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Finding maxima and minima of $$f(x,y) = x^{4} + y^{4} - x^{3}$$ I tried solving this question but need help. In this question, I calculated the partial derivatives of first and second order. I found $$f_{x} = 4x^{3} - 3x^{2}\\ f_{y}= 4y^{3} \\ f_{xx} = 12x^{2}-6x \\ f_{yy}= 0$$ On equating $$f_{x} = 0 \hspace{0.5cm} and\hspace{0.5cm} f_{y} = 0 $$ I get the critical points as (0,0) and (0.75,0) but I am not able to understand whether these points constitute a maxima or minima because $$f_{xx}f_{yy} - f_{xy}f_{yx}=0$$

Can somebody please guide?

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  • $\begingroup$ This function has no global maximum. $\endgroup$ – Mastermind817 Mar 24 at 10:55
  • $\begingroup$ I am not clear as to how to prove it mathematically. Can you please share your working? $\endgroup$ – User_in Mar 24 at 11:00
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The function does not have a global maximum because $y^4$ grows without bound as $y\to\pm\infty$; the same holds for the $x$ variable. This can be shown by just taking the limits separately, because $f$ splits nicely between $x$ and $y$. However, because of this behavior at infinity and due to its continuity, it must have a global minimum thanks to a variant of Weierstrass' theorem.

As for $(0,0)$ and $(3/4,0)$, which annihilate the gradient, the Hessian-determinant condition you mention implies that they are degenerate points (which means analyzing the Hessian is not sufficient to conclude that they are local maxima, local minima, or saddle points). Indeed, the Hessian is $$\mathbf D^2f(x,y) = \left(\begin{array}{cc} 12 x^{2}-6 x & 0 \\ 0 & 12 y^{2} \end{array}\right), $$ so at those points $$\mathbf D^2f(0,0) = \left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right), \qquad \mathbf D^2f(3/4,0) = \left(\begin{array}{cc} 9/4 & 0 \\ 0 & 0 \end{array}\right). $$ To decide their nature as critical points, we must take restrictions of the function near those points, for example along lines. Take $(0,0)$: we know $f(0,y) \geqslant 0$ in a neighborhood of $y=0$, whereas $f(x,0)\geqslant 0$ in a left-neighborhood of $x=0$ and $f(x,0) \leqslant 0$ in a right-neighborhood of $x=0$, so $f$ must have a (generalized?) saddle point at the origin.

As for $(3/4,0)$, it is sufficient to observe that $f(x,y) = g(x) + h(y)$ with $g(x)=x^4-x^3$ and $h(y)= y^4$, and the degeneracy of the Hessian at that point is only due to the fact that $h''(0) = 0$. Since we know $h \geqslant 0$ for all $y$, and $g''(3/4) > 0$ (it is the positive eigenvalue of $\mathbf D^2 f(3/4,0)$ that was written before), we obtain that $f$ has a local minimum at $(3/4,0)$. This is also a global minimum because there are no other options.

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  • $\begingroup$ Thank you for the clarification. For $$(3/{4},0) $$ $\endgroup$ – User_in Mar 24 at 11:41
  • $\begingroup$ My pleasure. I have added a discussion of the second critical point. $\endgroup$ – giobrach Mar 24 at 11:43
  • $\begingroup$ Thank you for the clarification. For the critical point(3/4,0) I found that for y <=0, f(3/4,y) is greater than -27/256 and for y<=0 f(3/4,y) is greater than -27/256. Hence (3/4,0) is a point of local minima. Is this reasoning also correct? I saw your updated discussion thanks for sharing it. $\endgroup$ – User_in Mar 24 at 11:53
  • $\begingroup$ To conclude, you need to explore the behavior of $f$ in the $x$ direction as well. In this case, it works because the restriction of $f$ to the $x$-axis is strictly convex in a neighborhood of $3/4$ $\endgroup$ – giobrach Mar 24 at 11:55
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In XY plane you'll observe that in some direction (neighborhood points)from critical point function is greater while in some other direction value of function is Less than that of value of function at critical point (0,0)and hence no local maxima or minima at (0,0)

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For the minimum, if $x\ge 1$ or $x\le 0$ we have $x^4-x^3+y^4=x^3(x-1)+y^4\ge 0$. If $0<x<1$ we have by AM-GM $$x^3(1-x)=27(x/3)^3(1-x)\le 27\left(\frac{3(x/3)+(1-x)}{4}\right)^4=27/256$$ so $x^4-x^3\ge -27/256$ and so $x^4-x^3+y^4\ge x^4-x^3\ge -27/256$. The minimum value is reached at $x=3/4$, $y=0$.

For the maximum, clearly it doesn't exist.

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$$f(x,y)=x^4+y^4-x^3\\\qquad\quad ~~ ~=x^3(x-1)+y^4$$

Apply $x\to\ +\infty$ and $y=0$, you get $f(x,y)\to +\infty.$

So, global maxima doesn't exist.


Now, we will prove that global minimum exists and is equal to

$$\text{min}[x^4-x^3+y^2]=-\frac{27}{256}.$$

In this case , of course, $y=0$ must be. Then observe that,

$$\begin{align}x^4-x^3+\frac{27}{256}=\frac{1}{256}(256x^4-256x^3+27)=\frac{1}{256}(4 x - 3)^2 (16 x^2 + 8 x + 3)≥0. \end{align}$$

Because,

$\Delta_{\text{half}}=16-16\times 3=-32<0$.

This means, $16x^2+8x+3>0$ for all $x\in\mathbb R.$

Finally, we deduce that

$$\text{min}\left[x^4-x^3+\frac{27}{256}\right]=0$$

at $x=\dfrac 34.$

which follows

$$\text{min}\left[x^4-x^3+y^4\right]=-\frac{27}{256}.$$

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