Mathematical Enigma - VIII

We can simplify the function:

$x^3y^3(x^3+y^3)$

= $(xy)^3(x+y)(x^2-xy+y^2)$

= $(xy)^3(x+y)((x+y)^2-3xy)$

We know from the question that $x+y=2$ :

= $(xy)^3 \cdot 2 \cdot (4-3xy)$

= $(xy)^3 \cdot (8-6xy)$

$x+y=2 \Rightarrow y=2-x$

Hence we can write the function $xy$ as $x(2-x)$ = $2-x^2$

$f(x) = 2-x^2 \Rightarrow$ The maximum of the function is when $f'(x) = 0$ because it is a negative quadratic

$f'(x) = 2-2x=0 \Rightarrow$ Hence the maximum value of the function is when $x=1$

So the maximum value of the function $= xy =1(2-1) = 1$

We can now substitute the maximum into the original equation:

$(1)^3 \cdot (8-6(1))$

= $1 \cdot 2$

= $\boxed{2}$

So $\boxed{2}$ is the maximum value of the function.

$f(x)=x^3*(2-x)^3*(x^3 + (2-x)^3).\\ \therefore~f '(x)=f(x) \left \{\dfrac {3x^2}{x^3} - \dfrac{3(2 - x)^2}{(2-x)^3} +\dfrac {3x^2 - 3(2 - x)^2}{x^3 + (2 - x)^3} \right \}= 0. \\ \therefore~f '(x)=3*f(x)*\left \{ \dfrac 1 x -\dfrac 1 {2 - x } + \dfrac{4 - 4x}{- 6x^2+12x - 8} \right \}=0\\ \therefore~f '(x)=3*f(x)*\left \{ \dfrac 1 {x(2 - x) } - \dfrac 1{3x^2 - 6x +4} \right \}*\color{#3D99F6}{\{2-2x \}}=0\\ \text{Since x,y>0, for minimum value 0f the function is x=0, x=1 is maximum. }\\ \therefore~f(x)=1*1*(1+1)= \Large ~~~~~~\color{#D61F06}{2}$

$Given\quad that\quad x+y=2\\ Using\quad the\quad AM\quad GM\quad Inequality\quad \\ \frac { x+y }{ 2 } \ge \sqrt [ 2 ]{ xy } \\ \Rrightarrow \quad 1\quad \ge \quad xy\\ Hence\quad the\quad maximum\quad Value\quad of\quad xy\quad is\quad 1.\\ Now\quad by\quad simplifying\quad ({ x }^{ 3\quad }{ y }^{ 3 })({ x }^{ 3 }+{ y }^{ 3 })\\ We\quad get\quad { (xy) }^{ 3 }(x+y)({ x }^{ 2 }+{ y }^{ 2 }-xy)\\ \Rrightarrow \quad { (xy) }^{ 3 }(x+y){ [(x+y) }^{ 2 }-3xy]\\ Now\quad putting\quad the\quad values\quad in\quad the\quad equation\quad we\quad get\\ ({ x }^{ 3\quad }{ y }^{ 3 })({ x }^{ 3 }+{ y }^{ 3 })=\quad 2$

$x^3y^3(x^3+y^3)$

= $(xy)^3(x+y)(x^2-xy+y^2)$

= $(xy)^3 \cdot 2 \cdot (4-3xy)$

Applying the AM-GM,we have:

$(xy) \cdot (xy) \cdot (xy) \cdot (4-3xy) \le (\frac{xy+xy+xy+4-3xy}{4}))^4$ =1

Then $x^3y^3(x^3+y^3) \le 2 \cdot 1=2$

In all the solution they have found greatest value of xy. All have ignored -3xy has a negative impact, to make the product greatest. I don't think that the process is correct. I have a better solution.
x+y=2=>x=1+k and y=1-k
So the expression becomes
2-2k⁴(3k⁴-8k²+6), after substitution.
Now, k is positive. Also 3k⁴-8k²+6 can be considered as a quadratic polynomial, if k²=p. It's lowest value is 0.666... (using the method of completing the square). Hence the maximum value of expression is when k=0. At that value of k the value of the expression is 2, which is it's maximum value.

Given that x+y=2 for maximum value apply AM greter than GM then we get xy=1 substituting in equation( xy^3 )(x^3+y^3)= 1×(x+y)(x^2+y^2 -xy )=1×2×((x+y)^2)-3xy)=1×2×(4-3)=2 So answer is 2

Take cube both sides then put the value that is give u have equation (8-6x³y³)x³y³, if we put the value of x and y =1then we easily get the ans.