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Solve function not solving simultaneous equations
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How do I solve a problem with 3 simultaneous equations? My code shown below is not giving the correct results.
I am attempting to find the maximum area (A) whilst both lengths x and y follow the following property: 2x + y = 960.
I have already looked at the documentation, and it seems that the format of my arguments is correct.
Solve[{2 x + y == 960, A == x*y, D[A] == 0}, {x, y}]
I am unsure of this, however it might be too complex for the Solve function to work, as it is getting the derivative of one of the variables (D[A]).
However I am able to do this question by hand:
Rearrange 1st equation so that y = 960 - 2x
Substitute y into 2nd equation so that A = x(960 - 2x) = 2x^2 + 960x
Get the derivative: 4x + 960 and solve for 4x + 960 = 0
x = 240
Substitute x = 240 into y = 960 - 2x
y = 960 - 2(240) = 960 - 480 = 480
Therefore dimensions are 240 x 480.
I expect the output to be {240, 480}. Thanks :)
EDIT: Here is what I have typed into mathematica:
Clear[x, y, A]
Solve[{2 x + y == 960, A == x*y, D[A, x] == 0}, {x, y}]
OUT: {{x -> 1/2 (480 - Sqrt[2] Sqrt[115200 - A]),
y -> 480 + Sqrt[2] Sqrt[115200 - A]}, {x ->
1/2 (480 + Sqrt[2] Sqrt[115200 - A]),
y -> 480 - Sqrt[2] Sqrt[115200 - A]}
NMaximize[{x*y, 2 x + y == 960}, {x, y}]
OUT: {115200., {x -> 240., y -> 480.}}
wolfram-mathematica
asked Jan 3 at 7:41
Jordan ZdimirovicJordan Zdimirovic
134
add a comment |
How do I solve a problem with 3 simultaneous equations? My code shown below is not giving the correct results.
I am attempting to find the maximum area (A) whilst both lengths x and y follow the following property: 2x + y = 960.
I have already looked at the documentation, and it seems that the format of my arguments is correct.
Solve[{2 x + y == 960, A == x*y, D[A] == 0}, {x, y}]
I am unsure of this, however it might be too complex for the Solve function to work, as it is getting the derivative of one of the variables (D[A]).
However I am able to do this question by hand:
Rearrange 1st equation so that y = 960 - 2x
Substitute y into 2nd equation so that A = x(960 - 2x) = 2x^2 + 960x
Get the derivative: 4x + 960 and solve for 4x + 960 = 0
x = 240
Substitute x = 240 into y = 960 - 2x
y = 960 - 2(240) = 960 - 480 = 480
Therefore dimensions are 240 x 480.
I expect the output to be {240, 480}. Thanks :)
EDIT: Here is what I have typed into mathematica:
Clear[x, y, A]
Solve[{2 x + y == 960, A == x*y, D[A, x] == 0}, {x, y}]
OUT: {{x -> 1/2 (480 - Sqrt[2] Sqrt[115200 - A]),
y -> 480 + Sqrt[2] Sqrt[115200 - A]}, {x ->
1/2 (480 + Sqrt[2] Sqrt[115200 - A]),
y -> 480 - Sqrt[2] Sqrt[115200 - A]}
NMaximize[{x*y, 2 x + y == 960}, {x, y}]
OUT: {115200., {x -> 240., y -> 480.}}
wolfram-mathematica
asked Jan 3 at 7:41
Jordan ZdimirovicJordan Zdimirovic
134
add a comment |
How do I solve a problem with 3 simultaneous equations? My code shown below is not giving the correct results.
I am attempting to find the maximum area (A) whilst both lengths x and y follow the following property: 2x + y = 960.
I have already looked at the documentation, and it seems that the format of my arguments is correct.
Solve[{2 x + y == 960, A == x*y, D[A] == 0}, {x, y}]
I am unsure of this, however it might be too complex for the Solve function to work, as it is getting the derivative of one of the variables (D[A]).
However I am able to do this question by hand:
Rearrange 1st equation so that y = 960 - 2x
Substitute y into 2nd equation so that A = x(960 - 2x) = 2x^2 + 960x
Get the derivative: 4x + 960 and solve for 4x + 960 = 0
x = 240
Substitute x = 240 into y = 960 - 2x
y = 960 - 2(240) = 960 - 480 = 480
Therefore dimensions are 240 x 480.
I expect the output to be {240, 480}. Thanks :)
EDIT: Here is what I have typed into mathematica:
Clear[x, y, A]
Solve[{2 x + y == 960, A == x*y, D[A, x] == 0}, {x, y}]
OUT: {{x -> 1/2 (480 - Sqrt[2] Sqrt[115200 - A]),
y -> 480 + Sqrt[2] Sqrt[115200 - A]}, {x ->
1/2 (480 + Sqrt[2] Sqrt[115200 - A]),
y -> 480 - Sqrt[2] Sqrt[115200 - A]}
NMaximize[{x*y, 2 x + y == 960}, {x, y}]
OUT: {115200., {x -> 240., y -> 480.}}
wolfram-mathematica
asked Jan 3 at 7:41
Jordan ZdimirovicJordan Zdimirovic
134
How do I solve a problem with 3 simultaneous equations? My code shown below is not giving the correct results.
