Maximising function of $n$ variables
$begingroup$
I am considering the following function
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), ,$$
where $(x_1,dots,x_n,y)in[d,1]^{n+1}$, $d>0$ and $alpha$, $beta$, $gamma$, $delta$, $k_1$, $k_2>0$. Moreover, $x_ileq y$ $forall$ $i=1,dots,$ $n$. I'm trying to calculate the maximum of this function on that domain.
Befor using "brute force" approach (i.e. by calculating derivatives, Hessian and so on), I wonder whether it's possible to obtain the absolute maximum in a more simple way. For example, I notice that $fleq0$...
multivariable-calculus optimization maxima-minima
asked Dec 24 '18 at 14:22
MarkMark
3,45251947
$endgroup$
add a comment |
$begingroup$
I am considering the following function
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), ,$$
where $(x_1,dots,x_n,y)in[d,1]^{n+1}$, $d>0$ and $alpha$, $beta$, $gamma$, $delta$, $k_1$, $k_2>0$. Moreover, $x_ileq y$ $forall$ $i=1,dots,$ $n$. I'm trying to calculate the maximum of this function on that domain.
Befor using "brute force" approach (i.e. by calculating derivatives, Hessian and so on), I wonder whether it's possible to obtain the absolute maximum in a more simple way. For example, I notice that $fleq0$...
multivariable-calculus optimization maxima-minima
asked Dec 24 '18 at 14:22
MarkMark
3,45251947
$endgroup$
$begingroup$
without squaring that final term, this problem is not easy
$endgroup$
– LinAlg
Dec 24 '18 at 16:10
$begingroup$
Your "brute force" method is really quite easy. You should try it before wasting a lot of time on alternatives. First hold $y$ constant. The derivatives for the $x_i$ are independent of other terms, so you can find the $x_i$ in terms of $y$ without any need for a Hessian. Then maximize for $y$.
$endgroup$
– Paul Sinclair
Dec 25 '18 at 3:11
$begingroup$
@PaulSinclair, thank you. Then I should try to maximize $g(x_1,dots,x_n):= f(x_1,dots,x_n,bar{y})$, where $bar{y}$ is assumed as constant at moment. But why "without any need for Hessian"? I mean, how do I know that the point that makes to zero the partial derivatives of $g$ is exactly a maximum point for $g$, without Hessian?
$endgroup$
– Mark
Dec 28 '18 at 8:05
$begingroup$
I was talking about finding the maximum, not proving that it was a maximum. However, I was also confusing the Jacobian (which identifies extrema) with the Hessian (which classifies the extrema when found), so the statement is not correct. However, I re-iterate that the Hessian and Jacobian here are easy to calculate.
$endgroup$
– Paul Sinclair
Dec 29 '18 at 15:34
add a comment |
$begingroup$
I am considering the following function
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), ,$$
where $(x_1,dots,x_n,y)in[d,1]^{n+1}$, $d>0$ and $alpha$, $beta$, $gamma$, $delta$, $k_1$, $k_2>0$. Moreover, $x_ileq y$ $forall$ $i=1,dots,$ $n$. I'm trying to calculate the maximum of this function on that domain.
Befor using "brute force" approach (i.e. by calculating derivatives, Hessian and so on), I wonder whether it's possible to obtain the absolute maximum in a more simple way. For example, I notice that $fleq0$...
multivariable-calculus optimization maxima-minima
asked Dec 24 '18 at 14:22
MarkMark
3,45251947
$endgroup$
I am considering the following function
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), ,$$
where $(x_1,dots,x_n,y)in[d,1]^{n+1}$, $d>0$ and $alpha$, $beta$, $gamma$, $delta$, $k_1$, $k_2>0$. Moreover, $x_ileq y$ $forall$ $i=1,dots,$ $n$. I'm trying to calculate the maximum of this function on that domain.
