Minimize function subject to constraint

Click For Summary
The discussion centers on minimizing the function f(x,y,z) = ax + by + cz under the constraint x + y = k. The original method presented involves using differential calculus, leading to the equation a - b = c dz/dx, but the author finds inconsistencies when deriving partial derivatives. A key point raised is that if a, b, and c are all non-zero, the function does not have a finite minimum, as it can approach negative infinity. The suggestion is made to use Lagrange multipliers for such problems, which could provide a more effective solution method. The overall conclusion is that the problem as stated may lack a proper solution due to its linear nature.
aaaa202
Messages
1,144
Reaction score
2
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?
 
Physics news on Phys.org
(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 ;)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

Replies
8
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
1K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K