Finding a bound maximum with multivariables

[makka]

I've been reading things for several days on multivariable calculus, but I just can't seem to be able to do this myself . Looking for any insight or complete solution I can get.

To put this in perspective about why everything is labelled as constants rather than using actual numbers, I'm doing this math for use in excel, so many of the constants are user inputs.

Please excuse any mathematical convention faux pas' .

f(x,y,z) = (a + bcx + cdy) * (1 + f(z - z[sup:3ggwggz6]2[/sup:3ggwggz6]))

Where:
x+y+z = r
0 < r <= 1
a, b, c, d, f = positive "user input" constants for the program

I want to find the maximum possible result of the function, and the corresponding values of x,y,z.

The maximum will never occur where x+y+z < r so the extrema will always be along the boundary condition (which is a cube shape).

I attempted to use the lagrange multiplier technique, but failed... I'm reasonably sure this is the technique that should solve my problem, I just can't seem to apply it properly.

---------------------------------------------

Note: For calculations sake, I'll input some actual values for the constants.
f(x,y,z) = (50000 + 50(800x) + 8(800y)) * (1+1.5(z - z[sup:3ggwggz6]2[/sup:3ggwggz6]))
f(x,y,z) = (50000 + 40000x + 6400y)) * (1+1.5(z - z[sup:3ggwggz6]2[/sup:3ggwggz6]))

Where:
x+y+z = 0.4

Any assistance would be greatly appreciated.

Cheers.

 

[Deleted member 4993]

Re: Finding a bounded maximum with multivariables

I've been reading things for several days on multivariable calculus, but I just can't seem to be able to do this myself . Looking for any insight or complete solution I can get.

To put this in perspective about why everything is labelled as constants rather than using actual numbers, I'm doing this math for use in excel, so many of the constants are user inputs.

Please excuse any mathematical convention faux pas' .

f(x,y,z) = (a + bcx + cdy) * (1+e(z-z^2))

Where:
x+y+z = r
0 < r <= 1<<< this is also a constant - "user input" - correct?
a, b, c, d, e = positive "user input" constants for the program <<< do not use "e" as user input constant in any program - that one is generally reserved for special constant( = 2.718281828)

I want to find the maximum possible result of the function, and the corresponding values of x,y,z.

The maximum will never occur where x+y+z < r so the extrema will always be along the boundary condition (which is a cube shape).

I attempted to use the lagrange multiplier technique, but failed... I'm reasonably sure this is the technique that should solve my problem, I just can't seem to apply it properly.

---------------------------------------------

Note: For calculations sake, I'll input some actual values for the constants.
f(x,y,z) = (50000 + 50(800x) + 8(800y)) * (1+1.5(z-z^2))
f(x,y,z) = (50000 + 40000x + 6400y)) * (1+1.5(z-z^2))

Where:
x+y+z = 0.4

Any assistance would be greatly appreciated.

Cheers.

Please show us the equations you obtained for optimization (df/dx = 0, etc.)

 

[makka]

Re: Finding a bounded maximum with multivariables

f(x,y,z) = (50000 + 40000x + 6400y)) * (1+1.5(z-z[sup:2rlsttkj]2[/sup:2rlsttkj]))
x+y+z = 0.4

Expanded Function f(x,y,z)
f(x,y,z) = -75000z[sup:2rlsttkj]2[/sup:2rlsttkj] - 60000xz[sup:2rlsttkj]2[/sup:2rlsttkj] - 9600yz[sup:2rlsttkj]2[/sup:2rlsttkj] + 75000z + 60000xz + 9600yz + 40000x + 6400y + 50000

Partial Derivatives
(?f/?z) = -150000z - 120000xz - 19200yz + 60000x + 9600y + 75000
(?f/?z) = -600(100x + 16y + 125)(2z - 1)

(?f/?y) = -9600z[sup:2rlsttkj]2[/sup:2rlsttkj] + 9600z + 6400
(?f/?y) = -3200(3z[sup:2rlsttkj]2[/sup:2rlsttkj] - 3z - 2)

(?f/?x) = -60000z[sup:2rlsttkj]2[/sup:2rlsttkj] + 60000z + 40000
(?f/?x) = -20000(3z[sup:2rlsttkj]2[/sup:2rlsttkj] - 3z - 2)

I don't really know if I'm on the right track with this, and if I am, I don't know what to do next... So yeah, pretty lost right now hehe.

 

[Deleted member 4993]

Re: Finding a bounded maximum with multivariables

f(x,y,z) = (50000 + 40000x + 6400y)) * (1+1.5(z-z[sup:194t64p0]2[/sup:194t64p0]))
x+y+z = 0.4

Expanded Function f(x,y,z)
f(x,y,z,?) = -75000z[sup:194t64p0]2[/sup:194t64p0] - 60000xz[sup:194t64p0]2[/sup:194t64p0] - 9600yz[sup:194t64p0]2[/sup:194t64p0] + 75000z + 60000xz + 9600yz + 40000x + 6400y + 50000 <<< Where did "lambda" go?

Partial Derivatives
(?f/?z) = -150000z - 120000xz - 19200yz + 60000x + 9600y + 75000
(?f/?z) = -600(100x + 16y + 125)(2z - 1)

(?f/?y) = -9600z[sup:194t64p0]2[/sup:194t64p0] + 9600z + 6400
(?f/?y) = -3200(3z[sup:194t64p0]2[/sup:194t64p0] - 3z - 2)

(?f/?x) = -60000z[sup:194t64p0]2[/sup:194t64p0] + 60000z + 40000
(?f/?x) = -20000(3z[sup:194t64p0]2[/sup:194t64p0] - 3z - 2)

I don't really know if I'm on the right track with this, and if I am, I don't know what to do next... So yeah, pretty lost right now hehe.

 

[makka]

Sorry, that was left over from an earlier attempt to use lagrange multipliers to solve it. I obviously didn't do it properly, so I decided I'd leave that out of the working I posted and didn't clean that part out of it.

 

[Deleted member 4993]

Re: Finding a bounded maximum with multivariables

f(x,y,z) = (50000 + 40000x + 6400y)) * (1+1.5(z-z[sup:3456bu4a]2[/sup:3456bu4a]))
x+y+z = 0.4

Expanded Function g(x,y,z,?)

g(x,y,z,?) = f(x,y,z) - ?(x+y+z-0.4)

then, for min/max

(?g/?x) = 0

(?g/?y) = 0

(?g/?z) = 0

(?g/??) = 0

The above will give you four equations and four unknowns (?,x,y,z) - solve by your favorite method.

-75000z[sup:3456bu4a]2[/sup:3456bu4a] - 60000xz[sup:3456bu4a]2[/sup:3456bu4a] - 9600yz[sup:3456bu4a]2[/sup:3456bu4a] + 75000z + 60000xz + 9600yz + 40000x + 6400y + 50000 <<< Where did "lambda" go?

Partial Derivatives
(?f/?z) = -150000z - 120000xz - 19200yz + 60000x + 9600y + 75000
(?f/?z) = -600(100x + 16y + 125)(2z - 1)

(?f/?y) = -9600z[sup:3456bu4a]2[/sup:3456bu4a] + 9600z + 6400
(?f/?y) = -3200(3z[sup:3456bu4a]2[/sup:3456bu4a] - 3z - 2)

(?f/?x) = -60000z[sup:3456bu4a]2[/sup:3456bu4a] + 60000z + 40000
(?f/?x) = -20000(3z[sup:3456bu4a]2[/sup:3456bu4a] - 3z - 2)

I don't really know if I'm on the right track with this, and if I am, I don't know what to do next... So yeah, pretty lost right now hehe.