b-2= uRightarrow Lto a = 2-bfracb+1u
$$
So $L$ should be tangent to $a^2+b^2=1;$ then substituting we have the condition
$$
(2b^2-4b+3)u^2+(4+2b-2b^2)u+(b+1)^2 = 0
$$
and solving for $u$
$$
u = frac(b+1)^2b^2-bpmsqrt(1-b) (b+1)^3-2
$$
but tangency implies on $sqrt(1-b) (b+1)^3=0;$ hence the solutions for tangency are $b = pm 1$ etc.
answered Jul 15 at 11:28
Cesareo
5,7922412
Didn't understand why L should be a tangent and why it should tend to a
– Abcd
Jul 15 at 11:40
1
Interesting method, though please explain the concepts used
– Abcd
Jul 15 at 11:40
@Abcd The problem $minmax u=f(a,b)$ s. t. $a^2+b^2-1=0$ when $f(a,b)$ is regular, is defined at the tangency points for $u=f(a,b)$ and $a^2+b^2-1=0$ At those points we have $u = u_min,cdots,u_max$. This is the idea of Lagrange multipliers method.
– Cesareo
Jul 15 at 13:43
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up vote
2
down vote
Pulling a small rabbit from a hat, consider
$$f(a,b)=3+b+1over a+b-2=3a+4b-5over a+b-2$$
It's clear that $a^2+b^2=1$ implies $a+b-2lt0$. By Cauchy-Schwartz, we have
$$(3a+4b)^2le(a^2+b^2)(3^2+4^2)=25=5^2$$
and therefore $3a+4b-5le0$ if $a^2+b^2=1$. Thus $f(a,b)ge0$ for all $a$ and $b$ for which $a^2+b^2=1$. But also $f(3over5,4over5)=0$, so the minimum value of $(b+1)/(a+b-2)=f(a,b)-3$ with $a^2+b^2=1$ is $-3$, the square of which is $9$.
answered Jul 15 at 14:47
Barry Cipra
56.5k652118
add a comment |Â
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Note that $$u=fracb+1sqrt1-b^2+b-2$$ so $$fracdudb=frac1big(sqrt1-b^2+b-2big)-(b+1)left(-frac2bsqrt1-b^2+1right)big(sqrt1-b^2+b-2big)^2$$ and setting to zero gives $$-3sqrt1-b^2+b+1=0implies 1-b^2=fracb^2+2b+19implies 5b^2+b-4=0$$ and we see that $b=4/5,-1$ are roots.
Checking second derivatives, we have that $4/5$ is a minimum.
Hence $$u^2=left(fracfrac45+1frac35+frac45-2right)^2=9.$$
Note that the negative root ($-3/5$) is also possible, but that yields a lower value of $u^2$ since $$bigg|-frac35+frac45-2bigg|>bigg|frac35+frac45-2bigg|$$
answered Jul 15 at 11:19
TheSimpliFire
9,69261951
4
You may need to consider $v=fracb+1-sqrt1-b^2+b-2$ as well
– Henry
Jul 15 at 11:24
@Henry Of course! Edited.
– TheSimpliFire
Jul 15 at 18:39
add a comment |Â
up vote
9
down vote
accepted
Note that $$u=fracb+1sqrt1-b^2+b-2$$ so $$fracdudb=frac1big(sqrt1-b^2+b-2big)-(b+1)left(-frac2bsqrt1-b^2+1right)big(sqrt1-b^2+b-2big)^2$$ and setting to zero gives $$-3sqrt1-b^2+b+1=0implies 1-b^2=fracb^2+2b+19implies 5b^2+b-4=0$$ and we see that $b=4/5,-1$ are roots.
Checking second derivatives, we have that $4/5$ is a minimum.
Hence $$u^2=left(fracfrac45+1frac35+frac45-2right)^2=9.$$
Note that the negative root ($-3/5$) is also possible, but that yields a lower value of $u^2$ since $$bigg|-frac35+frac45-2bigg|>bigg|frac35+frac45-2bigg|$$
answered Jul 15 at 11:19
TheSimpliFire
9,69261951
4
You may need to consider $v=fracb+1-sqrt1-b^2+b-2$ as well
– Henry
Jul 15 at 11:24
@Henry Of course! Edited.
– TheSimpliFire
Jul 15 at 18:39
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Note that $$u=fracb+1sqrt1-b^2+b-2$$ so $$fracdudb=frac1big(sqrt1-b^2+b-2big)-(b+1)left(-frac2bsqrt1-b^2+1right)big(sqrt1-b^2+b-2big)^2$$ and setting to zero gives $$-3sqrt1-b^2+b+1=0implies 1-b^2=fracb^2+2b+19implies 5b^2+b-4=0$$ and we see that $b=4/5,-1$ are roots.
