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Maximizing the value of a two variable function along any curve
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I read that, of all the points on an origin-centered circle in the x-y plane, the function $z=ax+by$ is maximum (or minimum) at the point where $fracxy=fracab$
I think this is too specific. I thought of maximizing any function $z(x,y)$ along any (well behaved: continuous, differentiable, etc) curve $y=f(x)$.
What I thought of is:
Suppose we're currently at a point $(x_1,f(x_1))$. Now, what direction we'd have to take to get to this point's nearest neighbour point along the curve $y=f(x_1)$? Suppose we change $x_1$ by an infinitesimally small amount $dx$. Then change in $y$ would be $f'(x_1)dx$.
So, we'd have to change by the vector $dx hati+ f'(x_1)dx hatj$ to go to a neighbouring point along the curve.
The change in $z(x,y)$ as we move from $(x_1, f(x_1))$ to this neighbour point on the curve is given by its gradient dotted with the change vector ($dx hati+ f'(x_1)dx hatj$)
So, change in $z$ is:
$$left( fracpartial zpartial x hati + fracpartial z partial y hatjright)cdot left(dx hati+ f'(x_1)dx hatjright) $$
$$=frac partial zpartial x dx + fracpartial zpartial y f'(x_1)dx$$
If the function is maximum (or minimum) at $(x_1, f(x_1))$, then this change should be 0.
So, $$frac partial zpartial x dx + fracpartial zpartial y f'(x_1)dx=0$$
which gives
$$frac partial zpartial x + fracpartial zpartial y f'(x_1)=0$$
Have I got this condition right? The condition for the function to be maximum (or minimum) along the curve? In the special case that $y=f(x)$ is a circle and $z=ax+by$, my equation gives us the familiar condition $fracx_1f(x_1)=fracab$.
multivariable-calculus vector-analysis maxima-minima infinitesimals
asked Jul 17 at 7:25
Ryder Rude
354110
 |Â
show 5 more comments
up vote
1
down vote
favorite
I read that, of all the points on an origin-centered circle in the x-y plane, the function $z=ax+by$ is maximum (or minimum) at the point where $fracxy=fracab$
I think this is too specific. I thought of maximizing any function $z(x,y)$ along any (well behaved: continuous, differentiable, etc) curve $y=f(x)$.
What I thought of is:
Suppose we're currently at a point $(x_1,f(x_1))$. Now, what direction we'd have to take to get to this point's nearest neighbour point along the curve $y=f(x_1)$? Suppose we change $x_1$ by an infinitesimally small amount $dx$. Then change in $y$ would be $f'(x_1)dx$.
So, we'd have to change by the vector $dx hati+ f'(x_1)dx hatj$ to go to a neighbouring point along the curve.
The change in $z(x,y)$ as we move from $(x_1, f(x_1))$ to this neighbour point on the curve is given by its gradient dotted with the change vector ($dx hati+ f'(x_1)dx hatj$)
So, change in $z$ is:
$$left( fracpartial zpartial x hati + fracpartial z partial y hatjright)cdot left(dx hati+ f'(x_1)dx hatjright) $$
$$=frac partial zpartial x dx + fracpartial zpartial y f'(x_1)dx$$
If the function is maximum (or minimum) at $(x_1, f(x_1))$, then this change should be 0.
So, $$frac partial zpartial x dx + fracpartial zpartial y f'(x_1)dx=0$$
which gives
$$frac partial zpartial x + fracpartial zpartial y f'(x_1)=0$$
Have I got this condition right? The condition for the function to be maximum (or minimum) along the curve? In the special case that $y=f(x)$ is a circle and $z=ax+by$, my equation gives us the familiar condition $fracx_1f(x_1)=fracab$.
multivariable-calculus vector-analysis maxima-minima infinitesimals
asked Jul 17 at 7:25
Ryder Rude
354110
These types of problems are best handled using Lagrange multipliers.
– Oliver Jones
Jul 17 at 7:56
@OliverJones I googled it. It seems to be above my high-school math level. I'll try to understand it later. Is my method any good?
– Ryder Rude
Jul 17 at 8:06
You say that Lagrange multipliers is too advanced for you but you use terms like gradient and partial derivatives. I advise you to leave this until you've learned multivariable calculus.
– Oliver Jones
Jul 17 at 8:15
I should point out that a circle can't be written in the form $y=f(x)$.
– Oliver Jones
Jul 17 at 8:17
1
Your equation seems correct for this reason. You can substitute your constraint equation $y=f(x)$ into $z(x,y)$ to get a function of one variable $z(x,f(x))$. Differentiating this with respect to $x$ using the chain rule gives your equation.