Befor using "brute force" approach (i.e. by calculating derivatives, Hessian and so on), I wonder whether it's possible to obtain the absolute maximum in a more simple way. For example, I notice that $fleq0$...
multivariable-calculus optimization maxima-minima
multivariable-calculus optimization maxima-minima
asked Dec 24 '18 at 14:22
MarkMark
3,45251947
asked Dec 24 '18 at 14:22
MarkMark
3,45251947
asked Dec 24 '18 at 14:22
MarkMark
3,45251947
asked Dec 24 '18 at 14:22
MarkMark
3,45251947
asked Dec 24 '18 at 14:22
MarkMark
3,45251947
3,45251947
$begingroup$
without squaring that final term, this problem is not easy
$endgroup$
– LinAlg
Dec 24 '18 at 16:10
$begingroup$
Your "brute force" method is really quite easy. You should try it before wasting a lot of time on alternatives. First hold $y$ constant. The derivatives for the $x_i$ are independent of other terms, so you can find the $x_i$ in terms of $y$ without any need for a Hessian. Then maximize for $y$.
$endgroup$
– Paul Sinclair
Dec 25 '18 at 3:11
$begingroup$
@PaulSinclair, thank you. Then I should try to maximize $g(x_1,dots,x_n):= f(x_1,dots,x_n,bar{y})$, where $bar{y}$ is assumed as constant at moment. But why "without any need for Hessian"? I mean, how do I know that the point that makes to zero the partial derivatives of $g$ is exactly a maximum point for $g$, without Hessian?
$endgroup$
– Mark
Dec 28 '18 at 8:05
$begingroup$
I was talking about finding the maximum, not proving that it was a maximum. However, I was also confusing the Jacobian (which identifies extrema) with the Hessian (which classifies the extrema when found), so the statement is not correct. However, I re-iterate that the Hessian and Jacobian here are easy to calculate.
$endgroup$
– Paul Sinclair
Dec 29 '18 at 15:34
add a comment |
$begingroup$
without squaring that final term, this problem is not easy
$endgroup$
– LinAlg
Dec 24 '18 at 16:10
$begingroup$
Your "brute force" method is really quite easy. You should try it before wasting a lot of time on alternatives. First hold $y$ constant. The derivatives for the $x_i$ are independent of other terms, so you can find the $x_i$ in terms of $y$ without any need for a Hessian. Then maximize for $y$.
$endgroup$
– Paul Sinclair
Dec 25 '18 at 3:11
$begingroup$
@PaulSinclair, thank you. Then I should try to maximize $g(x_1,dots,x_n):= f(x_1,dots,x_n,bar{y})$, where $bar{y}$ is assumed as constant at moment. But why "without any need for Hessian"? I mean, how do I know that the point that makes to zero the partial derivatives of $g$ is exactly a maximum point for $g$, without Hessian?
$endgroup$
– Mark
Dec 28 '18 at 8:05
$begingroup$
I was talking about finding the maximum, not proving that it was a maximum. However, I was also confusing the Jacobian (which identifies extrema) with the Hessian (which classifies the extrema when found), so the statement is not correct. However, I re-iterate that the Hessian and Jacobian here are easy to calculate.
$endgroup$
– Paul Sinclair
Dec 29 '18 at 15:34
$begingroup$
without squaring that final term, this problem is not easy
$endgroup$
– LinAlg
Dec 24 '18 at 16:10
$begingroup$
without squaring that final term, this problem is not easy
$endgroup$
– LinAlg
Dec 24 '18 at 16:10
$begingroup$
Your "brute force" method is really quite easy. You should try it before wasting a lot of time on alternatives. First hold $y$ constant. The derivatives for the $x_i$ are independent of other terms, so you can find the $x_i$ in terms of $y$ without any need for a Hessian. Then maximize for $y$.
$endgroup$
– Paul Sinclair
Dec 25 '18 at 3:11
$begingroup$
Your "brute force" method is really quite easy. You should try it before wasting a lot of time on alternatives. First hold $y$ constant. The derivatives for the $x_i$ are independent of other terms, so you can find the $x_i$ in terms of $y$ without any need for a Hessian. Then maximize for $y$.
