The discussion revolves around finding the maximum value of the expression $$x^4y+x^3y+x^2y+xy+xy^2+xy^3+xy^4$$ under the constraint $$x+y=2$$, utilizing both algebraic and calculus-based approaches. Participants explore different methods, including substitution and differentiation, to analyze the problem.
Participants agree on the maximum value being $$\frac{225}{32}$$, but they present different methods to arrive at this conclusion. There is no disagreement on the maximum value itself, but the approaches to find it vary.
The discussion includes various mathematical manipulations and substitutions, which may depend on specific assumptions about the variables involved. The methods presented do not explore potential limitations or alternative constraints that could affect the results.
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 $$x+y=2$$.