Mixing
Minimum value of $frac{4}{4-x^2}+frac{9}{9-y^2}$
$begingroup$
Given $x,y in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=frac{4}{4-x^2}+frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=frac{4}{4-x^2}+frac{9x^2}{9x^2-1}$$
$$g(x)=frac{4}{4-x^2}+1+frac{1}{9x^2-1}$$
Using Differentiation we get:
$$g'(x)=frac{8x}{(4-x^2)^2}-frac{18x}{(9x^2-1)^2}$$
$$g'(x)=2xleft(frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}right)$$
$$g'(x)=70xfrac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$
So the critical points are:
$x=0, x=pm sqrt{frac{2}{3}}$
But $x ne 0$ since $xy=-1$
$$g'(x)=70x frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$
By using derivative test we get Minimum occurs when $x=pm sqrt{frac{2}{3}}$
Hence $$x^2=frac{2}{3}, y^2=frac{3}{2}$$
Min value is $$frac{4}{4-frac{2}{3}}+frac{9}{9-frac{3}{2}}=frac{12}{5}$$
Is there any other approach?
algebra-precalculus derivatives maxima-minima
asked Jan 10 at 14:25
Umesh shankarUmesh shankar
2,68031219
$endgroup$
add a comment |
$begingroup$
Given $x,y in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=frac{4}{4-x^2}+frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=frac{4}{4-x^2}+frac{9x^2}{9x^2-1}$$
$$g(x)=frac{4}{4-x^2}+1+frac{1}{9x^2-1}$$
Using Differentiation we get:
$$g'(x)=frac{8x}{(4-x^2)^2}-frac{18x}{(9x^2-1)^2}$$
$$g'(x)=2xleft(frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}right)$$
$$g'(x)=70xfrac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$
So the critical points are:
$x=0, x=pm sqrt{frac{2}{3}}$
But $x ne 0$ since $xy=-1$
$$g'(x)=70x frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$
By using derivative test we get Minimum occurs when $x=pm sqrt{frac{2}{3}}$
Hence $$x^2=frac{2}{3}, y^2=frac{3}{2}$$
Min value is $$frac{4}{4-frac{2}{3}}+frac{9}{9-frac{3}{2}}=frac{12}{5}$$
Is there any other approach?
algebra-precalculus derivatives maxima-minima
asked Jan 10 at 14:25
Umesh shankarUmesh shankar
2,68031219
$endgroup$
$begingroup$
You also need to check the edges of the interval. They don't yield a smaller value here, but they could in other cases.
$endgroup$
– Ross Millikan
Jan 10 at 15:16
add a comment |
$begingroup$
Given $x,y in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=frac{4}{4-x^2}+frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=frac{4}{4-x^2}+frac{9x^2}{9x^2-1}$$
$$g(x)=frac{4}{4-x^2}+1+frac{1}{9x^2-1}$$
Using Differentiation we get:
$$g'(x)=frac{8x}{(4-x^2)^2}-frac{18x}{(9x^2-1)^2}$$
$$g'(x)=2xleft(frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}right)$$
$$g'(x)=70xfrac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$
So the critical points are:
$x=0, x=pm sqrt{frac{2}{3}}$
But $x ne 0$ since $xy=-1$
$$g'(x)=70x frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$
By using derivative test we get Minimum occurs when $x=pm sqrt{frac{2}{3}}$
Hence $$x^2=frac{2}{3}, y^2=frac{3}{2}$$
Min value is $$frac{4}{4-frac{2}{3}}+frac{9}{9-frac{3}{2}}=frac{12}{5}$$
Is there any other approach?