$endgroup$
– Paul Sinclair
Dec 25 '18 at 3:11
$begingroup$
@PaulSinclair, thank you. Then I should try to maximize $g(x_1,dots,x_n):= f(x_1,dots,x_n,bar{y})$, where $bar{y}$ is assumed as constant at moment. But why "without any need for Hessian"? I mean, how do I know that the point that makes to zero the partial derivatives of $g$ is exactly a maximum point for $g$, without Hessian?
$endgroup$
– Mark
Dec 28 '18 at 8:05
$begingroup$
@PaulSinclair, thank you. Then I should try to maximize $g(x_1,dots,x_n):= f(x_1,dots,x_n,bar{y})$, where $bar{y}$ is assumed as constant at moment. But why "without any need for Hessian"? I mean, how do I know that the point that makes to zero the partial derivatives of $g$ is exactly a maximum point for $g$, without Hessian?
$endgroup$
– Mark
Dec 28 '18 at 8:05
$begingroup$
I was talking about finding the maximum, not proving that it was a maximum. However, I was also confusing the Jacobian (which identifies extrema) with the Hessian (which classifies the extrema when found), so the statement is not correct. However, I re-iterate that the Hessian and Jacobian here are easy to calculate.
$endgroup$
– Paul Sinclair
Dec 29 '18 at 15:34
$begingroup$
I was talking about finding the maximum, not proving that it was a maximum. However, I was also confusing the Jacobian (which identifies extrema) with the Hessian (which classifies the extrema when found), so the statement is not correct. However, I re-iterate that the Hessian and Jacobian here are easy to calculate.
$endgroup$
– Paul Sinclair
Dec 29 '18 at 15:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Consider $y$ fixed as Paul Sinclair suggested. Your objective function is now separable: you can optimize each $x_i$ independently. It is also concave in $x_i$ (since the second derivative is negative), so it is maximized when the derivative is 0:
$$-2beta (k_2-x_i) - 2gamma(y-x_i) - frac{delta}{y-d} = 0$$
which can be written as
$$2(beta + gamma)x_i = 2beta k_2 + 2gamma y + frac{delta}{y-d}$$
so the solution is
$$x_i = frac{beta k_2 + gamma y}{beta + gamma} + frac{delta}{2(beta + gamma)(y-d)}$$
You can plug this in and maximize over just $y$. Since the last term in your objective is not squared, maximizing over $y$ is not simple: there are probably multiple local optima. You could perform grid search for $y$, or apply a gradient based optimization algorithm and try multiple starting points.
answered Dec 29 '18 at 19:37
LinAlgLinAlg
10k1521
$endgroup$
$begingroup$
Note that all the $x_i$ are the same at any stationary point. Plugging the expression for $x_i$ into $f$ gives a function of $y$ only that is a rational function of degree $2$. It's maxima are not hard to find.
$endgroup$
– Paul Sinclair
Dec 30 '18 at 18:42
add a comment |
$begingroup$
Are really the derivatives way the bruit force?
The derivates of
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), $$
are zeros in the stationary points,
$$begin{cases}
f''_{x_j}(x_1,dots,x_n,y)=2beta left(k_2-x_jright)+2gamma left(y-x_jright) - dfrac{delta}{y-d}, =0\
f'_y(x_1,dots,x_n,y)=-2alpha left(y-k_1right)-2gamma sumlimits_{i=1}^nleft(y-x_iright) - dfrac{delta}{y-d} sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
or, for $ynot=d,$
$$begin{cases}
2beta (y-d)left(k_2-x_jright)+2gamma (y-d) left(y-x_jright) - delta, =0, quad j=1dots n\
-2alpha (y-d)^2left(y-k_1right)-gamma (y-d)^2sumlimits_{i=1}^nleft(y-x_iright) - delta sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
so
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
-2(beta+gamma) (y-d)sumlimits_{i=1}^n x_i + 2n(y-d)(beta k_2+gamma y) - ndelta, =0\
(gamma-delta)(y-d)^2sumlimits_{i=1}^nx_i + (2alpha k_1 - gamma y)n(y-d)^2 +ndelta d, =0.