Checking second derivatives, we have that $4/5$ is a minimum.
Hence $$u^2=left(fracfrac45+1frac35+frac45-2right)^2=9.$$
Note that the negative root ($-3/5$) is also possible, but that yields a lower value of $u^2$ since $$bigg|-frac35+frac45-2bigg|>bigg|frac35+frac45-2bigg|$$
answered Jul 15 at 11:19
TheSimpliFire
9,69261951
Note that $$u=fracb+1sqrt1-b^2+b-2$$ so $$fracdudb=frac1big(sqrt1-b^2+b-2big)-(b+1)left(-frac2bsqrt1-b^2+1right)big(sqrt1-b^2+b-2big)^2$$ and setting to zero gives $$-3sqrt1-b^2+b+1=0implies 1-b^2=fracb^2+2b+19implies 5b^2+b-4=0$$ and we see that $b=4/5,-1$ are roots.
Checking second derivatives, we have that $4/5$ is a minimum.
Hence $$u^2=left(fracfrac45+1frac35+frac45-2right)^2=9.$$
Note that the negative root ($-3/5$) is also possible, but that yields a lower value of $u^2$ since $$bigg|-frac35+frac45-2bigg|>bigg|frac35+frac45-2bigg|$$
answered Jul 15 at 11:19
TheSimpliFire
9,69261951
edited Jul 15 at 18:39
answered Jul 15 at 11:19
TheSimpliFire
9,69261951
answered Jul 15 at 11:19
TheSimpliFire
9,69261951
answered Jul 15 at 11:19
TheSimpliFire
9,69261951
9,69261951
4
You may need to consider $v=fracb+1-sqrt1-b^2+b-2$ as well
– Henry
Jul 15 at 11:24
@Henry Of course! Edited.
– TheSimpliFire
Jul 15 at 18:39
add a comment |Â
4
You may need to consider $v=fracb+1-sqrt1-b^2+b-2$ as well
– Henry
Jul 15 at 11:24
@Henry Of course! Edited.
– TheSimpliFire
Jul 15 at 18:39
4
4
You may need to consider $v=fracb+1-sqrt1-b^2+b-2$ as well
– Henry
Jul 15 at 11:24
You may need to consider $v=fracb+1-sqrt1-b^2+b-2$ as well
– Henry
Jul 15 at 11:24
@Henry Of course! Edited.
– TheSimpliFire
Jul 15 at 18:39
@Henry Of course! Edited.
– TheSimpliFire
Jul 15 at 18:39
add a comment |Â
up vote
13
down vote
Try with $b=cos 2x$ and $a= sin 2x$.
begineqnarrayb+1over a+b-2&=& 2cos^2 xover -cos^2x+2sin x cos x -3sin^2x\
&=& 2over -1+2tan x -3tan^2x\
&=& 2over -1+2t -3t^2
endeqnarray
where $t= tan x $. So the expression will take a minimum when quadratic function $g(t)=-3t^2+2t-1$ will take a maximum. Note that $g(t)<0$ for all $t in mathbbR$
So $$ u= 2over -2over 3 = -3implies ....$$
answered Jul 15 at 11:04
greedoid
26.6k93574
2
This was the first thing that came to my mind.
– prog_SAHIL
Jul 15 at 11:33
add a comment |Â
up vote
13
down vote
Try with $b=cos 2x$ and $a= sin 2x$.
begineqnarrayb+1over a+b-2&=& 2cos^2 xover -cos^2x+2sin x cos x -3sin^2x\
&=& 2over -1+2tan x -3tan^2x\
&=& 2over -1+2t -3t^2
endeqnarray
where $t= tan x $. So the expression will take a minimum when quadratic function $g(t)=-3t^2+2t-1$ will take a maximum. Note that $g(t)<0$ for all $t in mathbbR$
So $$ u= 2over -2over 3 = -3implies ....$$
answered Jul 15 at 11:04
greedoid
26.6k93574
2
This was the first thing that came to my mind.
