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.
The expression is of the form f(x,y,z). I will use pf/px for the partial of f wrt x. Same for y and z. For a stationary point pf/px, pf/py and pf/pz must all =0. It will be a minimum value since all the terms are positive and f does not have an upper bound.
1) Find pf/px. This will be a radical expression containing only x and y. Set = 0, re-arrange, square both sides and this will simplify to: (3x-2y)(x-2y) = 0. x = 2y will not verify as the LHS and the RHS will be negatives of each other. So we have x = 2y/3
2) Find pf/py. This will be a radical expression in x,y,z. So substitute x = 2y/3 and pf/py will now be a radical expression in only y and z. Set = 0, re-arrange, square both sides and simplify to (7y-2z)(3y+2z) = 0. y = -2z/3 will not verify (see 1) so we have y = 2z/7
3) Find pf/pz. This will be a radical expression in y and z. Substitute y=2z/7 and pf/pz now becomes a radical expression in z only. Set = 0, re-arrange, square both sides and solve for z.
This gets you (5z-36)(z-36) = 0. Again z = 36 will not verify (see 1) so z=36/5 = 252/35.
4) y = 2z/7 = (2/7)(36/5) . y = 72/35.
5) x = 2y/3 = (2/3)(72/35). x = 48/35.
6) Substitute x= 48/35, y=72/35 and z=252/35 back into f(x,y,z) to get the minimum value.
f(48/35,72/35,252/35)
=74/35 + 555/35 + 148/35 + 518/35
=1295/35
=37
I did not post all the gory details (I can take a photo if you need it) but try the algebra yourself. It's long and tedious but really only at a high school level. Let me know how you make out - Ian
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.
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.
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.
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 was thinking along those same lines. BUT:
x = y = z and z = 12 forces x = 12 and y = 12.
So f(12,12,12)
= 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
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.