end{cases}$$
Summation of the second and third equations factors $2(gamma-delta)(y-d)$ and $(beta+gamma)$ gives
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
2(gamma-delta)(2n(y-d)(beta k_2+gamma y) - ndelta) + ((2alpha k_1 - gamma y)n(y-d)^2 +ndelta d)(beta-gamma), =0,
end{cases}$$
and this leads to the cubic equation for $y$ and explicit expressions for $x_j.$
answered Jan 4 at 17:59
Yuri NegometyanovYuri Negometyanov
11.9k1729
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
Consider $y$ fixed as Paul Sinclair suggested. Your objective function is now separable: you can optimize each $x_i$ independently. It is also concave in $x_i$ (since the second derivative is negative), so it is maximized when the derivative is 0:
$$-2beta (k_2-x_i) - 2gamma(y-x_i) - frac{delta}{y-d} = 0$$
which can be written as
$$2(beta + gamma)x_i = 2beta k_2 + 2gamma y + frac{delta}{y-d}$$
so the solution is
$$x_i = frac{beta k_2 + gamma y}{beta + gamma} + frac{delta}{2(beta + gamma)(y-d)}$$
You can plug this in and maximize over just $y$. Since the last term in your objective is not squared, maximizing over $y$ is not simple: there are probably multiple local optima. You could perform grid search for $y$, or apply a gradient based optimization algorithm and try multiple starting points.
answered Dec 29 '18 at 19:37
LinAlgLinAlg
10k1521
$endgroup$
$begingroup$
Note that all the $x_i$ are the same at any stationary point. Plugging the expression for $x_i$ into $f$ gives a function of $y$ only that is a rational function of degree $2$. It's maxima are not hard to find.
$endgroup$
– Paul Sinclair
Dec 30 '18 at 18:42
add a comment |
$begingroup$
Consider $y$ fixed as Paul Sinclair suggested. Your objective function is now separable: you can optimize each $x_i$ independently. It is also concave in $x_i$ (since the second derivative is negative), so it is maximized when the derivative is 0:
$$-2beta (k_2-x_i) - 2gamma(y-x_i) - frac{delta}{y-d} = 0$$
which can be written as
$$2(beta + gamma)x_i = 2beta k_2 + 2gamma y + frac{delta}{y-d}$$
so the solution is
$$x_i = frac{beta k_2 + gamma y}{beta + gamma} + frac{delta}{2(beta + gamma)(y-d)}$$
You can plug this in and maximize over just $y$. Since the last term in your objective is not squared, maximizing over $y$ is not simple: there are probably multiple local optima. You could perform grid search for $y$, or apply a gradient based optimization algorithm and try multiple starting points.
answered Dec 29 '18 at 19:37
LinAlgLinAlg
10k1521
$endgroup$
$begingroup$
Note that all the $x_i$ are the same at any stationary point. Plugging the expression for $x_i$ into $f$ gives a function of $y$ only that is a rational function of degree $2$. It's maxima are not hard to find.
$endgroup$
– Paul Sinclair
Dec 30 '18 at 18:42
add a comment |
$begingroup$
Consider $y$ fixed as Paul Sinclair suggested. Your objective function is now separable: you can optimize each $x_i$ independently. It is also concave in $x_i$ (since the second derivative is negative), so it is maximized when the derivative is 0:
$$-2beta (k_2-x_i) - 2gamma(y-x_i) - frac{delta}{y-d} = 0$$
which can be written as
$$2(beta + gamma)x_i = 2beta k_2 + 2gamma y + frac{delta}{y-d}$$
so the solution is
$$x_i = frac{beta k_2 + gamma y}{beta + gamma} + frac{delta}{2(beta + gamma)(y-d)}$$
You can plug this in and maximize over just $y$. Since the last term in your objective is not squared, maximizing over $y$ is not simple: there are probably multiple local optima. You could perform grid search for $y$, or apply a gradient based optimization algorithm and try multiple starting points.
