Lagrange Multipliers with 4 Variables
$begingroup$
I was wondering if there is a way to maximize the function $f(a,b,c,d) = frac{1}{2}(ac + bd)$ given the following constraints: $C_1 (a,b,c,d) := a^2 + c^2 = b^2 + d^2, C_2 (a,c) := a+c = 12,$ and $C_3(b,d) := b+d = 13$. Also, all $a,b,c,d > 0$ strictly.
I'm familiar with using Lagrange Multipliers for two, or even three variables, but is it possible to use that technique on this function and with those constraints? If not, what would is a feasible optimization strategy in this problem?
calculus optimization
asked Dec 22 '18 at 3:03
OmicronGammaOmicronGamma
336
$endgroup$
locked by quid♦ Dec 22 '18 at 15:20
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
Read more about locked posts here.
comments disabled on deleted / locked posts / reviews |
$begingroup$
I was wondering if there is a way to maximize the function $f(a,b,c,d) = frac{1}{2}(ac + bd)$ given the following constraints: $C_1 (a,b,c,d) := a^2 + c^2 = b^2 + d^2, C_2 (a,c) := a+c = 12,$ and $C_3(b,d) := b+d = 13$. Also, all $a,b,c,d > 0$ strictly.
I'm familiar with using Lagrange Multipliers for two, or even three variables, but is it possible to use that technique on this function and with those constraints? If not, what would is a feasible optimization strategy in this problem?
calculus optimization
asked Dec 22 '18 at 3:03
OmicronGammaOmicronGamma
336
$endgroup$
locked by quid♦ Dec 22 '18 at 15:20
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
Read more about locked posts here.
$begingroup$
Alright, so I substituted $c = 12-a$ and $d = 13-b$ to get a new constraining function $2a^2 - 24a - 2b^2 + 26b - 25 = 0$. But When I use Lagrange Multipliers on the original $f(a,b,c,d)$ and this new constraining function, I can't solve the resulting system of equations...
$endgroup$
– OmicronGamma
Dec 22 '18 at 3:44
4
$begingroup$
This is question 3 of the current USAMTS round. As this contest is open through 1/2/2019, please do not answer this question until then. $$ $$ @OmicronGamma , please delete this question immediately.
$endgroup$
– Carl Schildkraut
Dec 22 '18 at 4:37
comments disabled on deleted / locked posts / reviews |
$begingroup$
I was wondering if there is a way to maximize the function $f(a,b,c,d) = frac{1}{2}(ac + bd)$ given the following constraints: $C_1 (a,b,c,d) := a^2 + c^2 = b^2 + d^2, C_2 (a,c) := a+c = 12,$ and $C_3(b,d) := b+d = 13$. Also, all $a,b,c,d > 0$ strictly.
I'm familiar with using Lagrange Multipliers for two, or even three variables, but is it possible to use that technique on this function and with those constraints? If not, what would is a feasible optimization strategy in this problem?
calculus optimization
asked Dec 22 '18 at 3:03
OmicronGammaOmicronGamma
336
$endgroup$
I was wondering if there is a way to maximize the function $f(a,b,c,d) = frac{1}{2}(ac + bd)$ given the following constraints: $C_1 (a,b,c,d) := a^2 + c^2 = b^2 + d^2, C_2 (a,c) := a+c = 12,$ and $C_3(b,d) := b+d = 13$. Also, all $a,b,c,d > 0$ strictly.
I'm familiar with using Lagrange Multipliers for two, or even three variables, but is it possible to use that technique on this function and with those constraints? If not, what would is a feasible optimization strategy in this problem?
calculus optimization
calculus optimization
asked Dec 22 '18 at 3:03
OmicronGammaOmicronGamma
336
asked Dec 22 '18 at 3:03
OmicronGammaOmicronGamma
336
asked Dec 22 '18 at 3:03
OmicronGammaOmicronGamma
336
asked Dec 22 '18 at 3:03
OmicronGammaOmicronGamma
336
asked Dec 22 '18 at 3:03
OmicronGammaOmicronGamma
336
336
locked by quid♦ Dec 22 '18 at 15:20
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
Read more about locked posts here.
locked by quid♦ Dec 22 '18 at 15:20
This question is locked in view of our policy about contest questions. Questions originating from active contests are locked for the duration of the contest, with answers hidden from view by soft-deletion. Please see the comments below for references to the originating contest.
