Question 1086674
If x+y+z=7 and xy+yz+zx=11,then least and largest value of z are ?
<pre><font size=4><b>
Since z=7-(x+y)

z will be smallest when x+y is large as possible, and
z will be largest when x+y is small as possible.

So let x+y = s 
then y = s-x
and z = 7-s

z will be smallest when s is large as possible, and
z will be largest when s is small as possible.

We substitute in 

xy+yz+zx = 11

x(s-x)+(s-x)(7-s)+(7-s)x = 11

sx-x²+7s-s²-7x+sx+7x-sx = 11
-x²+sx+7s-s²= 11
-x²+sx+7s-s²-11 = 0
-x²+sx-s²+7s-11 = 0
 x²-sx+s²-7s+11 = 0

The discriminant 

{{{b^2-4ac=(-s)^2-4(1)(s^2-7s+11)=s^2-4s^2+28s-44=-3s^2+28s-44}}} 

must not be negative, so

{{{-3s^2+28s-44>=0}}}
{{{3s^2-28s+44<=0}}}
{{{(s-2)(3s-22)<=0}}}

We solve this inequality:

Critical numbers are s=2 and s=22/3, which are possible
values for s.

For the interval {{{(matrix(1,3,-infinity,",",2))}}}
we choose test value 0, the value is +44, so we do not
include that interval for s.

For the interval {{{(matrix(1,3,2,",",22/3))}}}
we choose test value 3, the value is -13, so we do 
include that interval for s.

For the interval {{{(matrix(1,3,22/3,",",infinity))}}}
we choose test value 8, the value is +12, so we do not
include that interval for s.

So {{{2<=s<=22/3}}}

z is smallest when s=22/3, and since z=7-s, the smallest
value of z is 7-22/3 = 21/3-22/3 = -1/3

z is largest when s=2, and since z=7-s, the largest
value of z is 7-2 = 5.

Edwin</pre></font></b>