answered Dec 29 '18 at 19:37
LinAlgLinAlg
10k1521
$endgroup$
Consider $y$ fixed as Paul Sinclair suggested. Your objective function is now separable: you can optimize each $x_i$ independently. It is also concave in $x_i$ (since the second derivative is negative), so it is maximized when the derivative is 0:
$$-2beta (k_2-x_i) - 2gamma(y-x_i) - frac{delta}{y-d} = 0$$
which can be written as
$$2(beta + gamma)x_i = 2beta k_2 + 2gamma y + frac{delta}{y-d}$$
so the solution is
$$x_i = frac{beta k_2 + gamma y}{beta + gamma} + frac{delta}{2(beta + gamma)(y-d)}$$
You can plug this in and maximize over just $y$. Since the last term in your objective is not squared, maximizing over $y$ is not simple: there are probably multiple local optima. You could perform grid search for $y$, or apply a gradient based optimization algorithm and try multiple starting points.
answered Dec 29 '18 at 19:37
LinAlgLinAlg
10k1521
answered Dec 29 '18 at 19:37
LinAlgLinAlg
10k1521
answered Dec 29 '18 at 19:37
LinAlgLinAlg
10k1521
answered Dec 29 '18 at 19:37
LinAlgLinAlg
10k1521
10k1521
$begingroup$
Note that all the $x_i$ are the same at any stationary point. Plugging the expression for $x_i$ into $f$ gives a function of $y$ only that is a rational function of degree $2$. It's maxima are not hard to find.
$endgroup$
– Paul Sinclair
Dec 30 '18 at 18:42
add a comment |
$begingroup$
Note that all the $x_i$ are the same at any stationary point. Plugging the expression for $x_i$ into $f$ gives a function of $y$ only that is a rational function of degree $2$. It's maxima are not hard to find.
$endgroup$
– Paul Sinclair
Dec 30 '18 at 18:42
$begingroup$
Note that all the $x_i$ are the same at any stationary point. Plugging the expression for $x_i$ into $f$ gives a function of $y$ only that is a rational function of degree $2$. It's maxima are not hard to find.
$endgroup$
– Paul Sinclair
Dec 30 '18 at 18:42
$begingroup$
Note that all the $x_i$ are the same at any stationary point. Plugging the expression for $x_i$ into $f$ gives a function of $y$ only that is a rational function of degree $2$. It's maxima are not hard to find.
$endgroup$
– Paul Sinclair
Dec 30 '18 at 18:42
add a comment |
$begingroup$
Are really the derivatives way the bruit force?
The derivates of
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), $$
are zeros in the stationary points,
$$begin{cases}
f''_{x_j}(x_1,dots,x_n,y)=2beta left(k_2-x_jright)+2gamma left(y-x_jright) - dfrac{delta}{y-d}, =0\
f'_y(x_1,dots,x_n,y)=-2alpha left(y-k_1right)-2gamma sumlimits_{i=1}^nleft(y-x_iright) - dfrac{delta}{y-d} sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
or, for $ynot=d,$
$$begin{cases}
2beta (y-d)left(k_2-x_jright)+2gamma (y-d) left(y-x_jright) - delta, =0, quad j=1dots n\
-2alpha (y-d)^2left(y-k_1right)-gamma (y-d)^2sumlimits_{i=1}^nleft(y-x_iright) - delta sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
so
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
-2(beta+gamma) (y-d)sumlimits_{i=1}^n x_i + 2n(y-d)(beta k_2+gamma y) - ndelta, =0\
(gamma-delta)(y-d)^2sumlimits_{i=1}^nx_i + (2alpha k_1 - gamma y)n(y-d)^2 +ndelta d, =0.
end{cases}$$
Summation of the second and third equations factors $2(gamma-delta)(y-d)$ and $(beta+gamma)$ gives
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
2(gamma-delta)(2n(y-d)(beta k_2+gamma y) - ndelta) + ((2alpha k_1 - gamma y)n(y-d)^2 +ndelta d)(beta-gamma), =0,
end{cases}$$
and this leads to the cubic equation for $y$ and explicit expressions for $x_j.$
answered Jan 4 at 17:59
Yuri NegometyanovYuri Negometyanov
11.9k1729
$endgroup$
add a comment |
$begingroup$
Are really the derivatives way the bruit force?