algebra-precalculus derivatives maxima-minima
asked Jan 10 at 14:25
Umesh shankarUmesh shankar
2,68031219
$endgroup$
Given $x,y in (-2,2)$ and $xy=-1$
Minimum value of $$f(x,y)=frac{4}{4-x^2}+frac{9}{9-y^2}$$
My try:
Converting the function into single variable we get:
$$g(x)=frac{4}{4-x^2}+frac{9x^2}{9x^2-1}$$
$$g(x)=frac{4}{4-x^2}+1+frac{1}{9x^2-1}$$
Using Differentiation we get:
$$g'(x)=frac{8x}{(4-x^2)^2}-frac{18x}{(9x^2-1)^2}$$
$$g'(x)=2xleft(frac{4(9x^2-1)^2-9(4-x^2)^2}{(4-x^2)^2(9x^2-1)^2}right)$$
$$g'(x)=70xfrac{9x^4-4}{(4-x^2)^2(9x^2-1)^2}=0$$
So the critical points are:
$x=0, x=pm sqrt{frac{2}{3}}$
But $x ne 0$ since $xy=-1$
$$g'(x)=70x frac{(3x^2-2)(3x^2+2)}{(4-x^2)^2(9x^2-1)^2}$$
By using derivative test we get Minimum occurs when $x=pm sqrt{frac{2}{3}}$
Hence $$x^2=frac{2}{3}, y^2=frac{3}{2}$$
Min value is $$frac{4}{4-frac{2}{3}}+frac{9}{9-frac{3}{2}}=frac{12}{5}$$
Is there any other approach?
algebra-precalculus derivatives maxima-minima
algebra-precalculus derivatives maxima-minima
asked Jan 10 at 14:25
Umesh shankarUmesh shankar
2,68031219
asked Jan 10 at 14:25
Umesh shankarUmesh shankar
2,68031219
asked Jan 10 at 14:25
Umesh shankarUmesh shankar
2,68031219
asked Jan 10 at 14:25
Umesh shankarUmesh shankar
2,68031219
asked Jan 10 at 14:25
Umesh shankarUmesh shankar
2,68031219
2,68031219
$begingroup$
You also need to check the edges of the interval. They don't yield a smaller value here, but they could in other cases.
$endgroup$
– Ross Millikan
Jan 10 at 15:16
add a comment |
$begingroup$
You also need to check the edges of the interval. They don't yield a smaller value here, but they could in other cases.
$endgroup$
– Ross Millikan
Jan 10 at 15:16
$begingroup$
You also need to check the edges of the interval. They don't yield a smaller value here, but they could in other cases.
$endgroup$
– Ross Millikan
Jan 10 at 15:16
$begingroup$
You also need to check the edges of the interval. They don't yield a smaller value here, but they could in other cases.
$endgroup$
– Ross Millikan
Jan 10 at 15:16
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Using $xy = -1$, we can put the fractions over a common denominator as
$${4over{4 - x^2}} + {9over{9 - y^2}} = {{72 - 9x^2 - 4y^2}over{37 - 9x^2 - 4y^2}} = 1 + {35over{37 - (9x^2 + 4y^2)}}$$
So we need to minimize $9x^2 + 4y^2$. Since $9x^2 times 4y^2$ is constant, the minimum occurs when $9x^2 = 4y^2$, i.e. $9x^2 = 4/x^2$, which gives the same result as you found.
answered Jan 10 at 15:07
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
From “harmonic mean $le$ arithmetic mean” one gets
$$
frac{2}{f(x, y)} le frac12 left(frac{4-x^2}{4}+ frac{9-y^2}{9} right)
= frac 12 left(2- frac{x^2}{4} - frac{y^2}{9} right)
$$
and from “arithmetic mean $ge$ geometric mean”
$$
frac{x^2}{4} + frac{y^2}{9}ge 2 frac{|xy|}{6} = frac 13
$$
so that
$$
f(x, y) ge frac{4}{2-frac 13} = frac{12}{5} , .
$$
In both estimates equality holds if $frac{x^2}{4} = frac{y^2}{9} iff x^2=frac{2}{3}, y^2=frac{3}{2},$ i.e. equality is attained the two points
$$
( sqrt frac 23, -sqrt frac 32) quad text{and} quad
( -sqrt frac 23, sqrt frac 32)
$$
which are both in the given domain $(-2, 2) times (-2, 2)$ of $f$.