– prog_SAHIL
Jul 15 at 11:33
add a comment |Â
up vote
13
down vote
up vote
13
down vote
Try with $b=cos 2x$ and $a= sin 2x$.
begineqnarrayb+1over a+b-2&=& 2cos^2 xover -cos^2x+2sin x cos x -3sin^2x\
&=& 2over -1+2tan x -3tan^2x\
&=& 2over -1+2t -3t^2
endeqnarray
where $t= tan x $. So the expression will take a minimum when quadratic function $g(t)=-3t^2+2t-1$ will take a maximum. Note that $g(t)<0$ for all $t in mathbbR$
So $$ u= 2over -2over 3 = -3implies ....$$
answered Jul 15 at 11:04
greedoid
26.6k93574
Try with $b=cos 2x$ and $a= sin 2x$.
begineqnarrayb+1over a+b-2&=& 2cos^2 xover -cos^2x+2sin x cos x -3sin^2x\
&=& 2over -1+2tan x -3tan^2x\
&=& 2over -1+2t -3t^2
endeqnarray
where $t= tan x $. So the expression will take a minimum when quadratic function $g(t)=-3t^2+2t-1$ will take a maximum. Note that $g(t)<0$ for all $t in mathbbR$
So $$ u= 2over -2over 3 = -3implies ....$$
answered Jul 15 at 11:04
greedoid
26.6k93574
edited Jul 15 at 11:28
answered Jul 15 at 11:04
greedoid
26.6k93574
answered Jul 15 at 11:04
greedoid
26.6k93574
answered Jul 15 at 11:04
greedoid
26.6k93574
26.6k93574
2
This was the first thing that came to my mind.
– prog_SAHIL
Jul 15 at 11:33
add a comment |Â
2
This was the first thing that came to my mind.
– prog_SAHIL
Jul 15 at 11:33
2
2
This was the first thing that came to my mind.
– prog_SAHIL
Jul 15 at 11:33
This was the first thing that came to my mind.
– prog_SAHIL
Jul 15 at 11:33
add a comment |Â
up vote
4
down vote
A bit geometry;
1)$x^2+y^2 = 1$, a circle, centre $(0,0)$, $r=1$.
2) Minimum of $C$:
$C:=dfracy+1x+y-2$
(Note: $x+y-2 not =0$).
$C(x+y-2) = y+1$, or
$Cx +y(C-1) -(2C+1)= 0$, a straight line.
The line touches or intersects the circle 1)
if the distance line-to-origin $le 1$ (radius).
Distance to $(0,0):$
$d =dfracsqrtC^2+(C-1)^2 le 1.$
$(2C+1)^2 le C^2 + (C-1)^2;$
$4C^2 +4C +1 le 2C^2 -2C+1;$
$2C(C+3) le 0.$
Hence: $-3 le C le 0$.
Minimum at $C =-3$.
Used: Line to point distance formula:
http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
answered Jul 15 at 13:44
Peter Szilas
7,9552617
What's the geometrical interpretation of minimum at C= -3 ?
– Abcd
Jul 15 at 14:17
C=-3 ; Line: -3x -4y +5=0 , y intercept 5/4 , slope: -3/4, distance from (0,0) : =1, hence a tangent to the circle. You can sketch it.
– Peter Szilas
Jul 15 at 16:10
add a comment |Â
up vote
4
down vote
A bit geometry;
1)$x^2+y^2 = 1$, a circle, centre $(0,0)$, $r=1$.
2) Minimum of $C$:
$C:=dfracy+1x+y-2$
(Note: $x+y-2 not =0$).
$C(x+y-2) = y+1$, or
$Cx +y(C-1) -(2C+1)= 0$, a straight line.
The line touches or intersects the circle 1)
if the distance line-to-origin $le 1$ (radius).
Distance to $(0,0):$
$d =dfracsqrtC^2+(C-1)^2 le 1.$
$(2C+1)^2 le C^2 + (C-1)^2;$
$4C^2 +4C +1 le 2C^2 -2C+1;$
$2C(C+3) le 0.$
Hence: $-3 le C le 0$.
Minimum at $C =-3$.
Used: Line to point distance formula:
http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
answered Jul 15 at 13:44
Peter Szilas
7,9552617
What's the geometrical interpretation of minimum at C= -3 ?
– Abcd
Jul 15 at 14:17
C=-3 ; Line: -3x -4y +5=0 , y intercept 5/4 , slope: -3/4, distance from (0,0) : =1, hence a tangent to the circle. You can sketch it.
– Peter Szilas
Jul 15 at 16:10
add a comment |Â
up vote
4
down vote
up vote
4
down vote
A bit geometry;
1)$x^2+y^2 = 1$, a circle, centre $(0,0)$, $r=1$.
2) Minimum of $C$:
$C:=dfracy+1x+y-2$
(Note: $x+y-2 not =0$).
$C(x+y-2) = y+1$, or
$Cx +y(C-1) -(2C+1)= 0$, a straight line.
The line touches or intersects the circle 1)
if the distance line-to-origin $le 1$ (radius).