The derivates of
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), $$
are zeros in the stationary points,
$$begin{cases}
f''_{x_j}(x_1,dots,x_n,y)=2beta left(k_2-x_jright)+2gamma left(y-x_jright) - dfrac{delta}{y-d}, =0\
f'_y(x_1,dots,x_n,y)=-2alpha left(y-k_1right)-2gamma sumlimits_{i=1}^nleft(y-x_iright) - dfrac{delta}{y-d} sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
or, for $ynot=d,$
$$begin{cases}
2beta (y-d)left(k_2-x_jright)+2gamma (y-d) left(y-x_jright) - delta, =0, quad j=1dots n\
-2alpha (y-d)^2left(y-k_1right)-gamma (y-d)^2sumlimits_{i=1}^nleft(y-x_iright) - delta sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
so
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
-2(beta+gamma) (y-d)sumlimits_{i=1}^n x_i + 2n(y-d)(beta k_2+gamma y) - ndelta, =0\
(gamma-delta)(y-d)^2sumlimits_{i=1}^nx_i + (2alpha k_1 - gamma y)n(y-d)^2 +ndelta d, =0.
end{cases}$$
Summation of the second and third equations factors $2(gamma-delta)(y-d)$ and $(beta+gamma)$ gives
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
2(gamma-delta)(2n(y-d)(beta k_2+gamma y) - ndelta) + ((2alpha k_1 - gamma y)n(y-d)^2 +ndelta d)(beta-gamma), =0,
end{cases}$$
and this leads to the cubic equation for $y$ and explicit expressions for $x_j.$
answered Jan 4 at 17:59
Yuri NegometyanovYuri Negometyanov
11.9k1729
$endgroup$
add a comment |
$begingroup$
Are really the derivatives way the bruit force?
The derivates of
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), $$
are zeros in the stationary points,
$$begin{cases}
f''_{x_j}(x_1,dots,x_n,y)=2beta left(k_2-x_jright)+2gamma left(y-x_jright) - dfrac{delta}{y-d}, =0\
f'_y(x_1,dots,x_n,y)=-2alpha left(y-k_1right)-2gamma sumlimits_{i=1}^nleft(y-x_iright) - dfrac{delta}{y-d} sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
or, for $ynot=d,$
$$begin{cases}
2beta (y-d)left(k_2-x_jright)+2gamma (y-d) left(y-x_jright) - delta, =0, quad j=1dots n\
-2alpha (y-d)^2left(y-k_1right)-gamma (y-d)^2sumlimits_{i=1}^nleft(y-x_iright) - delta sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
so
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
-2(beta+gamma) (y-d)sumlimits_{i=1}^n x_i + 2n(y-d)(beta k_2+gamma y) - ndelta, =0\
(gamma-delta)(y-d)^2sumlimits_{i=1}^nx_i + (2alpha k_1 - gamma y)n(y-d)^2 +ndelta d, =0.
end{cases}$$
Summation of the second and third equations factors $2(gamma-delta)(y-d)$ and $(beta+gamma)$ gives
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
2(gamma-delta)(2n(y-d)(beta k_2+gamma y) - ndelta) + ((2alpha k_1 - gamma y)n(y-d)^2 +ndelta d)(beta-gamma), =0,
end{cases}$$
and this leads to the cubic equation for $y$ and explicit expressions for $x_j.$
answered Jan 4 at 17:59
Yuri NegometyanovYuri Negometyanov
11.9k1729
$endgroup$
Are really the derivatives way the bruit force?