answered Jan 10 at 14:58
Martin RMartin R
28.1k33355
$endgroup$
add a comment |
$begingroup$
With the change of variable $t=x^2$, and differentiating the logarithm of the function, things should be a bit easier:
$$h(t)=frac4{4-t}+frac{9t}{9t-1},tinleft(frac14,4right)$$
Then
$$h(t)=frac{-9t^2+72t-4}{(4-t)(9t-1)}$$
$$log h(t)=log(-9t^2+72t-4)-log(4-t)-log(9t-1)$$
$$(log h(t))'=frac{-18t+72}{-9t^2+72t-4}+frac1{4-t}-frac9{9t-1}=frac{315t^2-140}{text{Irrelevant}}$$
This is $0$ for $t=frac23$
answered Jan 10 at 14:30
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
From $xy=-1$
$$ S = frac{4}{4-x^2} + frac{9}{9-y^2} = 1 + frac{35}{25 - left(3x - frac{2}{x}right)^2}geq frac{12}{5}$$
Equality hold when $$3x=frac{2}{x}$$
answered Jan 12 at 10:08
DXTDXT
5,6742630
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using $xy = -1$, we can put the fractions over a common denominator as
$${4over{4 - x^2}} + {9over{9 - y^2}} = {{72 - 9x^2 - 4y^2}over{37 - 9x^2 - 4y^2}} = 1 + {35over{37 - (9x^2 + 4y^2)}}$$
So we need to minimize $9x^2 + 4y^2$. Since $9x^2 times 4y^2$ is constant, the minimum occurs when $9x^2 = 4y^2$, i.e. $9x^2 = 4/x^2$, which gives the same result as you found.
answered Jan 10 at 15:07
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
Using $xy = -1$, we can put the fractions over a common denominator as
$${4over{4 - x^2}} + {9over{9 - y^2}} = {{72 - 9x^2 - 4y^2}over{37 - 9x^2 - 4y^2}} = 1 + {35over{37 - (9x^2 + 4y^2)}}$$
So we need to minimize $9x^2 + 4y^2$. Since $9x^2 times 4y^2$ is constant, the minimum occurs when $9x^2 = 4y^2$, i.e. $9x^2 = 4/x^2$, which gives the same result as you found.
answered Jan 10 at 15:07
Michael BehrendMichael Behrend
1,22746
$endgroup$
add a comment |
$begingroup$
Using $xy = -1$, we can put the fractions over a common denominator as
$${4over{4 - x^2}} + {9over{9 - y^2}} = {{72 - 9x^2 - 4y^2}over{37 - 9x^2 - 4y^2}} = 1 + {35over{37 - (9x^2 + 4y^2)}}$$
So we need to minimize $9x^2 + 4y^2$. Since $9x^2 times 4y^2$ is constant, the minimum occurs when $9x^2 = 4y^2$, i.e. $9x^2 = 4/x^2$, which gives the same result as you found.
answered Jan 10 at 15:07
Michael BehrendMichael Behrend
1,22746
$endgroup$
Using $xy = -1$, we can put the fractions over a common denominator as
$${4over{4 - x^2}} + {9over{9 - y^2}} = {{72 - 9x^2 - 4y^2}over{37 - 9x^2 - 4y^2}} = 1 + {35over{37 - (9x^2 + 4y^2)}}$$
So we need to minimize $9x^2 + 4y^2$. Since $9x^2 times 4y^2$ is constant, the minimum occurs when $9x^2 = 4y^2$, i.e. $9x^2 = 4/x^2$, which gives the same result as you found.
answered Jan 10 at 15:07
Michael BehrendMichael Behrend
1,22746
answered Jan 10 at 15:07
Michael BehrendMichael Behrend
1,22746
answered Jan 10 at 15:07
Michael BehrendMichael Behrend
1,22746
answered Jan 10 at 15:07
Michael BehrendMichael Behrend
1,22746
1,22746
add a comment |
add a comment |
$begingroup$
From “harmonic mean $le$ arithmetic mean” one gets
$$
frac{2}{f(x, y)} le frac12 left(frac{4-x^2}{4}+ frac{9-y^2}{9} right)
= frac 12 left(2- frac{x^2}{4} - frac{y^2}{9} right)
$$
and from “arithmetic mean $ge$ geometric mean”
$$
frac{x^2}{4} + frac{y^2}{9}ge 2 frac{|xy|}{6} = frac 13
$$
so that
$$
f(x, y) ge frac{4}{2-frac 13} = frac{12}{5} , .
$$
In both estimates equality holds if $frac{x^2}{4} = frac{y^2}{9} iff x^2=frac{2}{3}, y^2=frac{3}{2},$ i.e. equality is attained the two points
$$
( sqrt frac 23, -sqrt frac 32) quad text{and} quad
( -sqrt frac 23, sqrt frac 32)
$$
which are both in the given domain $(-2, 2) times (-2, 2)$ of $f$.
