How can I get the exact answer of the maximum of this function?
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I am trying to find the maximum of a fuction: $ f(x)=sqrtx^2+9-sqrtx^2-sqrt3 x+1 $.
I tried
Maximize[Sqrt[x^2 + 9] - Sqrt[x^2 - Sqrt[3]*x + 1], x]
and I got
[Sqrt](9 +
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]^2) - [Sqrt](1 -
Sqrt[3] Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2] +
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]^2), x ->
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]
The exact answer is $sqrt7$.
How can I get the exact answer of the maximum of this function?
calculus-and-analysis mathematical-optimization
add a comment |Â
up vote
4
down vote
favorite
I am trying to find the maximum of a fuction: $ f(x)=sqrtx^2+9-sqrtx^2-sqrt3 x+1 $.
I tried
Maximize[Sqrt[x^2 + 9] - Sqrt[x^2 - Sqrt[3]*x + 1], x]
and I got
[Sqrt](9 +
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]^2) - [Sqrt](1 -
Sqrt[3] Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2] +
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]^2), x ->
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]
The exact answer is $sqrt7$.
How can I get the exact answer of the maximum of this function?
calculus-and-analysis mathematical-optimization
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am trying to find the maximum of a fuction: $ f(x)=sqrtx^2+9-sqrtx^2-sqrt3 x+1 $.
I tried
Maximize[Sqrt[x^2 + 9] - Sqrt[x^2 - Sqrt[3]*x + 1], x]
and I got
[Sqrt](9 +
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]^2) - [Sqrt](1 -
Sqrt[3] Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2] +
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]^2), x ->
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]
The exact answer is $sqrt7$.
How can I get the exact answer of the maximum of this function?
calculus-and-analysis mathematical-optimization
I am trying to find the maximum of a fuction: $ f(x)=sqrtx^2+9-sqrtx^2-sqrt3 x+1 $.
I tried
Maximize[Sqrt[x^2 + 9] - Sqrt[x^2 - Sqrt[3]*x + 1], x]
and I got
[Sqrt](9 +
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]^2) - [Sqrt](1 -
Sqrt[3] Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2] +
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]^2), x ->
Root[-3 + #1^2 &,
243 - 567 #1 #2 + 1557 #2^2 - 702 #1 #2^3 + 485 #2^4 -
71 #1 #2^5 + 35 #2^6 &, 2, 2]
The exact answer is $sqrt7$.
How can I get the exact answer of the maximum of this function?
calculus-and-analysis mathematical-optimization
calculus-and-analysis mathematical-optimization
edited yesterday
ÃÂûÃÂþñýôÃÂÿàÃÂõóó
2,424725
2,424725
asked yesterday
minhthien_2016
469310
469310
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
First set the function definition
Clear[f]
f[x_] := Sqrt[x^2 + 9] - Sqrt[x^2 - Sqrt[3] x + 1]
Method I
The conventional method, by finding the stationary point first:
Solve[f'[x] == 0, x][[1]]
f[x] /. %
(f''[x] /. %%) < 0
x -> (3 Sqrt[3])/5
Sqrt[7]
True
Method II
By using ToRadicals
to transform Root
:
Maximize[f[x], x] // ToRadicals
Sqrt[7], x -> (3 Sqrt[3])/5
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
First set the function definition
Clear[f]
f[x_] := Sqrt[x^2 + 9] - Sqrt[x^2 - Sqrt[3] x + 1]
Method I
The conventional method, by finding the stationary point first:
Solve[f'[x] == 0, x][[1]]
f[x] /. %
(f''[x] /. %%) < 0
x -> (3 Sqrt[3])/5
Sqrt[7]
True
Method II
By using ToRadicals
to transform Root
:
Maximize[f[x], x] // ToRadicals
Sqrt[7], x -> (3 Sqrt[3])/5
add a comment |Â
up vote
7
down vote
accepted
First set the function definition
Clear[f]
f[x_] := Sqrt[x^2 + 9] - Sqrt[x^2 - Sqrt[3] x + 1]
Method I
The conventional method, by finding the stationary point first:
Solve[f'[x] == 0, x][[1]]
f[x] /. %
(f''[x] /. %%) < 0
x -> (3 Sqrt[3])/5
Sqrt[7]
True
Method II
By using ToRadicals
to transform Root
:
Maximize[f[x], x] // ToRadicals
Sqrt[7], x -> (3 Sqrt[3])/5
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
First set the function definition
Clear[f]
f[x_] := Sqrt[x^2 + 9] - Sqrt[x^2 - Sqrt[3] x + 1]
Method I
The conventional method, by finding the stationary point first:
Solve[f'[x] == 0, x][[1]]
f[x] /. %
(f''[x] /. %%) < 0
x -> (3 Sqrt[3])/5
Sqrt[7]
True
Method II
By using ToRadicals
to transform Root
:
Maximize[f[x], x] // ToRadicals
Sqrt[7], x -> (3 Sqrt[3])/5
First set the function definition
Clear[f]
f[x_] := Sqrt[x^2 + 9] - Sqrt[x^2 - Sqrt[3] x + 1]
Method I
The conventional method, by finding the stationary point first:
Solve[f'[x] == 0, x][[1]]
f[x] /. %
(f''[x] /. %%) < 0
x -> (3 Sqrt[3])/5
Sqrt[7]
True
Method II
By using ToRadicals
to transform Root
:
Maximize[f[x], x] // ToRadicals
Sqrt[7], x -> (3 Sqrt[3])/5
edited yesterday
answered yesterday
ÃÂûÃÂþñýôÃÂÿàÃÂõóó
2,424725
2,424725
add a comment |Â
add a comment |Â
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