– Oliver Jones
Jul 17 at 8:42
 |Â
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I read that, of all the points on an origin-centered circle in the x-y plane, the function $z=ax+by$ is maximum (or minimum) at the point where $fracxy=fracab$
I think this is too specific. I thought of maximizing any function $z(x,y)$ along any (well behaved: continuous, differentiable, etc) curve $y=f(x)$.
What I thought of is:
Suppose we're currently at a point $(x_1,f(x_1))$. Now, what direction we'd have to take to get to this point's nearest neighbour point along the curve $y=f(x_1)$? Suppose we change $x_1$ by an infinitesimally small amount $dx$. Then change in $y$ would be $f'(x_1)dx$.
So, we'd have to change by the vector $dx hati+ f'(x_1)dx hatj$ to go to a neighbouring point along the curve.
The change in $z(x,y)$ as we move from $(x_1, f(x_1))$ to this neighbour point on the curve is given by its gradient dotted with the change vector ($dx hati+ f'(x_1)dx hatj$)
So, change in $z$ is:
$$left( fracpartial zpartial x hati + fracpartial z partial y hatjright)cdot left(dx hati+ f'(x_1)dx hatjright) $$
$$=frac partial zpartial x dx + fracpartial zpartial y f'(x_1)dx$$
If the function is maximum (or minimum) at $(x_1, f(x_1))$, then this change should be 0.
So, $$frac partial zpartial x dx + fracpartial zpartial y f'(x_1)dx=0$$
which gives
$$frac partial zpartial x + fracpartial zpartial y f'(x_1)=0$$
Have I got this condition right? The condition for the function to be maximum (or minimum) along the curve? In the special case that $y=f(x)$ is a circle and $z=ax+by$, my equation gives us the familiar condition $fracx_1f(x_1)=fracab$.
multivariable-calculus vector-analysis maxima-minima infinitesimals
asked Jul 17 at 7:25
Ryder Rude
354110
I read that, of all the points on an origin-centered circle in the x-y plane, the function $z=ax+by$ is maximum (or minimum) at the point where $fracxy=fracab$
I think this is too specific. I thought of maximizing any function $z(x,y)$ along any (well behaved: continuous, differentiable, etc) curve $y=f(x)$.
What I thought of is:
Suppose we're currently at a point $(x_1,f(x_1))$. Now, what direction we'd have to take to get to this point's nearest neighbour point along the curve $y=f(x_1)$? Suppose we change $x_1$ by an infinitesimally small amount $dx$. Then change in $y$ would be $f'(x_1)dx$.
So, we'd have to change by the vector $dx hati+ f'(x_1)dx hatj$ to go to a neighbouring point along the curve.
The change in $z(x,y)$ as we move from $(x_1, f(x_1))$ to this neighbour point on the curve is given by its gradient dotted with the change vector ($dx hati+ f'(x_1)dx hatj$)
So, change in $z$ is:
$$left( fracpartial zpartial x hati + fracpartial z partial y hatjright)cdot left(dx hati+ f'(x_1)dx hatjright) $$
$$=frac partial zpartial x dx + fracpartial zpartial y f'(x_1)dx$$
If the function is maximum (or minimum) at $(x_1, f(x_1))$, then this change should be 0.
So, $$frac partial zpartial x dx + fracpartial zpartial y f'(x_1)dx=0$$
which gives
$$frac partial zpartial x + fracpartial zpartial y f'(x_1)=0$$
Have I got this condition right? The condition for the function to be maximum (or minimum) along the curve? In the special case that $y=f(x)$ is a circle and $z=ax+by$, my equation gives us the familiar condition $fracx_1f(x_1)=fracab$.
multivariable-calculus vector-analysis maxima-minima infinitesimals
asked Jul 17 at 7:25
Ryder Rude
354110
edited Jul 17 at 8:45
asked Jul 17 at 7:25
Ryder Rude
354110
asked Jul 17 at 7:25
Ryder Rude
354110
asked Jul 17 at 7:25
Ryder Rude
354110
354110
These types of problems are best handled using Lagrange multipliers.
– Oliver Jones
Jul 17 at 7:56
@OliverJones I googled it. It seems to be above my high-school math level. I'll try to understand it later. Is my method any good?
– Ryder Rude
Jul 17 at 8:06
You say that Lagrange multipliers is too advanced for you but you use terms like gradient and partial derivatives. I advise you to leave this until you've learned multivariable calculus.
– Oliver Jones
Jul 17 at 8:15
I should point out that a circle can't be written in the form $y=f(x)$.
