Minimize function subject to constraint

aaaa202
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Suppose given a function of the form:

f(x,y,z) = ax + by + cz
with the constrain x+y=k

My book minimizes this function by a way I am not completely familiar with:

dF = adx + bdy + cdz 0

and since dy=-dx we can write:

dF = (a-b)dx + cdz = 0
=>
a-b = c dz/dx (1)

How I would minimize is simply plug y=k-x into the definition of f:

f(x,z) = (a-b)x + bk + cz

And take partial derivatives

df/dx = a-b + cdz/dx

df/dz = (a-b) dx/dz + c

And seting both equal to zero yields a system of equations which does not reduce to (1).
What is wrong?
 
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(1) should be
a-b+c dz/dx=0
all the equations are equivalent
 
What about the dx/dz term i get?
 
aaaa202 said:
Suppose given a function of the form:

f(x,y,z) = ax + by + cz
with the constrain x+y=k

My book minimizes this function by a way I am not completely familiar with:

dF = adx + bdy + cdz 0

and since dy=-dx we can write:

dF = (a-b)dx + cdz = 0
=>
a-b = c dz/dx (1)

How I would minimize is simply plug y=k-x into the definition of f:

f(x,z) = (a-b)x + bk + cz

And take partial derivatives

df/dx = a-b + cdz/dx

df/dz = (a-b) dx/dz + c

And seting both equal to zero yields a system of equations which does not reduce to (1).
What is wrong?

I hope you have left out some important information, because if we take your problem exactly as you have written it, it has no solution; that is, if a, b, c are all non-zero, you can find a sequence of ##x_n, y_n, z_n## giving ##x_n + y_n = k## for all ##n##, but ##ax_n + by_n+ cz_n \to -\infty ## as ##n \to \infty##. In other words, there is no finite minimum.

Are you sure you have stated the problem completely and exactly?
 
Last edited:
aaaa202 said:
How I would minimize is simply plug y=k-x into the definition of f:

f(x,z) = (a-b)x + bk + cz

Which has level sets at z = Ax + B for some suitable constants A and B. I.e. the function is linear and does not have a minimum.

I would generally tackle these problems using Lagrange multipliers by the way, if you haven't seen them check them out, they are cool ;)
 

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