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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^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!

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 $-2le f(x,y)le 2$.

First suppose $f(x,y)=-2$.
beginalign*
textThen;;&f(x,y)=-2
qquad;;
\[4pt]
implies;&x^3+y^4=-2\[4pt]
implies;&x^3le -2\[4pt]
implies;&x^3le -1\[4pt]
implies;&xle -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]
endalign*
Next suppose $f(x,y)=2$.
beginalign*
textThen;;&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+1ge 2\[4pt]
implies;&x^3ge 1\[4pt]
implies;&xge -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]
endalign*
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 $(x,y)=(1,0)$.
  

The maximum and minimum is situated on the curve $$x^2+y^2=1$$
So you have to consider $$f(x,pmsqrt1-x^2)=2x^3+(1-x^2)^2$$
Solve the equation $$6 x^2-4 x left(1-x^2right)=0$$ for $x$, the first derivative.

I'm sorry I couldn't follow your answer in the attached image

$$f(x,y) = 2x^3 + y^4$$
the first derivative:
$$dfover dx = 6x^2 rightarrow 0; x=0$$
$$dfover dy = 4y^3 rightarrow 0; y=0$$
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.

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
$$g(x)=f(x,pmsqrt1-x^2)=2x^3+(1-x^2)^2$$
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 $-1leq x^3leq x^2$ and $0leq y^4leq 2y^2$. Therefore
$$f(-1,0)=-2leq 2x^3leq f(x,y)=2x^3 + y^4leq 2x^2+2y^2leq 2=f(1,0).$$

$2x^3+y^4le 2x^2+y^2 $=$x^2+x^2+y^2le x^2+1le 2$ . By the same way we get $2x^3+y^4ge 2x^3ge -2$. We finally check that on the points $(1, 0) $ and $(-1, 0)$ $f$ gets respectively the above mentioned values.