Distance to $(0,0):$
$d =dfracsqrtC^2+(C-1)^2 le 1.$
$(2C+1)^2 le C^2 + (C-1)^2;$
$4C^2 +4C +1 le 2C^2 -2C+1;$
$2C(C+3) le 0.$
Hence: $-3 le C le 0$.
Minimum at $C =-3$.
Used: Line to point distance formula:
http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
answered Jul 15 at 13:44
Peter Szilas
7,9552617
A bit geometry;
1)$x^2+y^2 = 1$, a circle, centre $(0,0)$, $r=1$.
2) Minimum of $C$:
$C:=dfracy+1x+y-2$
(Note: $x+y-2 not =0$).
$C(x+y-2) = y+1$, or
$Cx +y(C-1) -(2C+1)= 0$, a straight line.
The line touches or intersects the circle 1)
if the distance line-to-origin $le 1$ (radius).
Distance to $(0,0):$
$d =dfracsqrtC^2+(C-1)^2 le 1.$
$(2C+1)^2 le C^2 + (C-1)^2;$
$4C^2 +4C +1 le 2C^2 -2C+1;$
$2C(C+3) le 0.$
Hence: $-3 le C le 0$.
Minimum at $C =-3$.
Used: Line to point distance formula:
http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
answered Jul 15 at 13:44
Peter Szilas
7,9552617
edited Jul 16 at 7:22
answered Jul 15 at 13:44
Peter Szilas
7,9552617
answered Jul 15 at 13:44
Peter Szilas
7,9552617
answered Jul 15 at 13:44
Peter Szilas
7,9552617
7,9552617
What's the geometrical interpretation of minimum at C= -3 ?
– Abcd
Jul 15 at 14:17
C=-3 ; Line: -3x -4y +5=0 , y intercept 5/4 , slope: -3/4, distance from (0,0) : =1, hence a tangent to the circle. You can sketch it.
– Peter Szilas
Jul 15 at 16:10
add a comment |Â
What's the geometrical interpretation of minimum at C= -3 ?
– Abcd
Jul 15 at 14:17
C=-3 ; Line: -3x -4y +5=0 , y intercept 5/4 , slope: -3/4, distance from (0,0) : =1, hence a tangent to the circle. You can sketch it.
– Peter Szilas
Jul 15 at 16:10
What's the geometrical interpretation of minimum at C= -3 ?
– Abcd
Jul 15 at 14:17
What's the geometrical interpretation of minimum at C= -3 ?
– Abcd
Jul 15 at 14:17
C=-3 ; Line: -3x -4y +5=0 , y intercept 5/4 , slope: -3/4, distance from (0,0) : =1, hence a tangent to the circle. You can sketch it.
– Peter Szilas
Jul 15 at 16:10
C=-3 ; Line: -3x -4y +5=0 , y intercept 5/4 , slope: -3/4, distance from (0,0) : =1, hence a tangent to the circle. You can sketch it.
– Peter Szilas
Jul 15 at 16:10
add a comment |Â
up vote
4
down vote
Let $displaystyle u=fracb+1a+b-2Rightarrow ua+ub-2u=b+1$
So $ua+(u-1)b=1+2u$
Now Using Cauchy Schwarz Inequality
$displaystyle bigg[u^2+(u-1)^2bigg]cdot bigg[a^2+b^2bigg]geq bigg[ua+(u-1)bbigg]^2$
So $displaystyle 2u^2-2u+1geq (1+2u)^2Rightarrow 2u^2+6uleq 0$
So $displaystyle 2u(u+3)leq 0Rightarrow -3 leq uleq 0$
So $displaystyle u^2 geq 9.$
answered Jul 15 at 11:51
Durgesh Tiwari
4,7882426
add a comment |Â
up vote
4
down vote
Let $displaystyle u=fracb+1a+b-2Rightarrow ua+ub-2u=b+1$
So $ua+(u-1)b=1+2u$
Now Using Cauchy Schwarz Inequality
$displaystyle bigg[u^2+(u-1)^2bigg]cdot bigg[a^2+b^2bigg]geq bigg[ua+(u-1)bbigg]^2$
So $displaystyle 2u^2-2u+1geq (1+2u)^2Rightarrow 2u^2+6uleq 0$
So $displaystyle 2u(u+3)leq 0Rightarrow -3 leq uleq 0$
So $displaystyle u^2 geq 9.$
answered Jul 15 at 11:51
Durgesh Tiwari
4,7882426
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Let $displaystyle u=fracb+1a+b-2Rightarrow ua+ub-2u=b+1$
So $ua+(u-1)b=1+2u$
Now Using Cauchy Schwarz Inequality
$displaystyle bigg[u^2+(u-1)^2bigg]cdot bigg[a^2+b^2bigg]geq bigg[ua+(u-1)bbigg]^2$
So $displaystyle 2u^2-2u+1geq (1+2u)^2Rightarrow 2u^2+6uleq 0$