Read more about locked posts here.
$begingroup$
Alright, so I substituted $c = 12-a$ and $d = 13-b$ to get a new constraining function $2a^2 - 24a - 2b^2 + 26b - 25 = 0$. But When I use Lagrange Multipliers on the original $f(a,b,c,d)$ and this new constraining function, I can't solve the resulting system of equations...
$endgroup$
– OmicronGamma
Dec 22 '18 at 3:44
4
$begingroup$
This is question 3 of the current USAMTS round. As this contest is open through 1/2/2019, please do not answer this question until then. $$ $$ @OmicronGamma , please delete this question immediately.
$endgroup$
– Carl Schildkraut
Dec 22 '18 at 4:37
comments disabled on deleted / locked posts / reviews |
$begingroup$
Alright, so I substituted $c = 12-a$ and $d = 13-b$ to get a new constraining function $2a^2 - 24a - 2b^2 + 26b - 25 = 0$. But When I use Lagrange Multipliers on the original $f(a,b,c,d)$ and this new constraining function, I can't solve the resulting system of equations...
$endgroup$
– OmicronGamma
Dec 22 '18 at 3:44
4
$begingroup$
This is question 3 of the current USAMTS round. As this contest is open through 1/2/2019, please do not answer this question until then. $$ $$ @OmicronGamma , please delete this question immediately.
$endgroup$
– Carl Schildkraut
Dec 22 '18 at 4:37
$begingroup$
Alright, so I substituted $c = 12-a$ and $d = 13-b$ to get a new constraining function $2a^2 - 24a - 2b^2 + 26b - 25 = 0$. But When I use Lagrange Multipliers on the original $f(a,b,c,d)$ and this new constraining function, I can't solve the resulting system of equations...
$endgroup$
– OmicronGamma
Dec 22 '18 at 3:44
$begingroup$
Alright, so I substituted $c = 12-a$ and $d = 13-b$ to get a new constraining function $2a^2 - 24a - 2b^2 + 26b - 25 = 0$. But When I use Lagrange Multipliers on the original $f(a,b,c,d)$ and this new constraining function, I can't solve the resulting system of equations...
$endgroup$
– OmicronGamma
Dec 22 '18 at 3:44
4
4
$begingroup$
This is question 3 of the current USAMTS round. As this contest is open through 1/2/2019, please do not answer this question until then. $$ $$ @OmicronGamma , please delete this question immediately.
$endgroup$
– Carl Schildkraut
Dec 22 '18 at 4:37
$begingroup$
This is question 3 of the current USAMTS round. As this contest is open through 1/2/2019, please do not answer this question until then. $$ $$ @OmicronGamma , please delete this question immediately.
$endgroup$
– Carl Schildkraut
Dec 22 '18 at 4:37
comments disabled on deleted / locked posts / reviews |
0
active
oldest
votes
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Alright, so I substituted $c = 12-a$ and $d = 13-b$ to get a new constraining function $2a^2 - 24a - 2b^2 + 26b - 25 = 0$. But When I use Lagrange Multipliers on the original $f(a,b,c,d)$ and this new constraining function, I can't solve the resulting system of equations...
$endgroup$
– OmicronGamma
Dec 22 '18 at 3:44
4
$begingroup$
This is question 3 of the current USAMTS round. As this contest is open through 1/2/2019, please do not answer this question until then. $$ $$ @OmicronGamma , please delete this question immediately.
$endgroup$
– Carl Schildkraut
Dec 22 '18 at 4:37