I am attempting to find the maximum area (A) whilst both lengths x and y follow the following property: 2x + y = 960.
I have already looked at the documentation, and it seems that the format of my arguments is correct.
Solve[{2 x + y == 960, A == x*y, D[A] == 0}, {x, y}]
I am unsure of this, however it might be too complex for the Solve function to work, as it is getting the derivative of one of the variables (D[A]).
However I am able to do this question by hand:
Rearrange 1st equation so that y = 960 - 2x
Substitute y into 2nd equation so that A = x(960 - 2x) = 2x^2 + 960x
Get the derivative: 4x + 960 and solve for 4x + 960 = 0
x = 240
Substitute x = 240 into y = 960 - 2x
y = 960 - 2(240) = 960 - 480 = 480
Therefore dimensions are 240 x 480.
I expect the output to be {240, 480}. Thanks :)
EDIT: Here is what I have typed into mathematica:
Clear[x, y, A]
Solve[{2 x + y == 960, A == x*y, D[A, x] == 0}, {x, y}]
OUT: {{x -> 1/2 (480 - Sqrt[2] Sqrt[115200 - A]),
y -> 480 + Sqrt[2] Sqrt[115200 - A]}, {x ->
1/2 (480 + Sqrt[2] Sqrt[115200 - A]),
y -> 480 - Sqrt[2] Sqrt[115200 - A]}
NMaximize[{x*y, 2 x + y == 960}, {x, y}]
OUT: {115200., {x -> 240., y -> 480.}}
wolfram-mathematica
wolfram-mathematica
asked Jan 3 at 7:41
Jordan ZdimirovicJordan Zdimirovic
134
asked Jan 3 at 7:41
Jordan ZdimirovicJordan Zdimirovic
134
edited Jan 3 at 8:35
Jordan Zdimirovic
asked Jan 3 at 7:41
Jordan ZdimirovicJordan Zdimirovic
134
asked Jan 3 at 7:41
Jordan ZdimirovicJordan Zdimirovic
134
asked Jan 3 at 7:41
Jordan ZdimirovicJordan Zdimirovic
134
134
add a comment |
add a comment |
1 Answer
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Try this
NMaximize[{x*y,2x+y==960},{x,y}]
which is maximizing the area with your constraint expression and that instantly returns x->240, y->480
The difficulty you were having was with the use of D[A] when Mathematica needs to know what variable you are differentiating with respect to.
Perhaps something in this will help you understand what is happening with your derivative.
EDIT
Look at what Solve is going to be given:
Clear[x,y,A];
A == x*y;
D[A, x]
which gives 0. Why is that? You are taking the derivative of A with respect to x, but A has never been assigned any value, you have only declared that A and x*y are equal. Thus
Clear[x,y,A];
{2 x + y == 960, A == x*y, D[A, x] == 0}
is handing
{2*x + y == 960, A == x*y, True}
to Solve and that is perhaps less puzzling when Solve returns something with A in it.
When some function in Mathematica isn't giving you the result that you expect or that makes sense then checking exactly what is being given to that function as arguments is always a good first step.
There are always several ways of doing anything in Mathematica and some of those seem to make no sense at all
answered Jan 3 at 8:13
BillBill
2,396178
Thanks for that. I found it interesting however that even after fixing up the 'respect to' issue I got solutions that included A.
– Jordan Zdimirovic
Jan 3 at 8:19
Perhaps my edit will help explain a bit. If not then we can try to keep at it until it is clear
– Bill
Jan 3 at 17:40
add a comment |
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Try this
NMaximize[{x*y,2x+y==960},{x,y}]
which is maximizing the area with your constraint expression and that instantly returns x->240, y->480
The difficulty you were having was with the use of D[A] when Mathematica needs to know what variable you are differentiating with respect to.
Perhaps something in this will help you understand what is happening with your derivative.
EDIT
Look at what Solve is going to be given:
Clear[x,y,A];
A == x*y;
D[A, x]
which gives 0. Why is that? You are taking the derivative of A with respect to x, but A has never been assigned any value, you have only declared that A and x*y are equal. Thus
Clear[x,y,A];
{2 x + y == 960, A == x*y, D[A, x] == 0}
is handing
{2*x + y == 960, A == x*y, True}
to Solve and that is perhaps less puzzling when Solve returns something with A in it.
When some function in Mathematica isn't giving you the result that you expect or that makes sense then checking exactly what is being given to that function as arguments is always a good first step.
There are always several ways of doing anything in Mathematica and some of those seem to make no sense at all
answered Jan 3 at 8:13
BillBill
2,396178
Thanks for that. I found it interesting however that even after fixing up the 'respect to' issue I got solutions that included A.