The derivates of
$$f(x_1,dots,x_n,y)=-alpha left(y-k_1right)^2-beta sum_{i=1}^nleft(k_2-x_iright)^2-gamma sum_{i=1}^nleft(y-x_iright)^2 - frac{delta}{y-d} sum_{i=1}^n (x_i-d), $$
are zeros in the stationary points,
$$begin{cases}
f''_{x_j}(x_1,dots,x_n,y)=2beta left(k_2-x_jright)+2gamma left(y-x_jright) - dfrac{delta}{y-d}, =0\
f'_y(x_1,dots,x_n,y)=-2alpha left(y-k_1right)-2gamma sumlimits_{i=1}^nleft(y-x_iright) - dfrac{delta}{y-d} sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
or, for $ynot=d,$
$$begin{cases}
2beta (y-d)left(k_2-x_jright)+2gamma (y-d) left(y-x_jright) - delta, =0, quad j=1dots n\
-2alpha (y-d)^2left(y-k_1right)-gamma (y-d)^2sumlimits_{i=1}^nleft(y-x_iright) - delta sumlimits_{i=1}^n (x_i-d), =0,
end{cases}$$
so
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
-2(beta+gamma) (y-d)sumlimits_{i=1}^n x_i + 2n(y-d)(beta k_2+gamma y) - ndelta, =0\
(gamma-delta)(y-d)^2sumlimits_{i=1}^nx_i + (2alpha k_1 - gamma y)n(y-d)^2 +ndelta d, =0.
end{cases}$$
Summation of the second and third equations factors $2(gamma-delta)(y-d)$ and $(beta+gamma)$ gives
$$begin{cases}
-2(beta+gamma) (y-d)x_j + 2(y-d)(beta k_2+gamma y) - delta, =0, quad j=1dots n\
2(gamma-delta)(2n(y-d)(beta k_2+gamma y) - ndelta) + ((2alpha k_1 - gamma y)n(y-d)^2 +ndelta d)(beta-gamma), =0,
end{cases}$$
and this leads to the cubic equation for $y$ and explicit expressions for $x_j.$
answered Jan 4 at 17:59
Yuri NegometyanovYuri Negometyanov
11.9k1729
answered Jan 4 at 17:59
Yuri NegometyanovYuri Negometyanov
11.9k1729
answered Jan 4 at 17:59
Yuri NegometyanovYuri Negometyanov
11.9k1729
answered Jan 4 at 17:59
Yuri NegometyanovYuri Negometyanov
11.9k1729
11.9k1729
add a comment |
add a comment |
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$begingroup$
without squaring that final term, this problem is not easy
$endgroup$
– LinAlg
Dec 24 '18 at 16:10
$begingroup$
Your "brute force" method is really quite easy. You should try it before wasting a lot of time on alternatives. First hold $y$ constant. The derivatives for the $x_i$ are independent of other terms, so you can find the $x_i$ in terms of $y$ without any need for a Hessian. Then maximize for $y$.
$endgroup$
– Paul Sinclair
Dec 25 '18 at 3:11
$begingroup$
@PaulSinclair, thank you. Then I should try to maximize $g(x_1,dots,x_n):= f(x_1,dots,x_n,bar{y})$, where $bar{y}$ is assumed as constant at moment. But why "without any need for Hessian"? I mean, how do I know that the point that makes to zero the partial derivatives of $g$ is exactly a maximum point for $g$, without Hessian?
$endgroup$
– Mark
Dec 28 '18 at 8:05
$begingroup$
I was talking about finding the maximum, not proving that it was a maximum. However, I was also confusing the Jacobian (which identifies extrema) with the Hessian (which classifies the extrema when found), so the statement is not correct. However, I re-iterate that the Hessian and Jacobian here are easy to calculate.
$endgroup$
– Paul Sinclair
Dec 29 '18 at 15:34