Given that x, y, z are real numbers, what's the minimum of √(x²-2xy+y²+4) +√(y²-2yz+z²+225) +√(x²+16) +√(z²-24z+340)?
Partial Derivatives worked like a charm! The minimum value is 37. Let me know if you need more details. It's bloody marvelous! I suspected it was a set-up from the beginning. Everything just falls into place - perfect squares, pyth. triplets, .... I love it when a plan falls into place. Made my day.
I was thinking along those same lines. BUT:
x = y = z and z = 12 forces x = 12 and y = 12.
= 2 + 15 + 4sqrt(10) + 14
=31 + 4sqrt(10)
x = y does not "cancel" out the first term but, instead, gives it a value of 2 - the minimum value for that term. So we can determine the minimum value of each term but it is not true that: sum(mins) = min(sums) - the mins for each term do not all occur at the same values of x,y,z.
So we are back to
(die)f/(die)x = (die)f/(die)y = (die)f/(die)z = 0
Taking x = y = z will cancel out the first 2 terms. The other terms can be minimized by taking z = 12.
This leads to the answer 4*sqrt(10) + 2*sqrt(23).
I will use pf/px to be the partial derivative of f wrt x.
In principle f(x,y,z) will have a stationary point (in this case a min because all positive) when pf/px, pf/py, pf/pz are all 0 giving you 3 equations in x,y,z. I tried this but got bogged down with a quartic, so there might be a better approach.
Maybe something like: sqrt(x^2-2xy+y^2+4) reaches a min value of 2 when x=y. etc...
Not sure this helps because sqrt(z^2-24z+340) reaches a min value of 14 when z=12.
There may be a better approach - but I don't see it yet.