– Jordan Zdimirovic
Jan 3 at 8:19
Perhaps my edit will help explain a bit. If not then we can try to keep at it until it is clear
– Bill
Jan 3 at 17:40
add a comment |
Try this
NMaximize[{x*y,2x+y==960},{x,y}]
which is maximizing the area with your constraint expression and that instantly returns x->240, y->480
The difficulty you were having was with the use of D[A] when Mathematica needs to know what variable you are differentiating with respect to.
Perhaps something in this will help you understand what is happening with your derivative.
EDIT
Look at what Solve is going to be given:
Clear[x,y,A];
A == x*y;
D[A, x]
which gives 0. Why is that? You are taking the derivative of A with respect to x, but A has never been assigned any value, you have only declared that A and x*y are equal. Thus
Clear[x,y,A];
{2 x + y == 960, A == x*y, D[A, x] == 0}
is handing
{2*x + y == 960, A == x*y, True}
to Solve and that is perhaps less puzzling when Solve returns something with A in it.
When some function in Mathematica isn't giving you the result that you expect or that makes sense then checking exactly what is being given to that function as arguments is always a good first step.
There are always several ways of doing anything in Mathematica and some of those seem to make no sense at all
answered Jan 3 at 8:13
BillBill
2,396178
Thanks for that. I found it interesting however that even after fixing up the 'respect to' issue I got solutions that included A.
– Jordan Zdimirovic
Jan 3 at 8:19
Perhaps my edit will help explain a bit. If not then we can try to keep at it until it is clear
– Bill
Jan 3 at 17:40
add a comment |
Try this
NMaximize[{x*y,2x+y==960},{x,y}]
which is maximizing the area with your constraint expression and that instantly returns x->240, y->480
The difficulty you were having was with the use of D[A] when Mathematica needs to know what variable you are differentiating with respect to.
Perhaps something in this will help you understand what is happening with your derivative.
EDIT
Look at what Solve is going to be given:
Clear[x,y,A];
A == x*y;
D[A, x]
which gives 0. Why is that? You are taking the derivative of A with respect to x, but A has never been assigned any value, you have only declared that A and x*y are equal. Thus
Clear[x,y,A];
{2 x + y == 960, A == x*y, D[A, x] == 0}
is handing
{2*x + y == 960, A == x*y, True}
to Solve and that is perhaps less puzzling when Solve returns something with A in it.
When some function in Mathematica isn't giving you the result that you expect or that makes sense then checking exactly what is being given to that function as arguments is always a good first step.
There are always several ways of doing anything in Mathematica and some of those seem to make no sense at all
answered Jan 3 at 8:13
BillBill
2,396178
Try this
NMaximize[{x*y,2x+y==960},{x,y}]
which is maximizing the area with your constraint expression and that instantly returns x->240, y->480
The difficulty you were having was with the use of D[A] when Mathematica needs to know what variable you are differentiating with respect to.
Perhaps something in this will help you understand what is happening with your derivative.
EDIT
Look at what Solve is going to be given:
Clear[x,y,A];
A == x*y;
D[A, x]
which gives 0. Why is that? You are taking the derivative of A with respect to x, but A has never been assigned any value, you have only declared that A and x*y are equal. Thus
Clear[x,y,A];
{2 x + y == 960, A == x*y, D[A, x] == 0}
is handing
{2*x + y == 960, A == x*y, True}
to Solve and that is perhaps less puzzling when Solve returns something with A in it.
When some function in Mathematica isn't giving you the result that you expect or that makes sense then checking exactly what is being given to that function as arguments is always a good first step.
There are always several ways of doing anything in Mathematica and some of those seem to make no sense at all
answered Jan 3 at 8:13
BillBill
2,396178
edited Jan 3 at 17:39
answered Jan 3 at 8:13
BillBill
2,396178
answered Jan 3 at 8:13
BillBill
2,396178
answered Jan 3 at 8:13
BillBill
2,396178
2,396178
Thanks for that. I found it interesting however that even after fixing up the 'respect to' issue I got solutions that included A.
– Jordan Zdimirovic
Jan 3 at 8:19
Perhaps my edit will help explain a bit. If not then we can try to keep at it until it is clear
– Bill
Jan 3 at 17:40
add a comment |
Thanks for that. I found it interesting however that even after fixing up the 'respect to' issue I got solutions that included A.
– Jordan Zdimirovic
Jan 3 at 8:19
Perhaps my edit will help explain a bit. If not then we can try to keep at it until it is clear
– Bill
Jan 3 at 17:40
Thanks for that. I found it interesting however that even after fixing up the 'respect to' issue I got solutions that included A.
– Jordan Zdimirovic
Jan 3 at 8:19
Thanks for that. I found it interesting however that even after fixing up the 'respect to' issue I got solutions that included A.
– Jordan Zdimirovic
Jan 3 at 8:19
Perhaps my edit will help explain a bit. If not then we can try to keep at it until it is clear
– Bill
Jan 3 at 17:40
Perhaps my edit will help explain a bit. If not then we can try to keep at it until it is clear
– Bill
Jan 3 at 17:40
add a comment |
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