answered Jan 10 at 14:58
Martin RMartin R
28.1k33355
$endgroup$
add a comment |
$begingroup$
From “harmonic mean $le$ arithmetic mean” one gets
$$
frac{2}{f(x, y)} le frac12 left(frac{4-x^2}{4}+ frac{9-y^2}{9} right)
= frac 12 left(2- frac{x^2}{4} - frac{y^2}{9} right)
$$
and from “arithmetic mean $ge$ geometric mean”
$$
frac{x^2}{4} + frac{y^2}{9}ge 2 frac{|xy|}{6} = frac 13
$$
so that
$$
f(x, y) ge frac{4}{2-frac 13} = frac{12}{5} , .
$$
In both estimates equality holds if $frac{x^2}{4} = frac{y^2}{9} iff x^2=frac{2}{3}, y^2=frac{3}{2},$ i.e. equality is attained the two points
$$
( sqrt frac 23, -sqrt frac 32) quad text{and} quad
( -sqrt frac 23, sqrt frac 32)
$$
which are both in the given domain $(-2, 2) times (-2, 2)$ of $f$.
answered Jan 10 at 14:58
Martin RMartin R
28.1k33355
$endgroup$
add a comment |
$begingroup$
From “harmonic mean $le$ arithmetic mean” one gets
$$
frac{2}{f(x, y)} le frac12 left(frac{4-x^2}{4}+ frac{9-y^2}{9} right)
= frac 12 left(2- frac{x^2}{4} - frac{y^2}{9} right)
$$
and from “arithmetic mean $ge$ geometric mean”
$$
frac{x^2}{4} + frac{y^2}{9}ge 2 frac{|xy|}{6} = frac 13
$$
so that
$$
f(x, y) ge frac{4}{2-frac 13} = frac{12}{5} , .
$$
In both estimates equality holds if $frac{x^2}{4} = frac{y^2}{9} iff x^2=frac{2}{3}, y^2=frac{3}{2},$ i.e. equality is attained the two points
$$
( sqrt frac 23, -sqrt frac 32) quad text{and} quad
( -sqrt frac 23, sqrt frac 32)
$$
which are both in the given domain $(-2, 2) times (-2, 2)$ of $f$.
answered Jan 10 at 14:58
Martin RMartin R
28.1k33355
$endgroup$
From “harmonic mean $le$ arithmetic mean” one gets
$$
frac{2}{f(x, y)} le frac12 left(frac{4-x^2}{4}+ frac{9-y^2}{9} right)
= frac 12 left(2- frac{x^2}{4} - frac{y^2}{9} right)
$$
and from “arithmetic mean $ge$ geometric mean”
$$
frac{x^2}{4} + frac{y^2}{9}ge 2 frac{|xy|}{6} = frac 13
$$
so that
$$
f(x, y) ge frac{4}{2-frac 13} = frac{12}{5} , .
$$
In both estimates equality holds if $frac{x^2}{4} = frac{y^2}{9} iff x^2=frac{2}{3}, y^2=frac{3}{2},$ i.e. equality is attained the two points
$$
( sqrt frac 23, -sqrt frac 32) quad text{and} quad
( -sqrt frac 23, sqrt frac 32)
$$
which are both in the given domain $(-2, 2) times (-2, 2)$ of $f$.