So $displaystyle 2u(u+3)leq 0Rightarrow -3 leq uleq 0$
So $displaystyle u^2 geq 9.$
answered Jul 15 at 11:51
Durgesh Tiwari
4,7882426
Let $displaystyle u=fracb+1a+b-2Rightarrow ua+ub-2u=b+1$
So $ua+(u-1)b=1+2u$
Now Using Cauchy Schwarz Inequality
$displaystyle bigg[u^2+(u-1)^2bigg]cdot bigg[a^2+b^2bigg]geq bigg[ua+(u-1)bbigg]^2$
So $displaystyle 2u^2-2u+1geq (1+2u)^2Rightarrow 2u^2+6uleq 0$
So $displaystyle 2u(u+3)leq 0Rightarrow -3 leq uleq 0$
So $displaystyle u^2 geq 9.$
answered Jul 15 at 11:51
Durgesh Tiwari
4,7882426
edited Jul 18 at 5:00
answered Jul 15 at 11:51
Durgesh Tiwari
4,7882426
answered Jul 15 at 11:51
Durgesh Tiwari
4,7882426
answered Jul 15 at 11:51
Durgesh Tiwari
4,7882426
4,7882426
add a comment |Â
add a comment |Â
up vote
3
down vote
Just for a variation, using Lagrange’s method:
$$
f(a,b,t)=fracb+1a+b-2-t(a^2+b^2-1)
$$
Then
beginalign
fracpartial fpartial a&=-fracb+1(a+b-2)^2-2at \[6px]
fracpartial fpartial b&=fraca-3(a+b-2)^2-2bt
endalign
If these equal $0$, then
$$
-fracb+1a(a+b-2)^2=fraca-3b(a+b-2)^2
$$
so that $-b^2-b=a^2-3a$, that gives $3a-b=1$ and from $a^2+b^2=1$ we derive $a=0$ or $a=3/5$.
The critical points are thus $(0,-1)$ and $(3/5,4/5)$. We have
$$
f(0,-1,0)=frac23,quad
f(3/5,4/5,0)=-3,quad
$$
This also shows the maximum.
answered Jul 15 at 13:27
egreg
164k1180187
Why should the partial derivatives equal 0?
– Abcd
Jul 18 at 7:52
@Abcd Because we are looking for critical points.
– egreg
Jul 18 at 8:10
With respect to the maximum part, can't maximum be at the endpoints of the definition interval? How do we know if it is or not?
– Abcd
Jul 20 at 2:10
1
@Abcd There's no “definition intervalâ€Â: the function is restricted to the circle $a^2+b^2=1$, which is compact and has no “endpointsâ€Â.
– egreg
Jul 20 at 8:32
add a comment |Â
up vote
3
down vote
Just for a variation, using Lagrange’s method:
$$
f(a,b,t)=fracb+1a+b-2-t(a^2+b^2-1)
$$
Then
beginalign
fracpartial fpartial a&=-fracb+1(a+b-2)^2-2at \[6px]
fracpartial fpartial b&=fraca-3(a+b-2)^2-2bt
endalign
If these equal $0$, then
$$
-fracb+1a(a+b-2)^2=fraca-3b(a+b-2)^2
$$
so that $-b^2-b=a^2-3a$, that gives $3a-b=1$ and from $a^2+b^2=1$ we derive $a=0$ or $a=3/5$.
The critical points are thus $(0,-1)$ and $(3/5,4/5)$. We have
$$
f(0,-1,0)=frac23,quad
f(3/5,4/5,0)=-3,quad
$$
This also shows the maximum.
answered Jul 15 at 13:27
egreg
164k1180187
Why should the partial derivatives equal 0?
– Abcd
Jul 18 at 7:52
@Abcd Because we are looking for critical points.
– egreg
Jul 18 at 8:10
With respect to the maximum part, can't maximum be at the endpoints of the definition interval? How do we know if it is or not?
– Abcd
Jul 20 at 2:10
1
@Abcd There's no “definition intervalâ€Â: the function is restricted to the circle $a^2+b^2=1$, which is compact and has no “endpointsâ€Â.
– egreg
Jul 20 at 8:32
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Just for a variation, using Lagrange’s method:
$$
f(a,b,t)=fracb+1a+b-2-t(a^2+b^2-1)
$$
Then
beginalign
fracpartial fpartial a&=-fracb+1(a+b-2)^2-2at \[6px]
fracpartial fpartial b&=fraca-3(a+b-2)^2-2bt
endalign
If these equal $0$, then
$$
-fracb+1a(a+b-2)^2=fraca-3b(a+b-2)^2
$$
so that $-b^2-b=a^2-3a$, that gives $3a-b=1$ and from $a^2+b^2=1$ we derive $a=0$ or $a=3/5$.
