Where do I start with this?
#vwxyz=v+w+x+y+z#
Find the maximum possible value of #x#
where #v, w, x, y, z > 0#
Find the maximum possible value of
where
2 Answers
Maximum for natural numbers is
Explanation:
Given:
#vwxyz = v+w+x+y+z#
For solutions in natural numbers, we can look at possible combinations of small values.
Note that:
-
If
#v=w=x=y=z=1# then#vwxyz=1 < 5 =v+w+x+y+z# -
If
#v=w=x=y=z=2# then#vwxyz=32 > 10 = v+w+x+y+z#
The minimum possible values of
#x = x+4" "# which has no solutions.
Considering the case where exactly one of
#2x = x+5" "=>" "color(blue)(x = 5)#
Giving
Considering the case where exactly two of
#4x = x+6" "=>" "color(blue)(x = 2)#
Giving
Considering the case where exactly three of
#8x = x+7" "=>" "color(blue)(x=1)#
Giving
Footnote
If
#v^4x = x+4v" "=>" "x = (4v)/(v^4-1)#
Note that:
#lim_(v->1+) (4v)/(v^4-1) = +oo#
So
Maximum value
Explanation:
Another approach...
Given:
#vwxyz = v+w+x+y+z" "# with#v, w, x, y, z > 0#
Looking for solutions in positive integers.
Note that if
#x = x+4" "# which has no solutions.
So at least one of
Considering the given equation as expressing
#((wyz)v - 1)x = v+(w+y+z)#
So:
#x = (v+(w+y+z))/((wyz)v-1)#
#color(white)(x) = ((wyz)v-1+1+(wyz)(w+y+z))/((wyz)((wyz)v-1))#
#color(white)(x) = 1/(wyz)+(1+(wyz)(w+y+z))/((wyz)((wyz)v-1))#
So:
#(del x)/(del v) = -(1+(wyz)(w+y+z))/((wyz)v-1)^2#
Note that if
So increasing the value of
Similarly for
Hence the maximum value for
#2x = x+5" "=>" "color(blue)(x=5)#