The maximum value of the expression $$x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$$ under the constraint $$x+y=2$$ is $$\frac{225}{32}$$. This value occurs when $$x=\frac{3}{4}$$ or $$x=\frac{5}{4}$$. Two methods were discussed: one utilizing substitution with $$x=1+t$$ and the other employing calculus to derive the critical points of the function $$f(x)=-8x^4+32x^3-47x^2+30x$$. Both methods confirm the same maximum value.
PREREQUISITESMathematicians, students studying calculus, and anyone interested in optimization problems involving polynomial functions.
This may not be the quickest solution, but it avoids calculus. Let $x=1+t$, then $y=1-t$. Notice that $1+x+x^2+x^3 = \dfrac{x^4-1}{x-1} = \dfrac{(1+t)^4-1}{t}$, and similarly $1+y+y^2+y^3 = -\dfrac{(1-t)^4-1}{t}.$ Also $xy = 1-t^2.$ Then $$ \begin{aligned}x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4 &= xy\bigl((1+x+x^2+x^3) + (1+y+y^2+y^3) - 1\bigr) \\ &= (1-t^2)\Bigl(\frac{(1+t)^4-1}{t} - \frac{(1-t)^4-1}{t} - 1\Bigr) \\ &= (1-t^2)(7+8t^2) \\ &= 7+t^2 -8t^4 \\ &= \frac{225}{32} - 8\Bigl(t^2 - \frac1{16}\Bigr)^2\quad \text{(completing the square).}\end{aligned}$$ Thus the maximum value is $\frac{225}{32}$, which occurs when $t = \pm\frac14$, or when $x = \frac34$ or $\frac54.$anemone said:Find the maximum of the expression $$x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$$ if $$x,\;y$$ are real numbers with $$x+y=2$$.