The critical points are thus $(0,-1)$ and $(3/5,4/5)$. We have
$$
f(0,-1,0)=frac23,quad
f(3/5,4/5,0)=-3,quad
$$
This also shows the maximum.
answered Jul 15 at 13:27
egreg
164k1180187
Just for a variation, using Lagrange’s method:
$$
f(a,b,t)=fracb+1a+b-2-t(a^2+b^2-1)
$$
Then
beginalign
fracpartial fpartial a&=-fracb+1(a+b-2)^2-2at \[6px]
fracpartial fpartial b&=fraca-3(a+b-2)^2-2bt
endalign
If these equal $0$, then
$$
-fracb+1a(a+b-2)^2=fraca-3b(a+b-2)^2
$$
so that $-b^2-b=a^2-3a$, that gives $3a-b=1$ and from $a^2+b^2=1$ we derive $a=0$ or $a=3/5$.
The critical points are thus $(0,-1)$ and $(3/5,4/5)$. We have
$$
f(0,-1,0)=frac23,quad
f(3/5,4/5,0)=-3,quad
$$
This also shows the maximum.
answered Jul 15 at 13:27
egreg
164k1180187
answered Jul 15 at 13:27
egreg
164k1180187
answered Jul 15 at 13:27
egreg
164k1180187
answered Jul 15 at 13:27
egreg
164k1180187
164k1180187
Why should the partial derivatives equal 0?
– Abcd
Jul 18 at 7:52
@Abcd Because we are looking for critical points.
– egreg
Jul 18 at 8:10
With respect to the maximum part, can't maximum be at the endpoints of the definition interval? How do we know if it is or not?
– Abcd
Jul 20 at 2:10
1
@Abcd There's no “definition intervalâ€Â: the function is restricted to the circle $a^2+b^2=1$, which is compact and has no “endpointsâ€Â.
– egreg
Jul 20 at 8:32
add a comment |Â
Why should the partial derivatives equal 0?
– Abcd
Jul 18 at 7:52
@Abcd Because we are looking for critical points.
– egreg
Jul 18 at 8:10
With respect to the maximum part, can't maximum be at the endpoints of the definition interval? How do we know if it is or not?
– Abcd
Jul 20 at 2:10
1
@Abcd There's no “definition intervalâ€Â: the function is restricted to the circle $a^2+b^2=1$, which is compact and has no “endpointsâ€Â.
– egreg
Jul 20 at 8:32
Why should the partial derivatives equal 0?
– Abcd
Jul 18 at 7:52
Why should the partial derivatives equal 0?
– Abcd
Jul 18 at 7:52
@Abcd Because we are looking for critical points.
– egreg
Jul 18 at 8:10
@Abcd Because we are looking for critical points.
– egreg
Jul 18 at 8:10
With respect to the maximum part, can't maximum be at the endpoints of the definition interval? How do we know if it is or not?
– Abcd
Jul 20 at 2:10
With respect to the maximum part, can't maximum be at the endpoints of the definition interval? How do we know if it is or not?
– Abcd
Jul 20 at 2:10
1
1
@Abcd There's no “definition intervalâ€Â: the function is restricted to the circle $a^2+b^2=1$, which is compact and has no “endpointsâ€Â.
– egreg
Jul 20 at 8:32
@Abcd There's no “definition intervalâ€Â: the function is restricted to the circle $a^2+b^2=1$, which is compact and has no “endpointsâ€Â.
– egreg
Jul 20 at 8:32
add a comment |Â
up vote
2
down vote
From
$$
fracb+1a+b-2= uRightarrow Lto a = 2-bfracb+1u
$$
So $L$ should be tangent to $a^2+b^2=1;$ then substituting we have the condition
$$
(2b^2-4b+3)u^2+(4+2b-2b^2)u+(b+1)^2 = 0
$$
and solving for $u$
$$
u = frac(b+1)^2b^2-bpmsqrt(1-b) (b+1)^3-2
$$
but tangency implies on $sqrt(1-b) (b+1)^3=0;$ hence the solutions for tangency are $b = pm 1$ etc.
answered Jul 15 at 11:28
Cesareo
5,7922412
Didn't understand why L should be a tangent and why it should tend to a
– Abcd
Jul 15 at 11:40
1
Interesting method, though please explain the concepts used
– Abcd
Jul 15 at 11:40
@Abcd The problem $minmax u=f(a,b)$ s. t. $a^2+b^2-1=0$ when $f(a,b)$ is regular, is defined at the tangency points for $u=f(a,b)$ and $a^2+b^2-1=0$ At those points we have $u = u_min,cdots,u_max$. This is the idea of Lagrange multipliers method.
