Long maximum

$\begin{aligned} z & = x^4y + x^3y+x^2y+xy+xy^2+xy^3+xy^4 \\ & = xy \left(x^3+x^2+x+1+y+y^2+y^3 \right) \\ & = xy \left({\color{#3D99F6} x^3 + y^3}+{\color{#D61F06}x^2+y^2}+x+y+1 \right) & \small \color{#3D99F6} \text{See notes.} \\ & = xy \left({\color{#3D99F6}27-9xy}+{\color{#D61F06}9-2xy}+3+1 \right) \\ & = xy \left(40-11xy \right) \\ & = - 11 \left((xy)^2 - \frac {40}{11}xy \right) \\ & = \frac {20^2}{11} - 11 \color{#3D99F6} \left(xy - \frac {20}{11} \right)^2 \end{aligned}$

Note that $\color{#3D99F6} \left(xy - \frac {20}{11} \right)^2 \ge 0$ , therefore, $z$ is maximum when $\color{#3D99F6} \left(xy - \frac {20}{11} \right)^2 = 0$ and $z_{max} = M = \dfrac {400}{11}$ $\implies \lfloor M \rfloor = \lfloor 36.\dot 3 \dot 6 \rfloor = \boxed{36}$

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Notes:

$\small \begin{aligned} x^2+y^2 & = (x+y)^2 - 2xy = 9 - 2xy \\ x^3+y^3 & = (x+y)(x^2+y^2 - xy) = 27 - 9xy \end{aligned}$

By taking xy=t and the taking xy common from the expression.

we get expression Value as : t(40-11t) . Since we know that x+y=3 .

So maximum value of xy=2.25.

But you may get doubt that both need not be positive but they have to be positive because if one of x,y is negative then other one will be positive making their product negative by which the expression will be negative so both should be positive so I applied A.M.≥G.M..

For the expression t(40-11t) to be maximum t=40/22.so t<2 .

Which is possible so therefore maximum is at t=40/22.

So ,(40/22)(40-11(40/22))=800/22=36.36.. so answer is 36.36 .😎

$f(X,Y)=X^4Y+X^3Y+X^2Y+XY+XY^2+XY^3+XY^4\\ =XY \bigg(\qquad (X^3+Y^3)\qquad ~~~~~\qquad+\qquad~~~~~\qquad (X^2+Y^2)\qquad~~~~~+\qquad~~~~~(X+Y)~~+1 \bigg)\\ =XY \bigg \{ \quad \{X+Y\} *\{(X+Y)^2-3XY\}\quad~~~~~+\quad~~~~~(X+Y)^2-2XY\qquad+\qquad X+Y\quad~~+1 \bigg \}.\\ Let~A=XY. \\ \implies~f(A)=A \Big\{\qquad(3)(9-3A)\qquad~\qquad~+\qquad~\qquad~9-2A\quad~\quad~\qquad + \qquad~\qquad4 \Big \}\\ =A( 27 - 9A + 9 - 2A + 4)=\quad40A\quad-\quad11A^2.\\ \therefore~f'(A)=40-22A=0,~~\implies~XY=3X -X^2=A=20/11.\\ \qquad~\qquad \{ OR~vertex~of~parabola~ ~\quad40A-11A^2,~~~is~~\dfrac{-40}{2*11}=20/11,~as~above.\} \\ \therefore~~f(A)_{max}=f(20/11)=400/11=\Huge \color{#D61F06}{36.36....}\\$