answered Jan 10 at 14:58
Martin RMartin R
28.1k33355
edited Jan 10 at 15:19
answered Jan 10 at 14:58
Martin RMartin R
28.1k33355
answered Jan 10 at 14:58
Martin RMartin R
28.1k33355
answered Jan 10 at 14:58
Martin RMartin R
28.1k33355
28.1k33355
add a comment |
add a comment |
$begingroup$
With the change of variable $t=x^2$, and differentiating the logarithm of the function, things should be a bit easier:
$$h(t)=frac4{4-t}+frac{9t}{9t-1},tinleft(frac14,4right)$$
Then
$$h(t)=frac{-9t^2+72t-4}{(4-t)(9t-1)}$$
$$log h(t)=log(-9t^2+72t-4)-log(4-t)-log(9t-1)$$
$$(log h(t))'=frac{-18t+72}{-9t^2+72t-4}+frac1{4-t}-frac9{9t-1}=frac{315t^2-140}{text{Irrelevant}}$$
This is $0$ for $t=frac23$
answered Jan 10 at 14:30
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
With the change of variable $t=x^2$, and differentiating the logarithm of the function, things should be a bit easier:
$$h(t)=frac4{4-t}+frac{9t}{9t-1},tinleft(frac14,4right)$$
Then
$$h(t)=frac{-9t^2+72t-4}{(4-t)(9t-1)}$$
$$log h(t)=log(-9t^2+72t-4)-log(4-t)-log(9t-1)$$
$$(log h(t))'=frac{-18t+72}{-9t^2+72t-4}+frac1{4-t}-frac9{9t-1}=frac{315t^2-140}{text{Irrelevant}}$$
This is $0$ for $t=frac23$
answered Jan 10 at 14:30
ajotatxeajotatxe
53.8k23890
$endgroup$
add a comment |
$begingroup$
With the change of variable $t=x^2$, and differentiating the logarithm of the function, things should be a bit easier:
$$h(t)=frac4{4-t}+frac{9t}{9t-1},tinleft(frac14,4right)$$
Then
$$h(t)=frac{-9t^2+72t-4}{(4-t)(9t-1)}$$
$$log h(t)=log(-9t^2+72t-4)-log(4-t)-log(9t-1)$$
$$(log h(t))'=frac{-18t+72}{-9t^2+72t-4}+frac1{4-t}-frac9{9t-1}=frac{315t^2-140}{text{Irrelevant}}$$
This is $0$ for $t=frac23$
answered Jan 10 at 14:30
ajotatxeajotatxe
53.8k23890
$endgroup$
With the change of variable $t=x^2$, and differentiating the logarithm of the function, things should be a bit easier:
$$h(t)=frac4{4-t}+frac{9t}{9t-1},tinleft(frac14,4right)$$
Then
$$h(t)=frac{-9t^2+72t-4}{(4-t)(9t-1)}$$
$$log h(t)=log(-9t^2+72t-4)-log(4-t)-log(9t-1)$$
$$(log h(t))'=frac{-18t+72}{-9t^2+72t-4}+frac1{4-t}-frac9{9t-1}=frac{315t^2-140}{text{Irrelevant}}$$
This is $0$ for $t=frac23$
answered Jan 10 at 14:30
ajotatxeajotatxe
53.8k23890
edited Jan 10 at 14:57
answered Jan 10 at 14:30
ajotatxeajotatxe
53.8k23890
answered Jan 10 at 14:30
ajotatxeajotatxe
53.8k23890
answered Jan 10 at 14:30
ajotatxeajotatxe
53.8k23890
53.8k23890
add a comment |
add a comment |
$begingroup$
From $xy=-1$
$$ S = frac{4}{4-x^2} + frac{9}{9-y^2} = 1 + frac{35}{25 - left(3x - frac{2}{x}right)^2}geq frac{12}{5}$$
Equality hold when $$3x=frac{2}{x}$$
answered Jan 12 at 10:08
DXTDXT
5,6742630
$endgroup$
add a comment |
$begingroup$
From $xy=-1$
$$ S = frac{4}{4-x^2} + frac{9}{9-y^2} = 1 + frac{35}{25 - left(3x - frac{2}{x}right)^2}geq frac{12}{5}$$
Equality hold when $$3x=frac{2}{x}$$
answered Jan 12 at 10:08
DXTDXT
5,6742630
$endgroup$
add a comment |
$begingroup$
From $xy=-1$
$$ S = frac{4}{4-x^2} + frac{9}{9-y^2} = 1 + frac{35}{25 - left(3x - frac{2}{x}right)^2}geq frac{12}{5}$$
Equality hold when $$3x=frac{2}{x}$$
answered Jan 12 at 10:08
DXTDXT
5,6742630
$endgroup$
From $xy=-1$
$$ S = frac{4}{4-x^2} + frac{9}{9-y^2} = 1 + frac{35}{25 - left(3x - frac{2}{x}right)^2}geq frac{12}{5}$$
Equality hold when $$3x=frac{2}{x}$$
answered Jan 12 at 10:08
DXTDXT
5,6742630
answered Jan 12 at 10:08
DXTDXT
5,6742630
answered Jan 12 at 10:08
DXTDXT
5,6742630
answered Jan 12 at 10:08
DXTDXT
5,6742630
5,6742630
add a comment |
add a comment |
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To learn more, see our tips on writing great answers.
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$begingroup$
You also need to check the edges of the interval. They don't yield a smaller value here, but they could in other cases.
$endgroup$
– Ross Millikan
Jan 10 at 15:16