– Cesareo
Jul 15 at 13:43
add a comment |Â
up vote
2
down vote
From
$$
fracb+1a+b-2= uRightarrow Lto a = 2-bfracb+1u
$$
So $L$ should be tangent to $a^2+b^2=1;$ then substituting we have the condition
$$
(2b^2-4b+3)u^2+(4+2b-2b^2)u+(b+1)^2 = 0
$$
and solving for $u$
$$
u = frac(b+1)^2b^2-bpmsqrt(1-b) (b+1)^3-2
$$
but tangency implies on $sqrt(1-b) (b+1)^3=0;$ hence the solutions for tangency are $b = pm 1$ etc.
answered Jul 15 at 11:28
Cesareo
5,7922412
Didn't understand why L should be a tangent and why it should tend to a
– Abcd
Jul 15 at 11:40
1
Interesting method, though please explain the concepts used
– Abcd
Jul 15 at 11:40
@Abcd The problem $minmax u=f(a,b)$ s. t. $a^2+b^2-1=0$ when $f(a,b)$ is regular, is defined at the tangency points for $u=f(a,b)$ and $a^2+b^2-1=0$ At those points we have $u = u_min,cdots,u_max$. This is the idea of Lagrange multipliers method.
– Cesareo
Jul 15 at 13:43
add a comment |Â
up vote
2
down vote
up vote
2
down vote
From
$$
fracb+1a+b-2= uRightarrow Lto a = 2-bfracb+1u
$$
So $L$ should be tangent to $a^2+b^2=1;$ then substituting we have the condition
$$
(2b^2-4b+3)u^2+(4+2b-2b^2)u+(b+1)^2 = 0
$$
and solving for $u$
$$
u = frac(b+1)^2b^2-bpmsqrt(1-b) (b+1)^3-2
$$
but tangency implies on $sqrt(1-b) (b+1)^3=0;$ hence the solutions for tangency are $b = pm 1$ etc.
answered Jul 15 at 11:28
Cesareo
5,7922412
From
$$
fracb+1a+b-2= uRightarrow Lto a = 2-bfracb+1u
$$
So $L$ should be tangent to $a^2+b^2=1;$ then substituting we have the condition
$$
(2b^2-4b+3)u^2+(4+2b-2b^2)u+(b+1)^2 = 0
$$
and solving for $u$
$$
u = frac(b+1)^2b^2-bpmsqrt(1-b) (b+1)^3-2
$$
but tangency implies on $sqrt(1-b) (b+1)^3=0;$ hence the solutions for tangency are $b = pm 1$ etc.
answered Jul 15 at 11:28
Cesareo
5,7922412
edited Jul 15 at 13:31
answered Jul 15 at 11:28
Cesareo
5,7922412
answered Jul 15 at 11:28
Cesareo
5,7922412
answered Jul 15 at 11:28
Cesareo
5,7922412
5,7922412
Didn't understand why L should be a tangent and why it should tend to a
– Abcd
Jul 15 at 11:40
1
Interesting method, though please explain the concepts used
– Abcd
Jul 15 at 11:40
@Abcd The problem $minmax u=f(a,b)$ s. t. $a^2+b^2-1=0$ when $f(a,b)$ is regular, is defined at the tangency points for $u=f(a,b)$ and $a^2+b^2-1=0$ At those points we have $u = u_min,cdots,u_max$. This is the idea of Lagrange multipliers method.
– Cesareo
Jul 15 at 13:43
add a comment |Â
Didn't understand why L should be a tangent and why it should tend to a
– Abcd
Jul 15 at 11:40
1
Interesting method, though please explain the concepts used
– Abcd
Jul 15 at 11:40
@Abcd The problem $minmax u=f(a,b)$ s. t. $a^2+b^2-1=0$ when $f(a,b)$ is regular, is defined at the tangency points for $u=f(a,b)$ and $a^2+b^2-1=0$ At those points we have $u = u_min,cdots,u_max$. This is the idea of Lagrange multipliers method.
