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I am trying to find the absolute maxima and minima of $ f(x,y) = 2x^3 + y^4$ in the domain $D = \{(x,y)| x^2 + y^2 \le 1 \}$

I have made an attempt as shown in attached picture. Please ignore dashed areas. Attached picture. The answer is (1,0) and (-1,0). I do NOT want to use the Lagrange multiplier method to solve.

Is it legal to substitute $y^4 = (1-x^2)^2$ into $f(x,y)$ to make a single variable $f(x)$ and then find $f'(x) =0 $? Then, after finding the $x$ can I substitute into $x^2 +y^2 = 1$ to find y? Then can I plug it back into $f(x,y)$ to find critical points?

I want to know how to find max and min on the boundary $x^2 + y^2 = 1$ without the LM method?

Thanks for your help!

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  • $\begingroup$ It's not illicit to substitute, I would do the same. $\endgroup$ – Michael Hoppe Aug 18 at 10:01
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Since the origin $(0,0)$ is the only critical point inside $D$ and the origin is not a local maximum or a local minimum (because $f(0,0)=0$ and $f$ changes its sign in any neighbourhood of $(0,0)$), it follows that the absolute maxima and minima of $f$ in the compact set $D$ have to be attained at the boundary. Hence, you may consider the restriction of $f$ along such boundary $x^2+y^2=1$, that is the one-variable function $$g(x)=f(x,\pm\sqrt{1-x^2})=2x^3+(1-x^2)^2$$ in the domain $[-1,1]$. Its derivative is $$g'(x)=2x(x+2)(2x-1)$$ so we have to compare $g(0)=1$, $g(1/2)=13/16$, plus the values at the extreme points $g(-1)=-2$ and $g(1)=1$. We may conclude that the absolute maximum point is $(1,0)$ and the absolute minimum point is $(-1,0)$.

Alternative way: if $x^2 + y^2 \le 1$ then $-1\leq x^3\leq x^2$ and $0\leq y^4\leq 2y^2$. Therefore $$f(-1,0)=-2\leq 2x^3\leq f(x,y)=2x^3 + y^4\leq 2x^2+2y^2\leq 2=f(1,0).$$

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  • $\begingroup$ @PcumP_Ravenclaw I edited my answer. I hope it can help. $\endgroup$ – Robert Z Aug 18 at 10:41
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Using Calculus, it's routine:

  • Find the critical points in the interior of $D\;$by setting the partial derivatives of $f$ to zero, and solving for $x,y$. In this case, the only critical point in the interior of $D\;$is the origin, but at the origin, we have $f(0,0)=0$, which is not an absolute extreme value of $f$ on $D$.$\\[4pt]$
  • On the boundary of $D$, you can do what you suggested, namely, replace the $y^4$ term of $f(x,y)$ by $(1-x^2)^2$ and then, using standard methods from single-variable Calculus, proceed to find the absolute extrema of the function $g(x)=2x^3+(1-x^2)^2$ on the closed interval $[-1,1]$.

But Calculus is not really needed for this problem.

By inspection, we have

  • $f(-1,0)=-2$.$\\[4pt]$
  • $f(1,0)=2$.

For $(x,y)\in D$,

  • If $x \ge 0$, then $0 \le 2x^3+y^4 \le 2x^2 + y^2 \le 2(x^2+y^2)\le 2$.$\\[4pt]$
  • If $x < 0$, then $-2 \le 2x^3 \le 2x^3+y^4$.

hence for all $(x,y)\in D$, we have $-2\le f(x,y)\le 2$.

First suppose $f(x,y)=-2$. \begin{align*} \text{Then}\;\;&f(x,y)=-2 \qquad\;\; \\[4pt] \implies\;&x^3+y^4=-2\\[4pt] \implies\;&x^3\le -2\\[4pt] \implies\;&x^3\le -1\\[4pt] \implies\;&x\le -1\\[4pt] \implies\;&x=-1\\[4pt] \implies\;&-2+y^4=-2\\[4pt] \implies\;&y^4=0\\[4pt] \implies\;&y=0\\[4pt] \implies\;&(x,y)=(-1,0)\\[4pt] \end{align*} Next suppose $f(x,y)=2$. \begin{align*} \text{Then}\;\;&f(x,y)=2\\[4pt] \implies\;&2x^3+y^4=2\\[4pt] \implies\;&x^3+(x^3+y^4)=2\\[4pt] \implies\;&x^3+(x^2+y^2)\ge 2\\[4pt] \implies\;&x^3+1\ge 2\\[4pt] \implies\;&x^3\ge 1\\[4pt] \implies\;&x\ge -1\\[4pt] \implies\;&x=1\\[4pt] \implies\;&2+y^4=2\\[4pt] \implies\;&y^4=0\\[4pt] \implies\;&y=0\\[4pt] \implies\;&(x,y)=(1,0)\\[4pt] \end{align*} It follows that

  • The absolute minimum value of $f$ on $D\;$is $-2$ which occurs for $(x,y)=(-1,0)$.$\\[4pt]$
  • The absolute maximum value of $f$ on $D\;$is $2$ which occurs for $(x,y)=(1,0)$.
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The maximum and minimum is situated on the curve $$x^2+y^2=1$$ So you have to consider $$f(x,\pm\sqrt{1-x^2})=2x^3+(1-x^2)^2$$ Solve the equation $$6 x^2-4 x \left(1-x^2\right)=0$$ for $x$, the first derivative.

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    $\begingroup$ I know that! It could be any point on the unit circle. How do I find them? $\endgroup$ – PcumP_Ravenclaw Aug 18 at 10:06
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I'm sorry I couldn't follow your answer in the attached image

$$f(x,y) = 2x^3 + y^4$$ the first derivative: $${df\over dx} = 6x^2 \rightarrow 0; x=0$$ $${df\over dy} = 4y^3 \rightarrow 0; y=0$$ With: $$x^2 + y^2 = 1$$ as an end point. Set each variable to zero and solve for the other. $$x=\pm1$$ $$y=\pm1$$

∴ we have five critical points: $$(0,0) , (0,1), (0,-1), (1,0), (-1,0)$$ plug/substitute each point in the function to get the maximum and minimum.

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  • $\begingroup$ Why do you substitute x = 0 and y = 0 seperately into $x^2+y^2 = 1$ ? $\endgroup$ – PcumP_Ravenclaw Aug 18 at 10:22
  • $\begingroup$ To get the intercepts. $\endgroup$ – Ahmed Elhefnawy Aug 18 at 10:25
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$2x^3+y^4\le 2x^2+y^2 $=$x^2+x^2+y^2\le x^2+1\le 2$ . By the same way we get $2x^3+y^4\ge 2x^3\ge -2$. We finally check that on the points $(1, 0) $ and $(-1, 0)$ $f$ gets respectively the above mentioned values.

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