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If $0\le x\le 1$ and $0\le y\le 1$, find $\max\{(x^2y-y^2x)\}$

My work:
Though I could not approach the problem, I tried to find out a few facts. So,I defined the above expression as $f(x,y)=x^2y-y^2x$. Now,I see that whenever $(x,y)=(k,k)$ the expression is 0 which is not the maxiumum value. So, $x\ne y$. [N.B. I am not sure of the tag,please feel free to edit]

Ahaan S. Rungta
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Hawk
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4 Answers4

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Note that your objective function is $f(x,y)=xy(x-y)$ so $x$ must be bigger than $y$ and for any given $y$, $f$ is increasing in $x$, so $x^*=1$. That leaves you to maximize $y(1-y)$ with respect to $y$, such that $y^*=1/2$.

JPi
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  • Yes, but do I prove that it is the maxiumum case, and suppose,as I said in the question, $x=\sin A,y=\sin B$, then we get the maximum at $2$, how to justify it in this question? – Hawk Jan 29 '14 at 02:31
  • My answer is a proof. Given that you're optimizing subject to the constraints that $0\leq x,y\leq 1$, the maximum value cannot be 2. – JPi Jan 29 '14 at 02:36
  • Indeed, my answer implies that the maximum value is $1/4$. – JPi Jan 29 '14 at 02:37
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    Definitely a much more compact and intuitive answer then the traditional calculus grind. – John Habert Jan 29 '14 at 02:39
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    Sorry, I actually made a mistake in using the formulas, actually you are right! Thanks! – Hawk Jan 29 '14 at 02:43
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Use the $f(x,y)$ that you defined. The region $0\leq x \leq 1, 0 \leq y \leq 1$ is closed and bounded so the only places you need to consider are the boundary of the region and any critical points on the inside (which most satisfy $f_x = 0$ and $f_y = 0$ and where $f_x = \frac{\partial f}{\partial x}$. With equations $f_x = 2xy-y^2=y(2x-y)=0$ and $f_y = x^2-2xy=x(x-2y)=0$, the only critical point is $(0,0)$ which is on the boundary. So we need only consider the boundary lines of $x=0, y=0, x=1, y=1$. Clearly if $x=0$ or $y=0$ the function is 0 so those can't be the max. $f(x,1) = x^2-x$ which has max at $x=-\frac{1}{2}$ which is not in an region. That leaves $f(1,y) = y-y^2$. This has max at $y=\frac{1}{2}$. Since we only have to check $(1,\frac{1}{2})$ and $f(1,\frac{1}{2}) > 0$, it is the max.

John Habert
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given $0 \leq x,y \leq 1$ and finding $\max(x^2y-y^2x)$

applying $\displaystyle xy(x-y) \leq x\cdot \frac{(y+x-y)^2}{4}\leq \frac{x^3}{4}\leq \frac{1}{4}$

when $x=1$ and $\displaystyle y = \frac{x}{2} = \frac{1}{2}$

DXT
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Since $x\in[0,1]$ and $xy\geqslant0$, we have $x^2y\leqslant xy$. Hence, $$x^2y-y^2x\leqslant xy-y^2x=x(y-y^2)\leqslant y-y^2$$ since $x\leqslant 1$ and $y-y^2=y(1-y)\geqslant 0$.

Now, $y-y^2=\frac14-\left(y-\frac12\right)^2\leqslant\frac14$.

Hence the maximum possible value of $x^2y-y^2x$ is $\frac14$, and this is attained when $x=1,y=\frac12$ .

A. Goodier
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