Question 977299
If x and y are two positive real nos. such that their sum is 1, then find the maximum value of {{{xy^4 + yx^4}}}
(Please solve using basic algebra and logic and NOT using calculus)
<pre>
{{{let}}}{{{z=xy^4 + yx^4}}}
{{{z=xy(y^3+x^3)}}}
{{{z=xy(y+x)(y^2-xy+x^2)}}}
{{{z=xy(y+x)(y^2-xy+x^2)}}}

Since x+y=1, substitute 1-x for y

{{{z=x(1-x)((1-x)+x)((1-x)^2-x(1-x)+x^2)}}}
{{{z=x(1-x)(1)(1-2x+x^2-x+x^2+x^2)}}}
{{{z=x(1-x)(1-3x+3x^2)}}}
{{{z=(x-x^2)(1-3x+3x^2)}}}
{{{z=x-3x^2+3x^3-x^2+3x^3-3x^4}}}
{{{z=-3x^4+6x^3-4x^2+x}}}

If we can find A, B and nonnegative C such that

{{{z=-3(x^2+Ax+B)^2+C}}}

then it's easy to see that z will have its maximum value 
when the quadratic {{{x^2+Ax+B}}} equals 0, and that maximum
value will be C. 

Let's see if we can find such A, B, and C by equating
coefficients:

{{{-3x^4+6x^3-4x^2+x = -3(x^2+Ax+B)^2+C}}}
{{{-3x^4+6x^3-4x^2+x = -3(x^4+A^2x^2+B^2+2Ax^3+2Bx^2+2ABx)+C}}}
{{{-3x^4+6x^3-4x^2+x = -3x^4-3A^2x^2-3B^2-6Ax^3-6Bx^2-6ABx+C}}}

Equating coefficients of x³:  {{{6 = -6A}}}, or {{{A = -1}}}

Equating coefficients of x²:  {{{-4 = -3A^2-6B}}}
                              {{{-4 = -3(-1)^2-6B}}}
                              {{{-4 = -3(1)-6B}}}
                              {{{-4 = -3-6B}}}
                              {{{-1 = -6B}}}
                              {{{1/6 = B}}}

Equating coefficients of x:  {{{1 = -6AB}}}
                             {{{1 = -6(-1)(1/6)}}}
                             {{{1 = 1}}}  

                              Already met.


Equating constant terms:      {{{0 = -3B^2+C}}}
                              {{{0 = -3(1/6)^2+C}}}
                              {{{0 = -3(1/36)+C}}}
                              {{{0 = -3/36+C}}}
                              {{{3/36=C}}}
                              {{{1/12=C}}}

That's the answer {{{C=1/12}}}

z will take on that maximum value 1/12 when the quadratic {{{x^2+Ax+B}}}
takes on the value 0.  

{{{x^2+Ax+B = 0}}}
{{{x^2-1x+1/6 = 0}}}
{{{x = (-(-1) +- sqrt((-1)^2-4*(1)*(1/6) ))/(2*(1)) }}}
{{{x = (1 +- sqrt(1-2/3))/2 }}}
{{{x = (1 +- sqrt(1/3))/2 }}}
{{{x = (1 +- sqrt(3)/3)/2 }}}
Multiply top and bottom by 3

{{{x = (3 +- sqrt(3))/6}}}

So there are two points where z reaches the maximum value of {{{1/12}}}

They are {{{( matrix(1,3,(3 +- sqrt(3))/6,",",1/12))}}}

Edwin</pre>