– Cesareo
Jul 15 at 13:43
Didn't understand why L should be a tangent and why it should tend to a
– Abcd
Jul 15 at 11:40
Didn't understand why L should be a tangent and why it should tend to a
– Abcd
Jul 15 at 11:40
1
1
Interesting method, though please explain the concepts used
– Abcd
Jul 15 at 11:40
Interesting method, though please explain the concepts used
– Abcd
Jul 15 at 11:40
@Abcd The problem $minmax u=f(a,b)$ s. t. $a^2+b^2-1=0$ when $f(a,b)$ is regular, is defined at the tangency points for $u=f(a,b)$ and $a^2+b^2-1=0$ At those points we have $u = u_min,cdots,u_max$. This is the idea of Lagrange multipliers method.
– Cesareo
Jul 15 at 13:43
@Abcd The problem $minmax u=f(a,b)$ s. t. $a^2+b^2-1=0$ when $f(a,b)$ is regular, is defined at the tangency points for $u=f(a,b)$ and $a^2+b^2-1=0$ At those points we have $u = u_min,cdots,u_max$. This is the idea of Lagrange multipliers method.
– Cesareo
Jul 15 at 13:43
add a comment |Â
up vote
2
down vote
Pulling a small rabbit from a hat, consider
$$f(a,b)=3+b+1over a+b-2=3a+4b-5over a+b-2$$
It's clear that $a^2+b^2=1$ implies $a+b-2lt0$. By Cauchy-Schwartz, we have
$$(3a+4b)^2le(a^2+b^2)(3^2+4^2)=25=5^2$$
and therefore $3a+4b-5le0$ if $a^2+b^2=1$. Thus $f(a,b)ge0$ for all $a$ and $b$ for which $a^2+b^2=1$. But also $f(3over5,4over5)=0$, so the minimum value of $(b+1)/(a+b-2)=f(a,b)-3$ with $a^2+b^2=1$ is $-3$, the square of which is $9$.
answered Jul 15 at 14:47
Barry Cipra
56.5k652118
add a comment |Â
up vote
2
down vote
Pulling a small rabbit from a hat, consider
$$f(a,b)=3+b+1over a+b-2=3a+4b-5over a+b-2$$
It's clear that $a^2+b^2=1$ implies $a+b-2lt0$. By Cauchy-Schwartz, we have
$$(3a+4b)^2le(a^2+b^2)(3^2+4^2)=25=5^2$$
and therefore $3a+4b-5le0$ if $a^2+b^2=1$. Thus $f(a,b)ge0$ for all $a$ and $b$ for which $a^2+b^2=1$. But also $f(3over5,4over5)=0$, so the minimum value of $(b+1)/(a+b-2)=f(a,b)-3$ with $a^2+b^2=1$ is $-3$, the square of which is $9$.
answered Jul 15 at 14:47
Barry Cipra
56.5k652118
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Pulling a small rabbit from a hat, consider
$$f(a,b)=3+b+1over a+b-2=3a+4b-5over a+b-2$$
It's clear that $a^2+b^2=1$ implies $a+b-2lt0$. By Cauchy-Schwartz, we have
$$(3a+4b)^2le(a^2+b^2)(3^2+4^2)=25=5^2$$
and therefore $3a+4b-5le0$ if $a^2+b^2=1$. Thus $f(a,b)ge0$ for all $a$ and $b$ for which $a^2+b^2=1$. But also $f(3over5,4over5)=0$, so the minimum value of $(b+1)/(a+b-2)=f(a,b)-3$ with $a^2+b^2=1$ is $-3$, the square of which is $9$.
answered Jul 15 at 14:47
Barry Cipra
56.5k652118
Pulling a small rabbit from a hat, consider
$$f(a,b)=3+b+1over a+b-2=3a+4b-5over a+b-2$$
It's clear that $a^2+b^2=1$ implies $a+b-2lt0$. By Cauchy-Schwartz, we have
$$(3a+4b)^2le(a^2+b^2)(3^2+4^2)=25=5^2$$
and therefore $3a+4b-5le0$ if $a^2+b^2=1$. Thus $f(a,b)ge0$ for all $a$ and $b$ for which $a^2+b^2=1$. But also $f(3over5,4over5)=0$, so the minimum value of $(b+1)/(a+b-2)=f(a,b)-3$ with $a^2+b^2=1$ is $-3$, the square of which is $9$.
answered Jul 15 at 14:47
Barry Cipra
56.5k652118
answered Jul 15 at 14:47
Barry Cipra
56.5k652118
answered Jul 15 at 14:47
Barry Cipra
56.5k652118
answered Jul 15 at 14:47
Barry Cipra
56.5k652118
56.5k652118
add a comment |Â
add a comment |Â
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1
Just an idea: you can interpret the question as having the points $(a,b)$ on the unit circle. Perhaps you can try creating a bijection by expressing $a$ and $b$ in terms of the angle $theta$.
– user496634
Jul 15 at 11:27