– Oliver Jones
Jul 17 at 8:17
1
Your equation seems correct for this reason. You can substitute your constraint equation $y=f(x)$ into $z(x,y)$ to get a function of one variable $z(x,f(x))$. Differentiating this with respect to $x$ using the chain rule gives your equation.
– Oliver Jones
Jul 17 at 8:42
 |Â
show 5 more comments
These types of problems are best handled using Lagrange multipliers.
– Oliver Jones
Jul 17 at 7:56
@OliverJones I googled it. It seems to be above my high-school math level. I'll try to understand it later. Is my method any good?
– Ryder Rude
Jul 17 at 8:06
You say that Lagrange multipliers is too advanced for you but you use terms like gradient and partial derivatives. I advise you to leave this until you've learned multivariable calculus.
– Oliver Jones
Jul 17 at 8:15
I should point out that a circle can't be written in the form $y=f(x)$.
– Oliver Jones
Jul 17 at 8:17
1
Your equation seems correct for this reason. You can substitute your constraint equation $y=f(x)$ into $z(x,y)$ to get a function of one variable $z(x,f(x))$. Differentiating this with respect to $x$ using the chain rule gives your equation.
– Oliver Jones
Jul 17 at 8:42
These types of problems are best handled using Lagrange multipliers.
– Oliver Jones
Jul 17 at 7:56
These types of problems are best handled using Lagrange multipliers.
– Oliver Jones
Jul 17 at 7:56
@OliverJones I googled it. It seems to be above my high-school math level. I'll try to understand it later. Is my method any good?
– Ryder Rude
Jul 17 at 8:06
@OliverJones I googled it. It seems to be above my high-school math level. I'll try to understand it later. Is my method any good?
– Ryder Rude
Jul 17 at 8:06
You say that Lagrange multipliers is too advanced for you but you use terms like gradient and partial derivatives. I advise you to leave this until you've learned multivariable calculus.
– Oliver Jones
Jul 17 at 8:15
You say that Lagrange multipliers is too advanced for you but you use terms like gradient and partial derivatives. I advise you to leave this until you've learned multivariable calculus.
– Oliver Jones
Jul 17 at 8:15
I should point out that a circle can't be written in the form $y=f(x)$.
– Oliver Jones
Jul 17 at 8:17
I should point out that a circle can't be written in the form $y=f(x)$.
– Oliver Jones
Jul 17 at 8:17
1
1
Your equation seems correct for this reason. You can substitute your constraint equation $y=f(x)$ into $z(x,y)$ to get a function of one variable $z(x,f(x))$. Differentiating this with respect to $x$ using the chain rule gives your equation.
– Oliver Jones
Jul 17 at 8:42
Your equation seems correct for this reason. You can substitute your constraint equation $y=f(x)$ into $z(x,y)$ to get a function of one variable $z(x,f(x))$. Differentiating this with respect to $x$ using the chain rule gives your equation.
– Oliver Jones
Jul 17 at 8:42
 |Â
show 5 more comments
1 Answer
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If I understand correctly, your equation for the infinitesimal change in $z$ is essentially a differential taken for infinitesimal change in the direction $langle 1, f'(x_1) rangle$. And when you divide out the $dx$, the equation becomes something like a directional derivative of $z(x, y)$, only the "direction" $langle 1, f'(x_1) rangle$ may not be a unit vector. The magnitude of the direction does not matter though, if we are only considering points where the derivative is 0.
This sounds right to me. If a point were a maximum or minimum, the directional derivative cannot be nonzero, because that would imply the function increases immediately in some direction. So it must be 0 at the candidate extrema. Keep in mind that just because the derivative is 0 doesn't guarantee an actual minimum or maximum point; it could also be a saddle/inflection point.
As a side note, the original problem you're considering, maximizing/minimizing a function of two variables given a constraint that $y = f(x)$, sounds awfully like a Lagrange multipliers question, with the constraint being $0 = g(x, y) = f(x) - y$. That is another, more mainstream way of approaching the problem.
answered Jul 17 at 8:17
Craveable Banana
874
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If I understand correctly, your equation for the infinitesimal change in $z$ is essentially a differential taken for infinitesimal change in the direction $langle 1, f'(x_1) rangle$. And when you divide out the $dx$, the equation becomes something like a directional derivative of $z(x, y)$, only the "direction" $langle 1, f'(x_1) rangle$ may not be a unit vector. The magnitude of the direction does not matter though, if we are only considering points where the derivative is 0.
This sounds right to me. If a point were a maximum or minimum, the directional derivative cannot be nonzero, because that would imply the function increases immediately in some direction. So it must be 0 at the candidate extrema. Keep in mind that just because the derivative is 0 doesn't guarantee an actual minimum or maximum point; it could also be a saddle/inflection point.
