Divide $T between X,Y,Z so that X may have twice as much as Z and $m more than three times Y's share.?

Question

Give answers in terms of T and m.

Answer 1

X = 2Z
Z = X/2
X = m + 3Y
Y = (X-m)/3

X + Y + Z = T = X + X/2 + (X-m)/3
6T = 6X + 3X + 2(X-m) = 11X - 2m
X = (6T+2m)/11
Y = 6T -9m/33 = (2T-3m)/11
Z = (6T-2m)/22 = (3T-m)/11

Answer 2

Edit: Morewood meant separated by using 2pi/3, i.e., the cube roots of a million. If imaginary values are allowed, ideas greater proper than 4 are attainable. as an social amassing, x=a million, y=-a million/2 + i sqrt(7)/2 z=-a million/2 - i sqrt(7)/2 grant |x| +|y| +|z| = a million + sqrt(2) + sqrt(2) = 3.828... greater many situations, the two numbers y and z fulfill the quadratic (T-y)(T-z)=0 T^2-(y+z)T+yz=0 T^2+xT+2/x=0, making use of x+y+z=0 and xyz=2. From the quadratic formula, y,z = -x/2 +/- sqrt(x^2-8/x)/2. So cut back over all complicated x |x| + |-x/2 + sqrt(x^2-8/x)/2| + |-x/2 - sqrt(x^2-8/x)/2|. I surely have not been waiting to boost in this, yet writing x=a+bi could desire to assist.