23is it 1/4 ?
$Q:$\Rightarrow$ Find Max. value of The Expression $f(x,y)=x^2y-y^2x$\\\\. Where $0\leq x,y\leq1.$
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23consider it to be a quadratic in y,
ay-by2= f
hence it will be a downward opening parabola , whoz max occurs at y = -B/2A = -(-a/2b) = a/2b = x2 /2x =x/2
i.e at y = x/2
so f = x32 - x3/4 = x3/4
so fmax = 1/4 , as x ε [0,1]
1708plz check me if i wrong....
$\boldsymbol{Ans:}$\Rightarrow $Let $M=f(x,y)=x^2y-y^2x=-(y^2x-x^2y)....................(1)$\\\\ Now Completing Square in terms of $y$,We Get\\\\ $M=-x.(y^2-xy)=-x.\left \{ \underbrace{y^2-xy+(\frac{x}{2})^2}-(\frac{x}{2})^2\right \}=-x.\left\{ (y-\frac{x}{2})^2-\frac{x^2}{4}\right \}$.\\\\ So $M = \frac{x^3}{4}-x.(y-\frac{x}{2})^2$, for Max. Vlaue of $M$, square term i.e $(y-\frac{x}{2})^2$ must be 0.\\\\ So $M_{max}=\frac{x^3}{4}=\frac{1}{4}$. (bcz $\frac{x^3}{4}$ is an strictly Increasing function in $0\leq x\leq1$).\\\\ So $\boxed{M_{max} = \frac{1}{4}}$
1708Yes This is a simple answer.
$\boldsymbol{Ans:}$\Rightarrow$ We Can also solve This question using $x=cos\theta$ and $y=sin\theta$\\\\Where $0\leq\theta\leq\frac{\pi}{2}$
341On similar lines, by AM-GM
y(x-y) \le \left(\frac{y+x-y}{2}\right)^2 = \frac{x^2}{4}
Hence xy(x-y) \le \frac{x^3}{4} \le \frac{1}{4}
with equality occurring when x=1 and y=1/2