As a side note, the original problem you're considering, maximizing/minimizing a function of two variables given a constraint that $y = f(x)$, sounds awfully like a Lagrange multipliers question, with the constraint being $0 = g(x, y) = f(x) - y$. That is another, more mainstream way of approaching the problem.
answered Jul 17 at 8:17
Craveable Banana
874
add a comment |Â
up vote
2
down vote
accepted
If I understand correctly, your equation for the infinitesimal change in $z$ is essentially a differential taken for infinitesimal change in the direction $langle 1, f'(x_1) rangle$. And when you divide out the $dx$, the equation becomes something like a directional derivative of $z(x, y)$, only the "direction" $langle 1, f'(x_1) rangle$ may not be a unit vector. The magnitude of the direction does not matter though, if we are only considering points where the derivative is 0.
This sounds right to me. If a point were a maximum or minimum, the directional derivative cannot be nonzero, because that would imply the function increases immediately in some direction. So it must be 0 at the candidate extrema. Keep in mind that just because the derivative is 0 doesn't guarantee an actual minimum or maximum point; it could also be a saddle/inflection point.
As a side note, the original problem you're considering, maximizing/minimizing a function of two variables given a constraint that $y = f(x)$, sounds awfully like a Lagrange multipliers question, with the constraint being $0 = g(x, y) = f(x) - y$. That is another, more mainstream way of approaching the problem.
answered Jul 17 at 8:17
Craveable Banana
874
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If I understand correctly, your equation for the infinitesimal change in $z$ is essentially a differential taken for infinitesimal change in the direction $langle 1, f'(x_1) rangle$. And when you divide out the $dx$, the equation becomes something like a directional derivative of $z(x, y)$, only the "direction" $langle 1, f'(x_1) rangle$ may not be a unit vector. The magnitude of the direction does not matter though, if we are only considering points where the derivative is 0.
This sounds right to me. If a point were a maximum or minimum, the directional derivative cannot be nonzero, because that would imply the function increases immediately in some direction. So it must be 0 at the candidate extrema. Keep in mind that just because the derivative is 0 doesn't guarantee an actual minimum or maximum point; it could also be a saddle/inflection point.
As a side note, the original problem you're considering, maximizing/minimizing a function of two variables given a constraint that $y = f(x)$, sounds awfully like a Lagrange multipliers question, with the constraint being $0 = g(x, y) = f(x) - y$. That is another, more mainstream way of approaching the problem.
answered Jul 17 at 8:17
Craveable Banana
874
If I understand correctly, your equation for the infinitesimal change in $z$ is essentially a differential taken for infinitesimal change in the direction $langle 1, f'(x_1) rangle$. And when you divide out the $dx$, the equation becomes something like a directional derivative of $z(x, y)$, only the "direction" $langle 1, f'(x_1) rangle$ may not be a unit vector. The magnitude of the direction does not matter though, if we are only considering points where the derivative is 0.
This sounds right to me. If a point were a maximum or minimum, the directional derivative cannot be nonzero, because that would imply the function increases immediately in some direction. So it must be 0 at the candidate extrema. Keep in mind that just because the derivative is 0 doesn't guarantee an actual minimum or maximum point; it could also be a saddle/inflection point.
As a side note, the original problem you're considering, maximizing/minimizing a function of two variables given a constraint that $y = f(x)$, sounds awfully like a Lagrange multipliers question, with the constraint being $0 = g(x, y) = f(x) - y$. That is another, more mainstream way of approaching the problem.
answered Jul 17 at 8:17
Craveable Banana
874
answered Jul 17 at 8:17
Craveable Banana
874
answered Jul 17 at 8:17
Craveable Banana
874
answered Jul 17 at 8:17
Craveable Banana
874
874
add a comment |Â
add a comment |Â
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These types of problems are best handled using Lagrange multipliers.
– Oliver Jones
Jul 17 at 7:56
@OliverJones I googled it. It seems to be above my high-school math level. I'll try to understand it later. Is my method any good?
– Ryder Rude
Jul 17 at 8:06
You say that Lagrange multipliers is too advanced for you but you use terms like gradient and partial derivatives. I advise you to leave this until you've learned multivariable calculus.
– Oliver Jones
Jul 17 at 8:15
I should point out that a circle can't be written in the form $y=f(x)$.
– Oliver Jones
Jul 17 at 8:17
1
Your equation seems correct for this reason. You can substitute your constraint equation $y=f(x)$ into $z(x,y)$ to get a function of one variable $z(x,f(x))$. Differentiating this with respect to $x$ using the chain rule gives your equation.
– Oliver Jones
Jul 17 at 8:42