Maximum and minimum function on area Find maximum and minimum value of function $f(x,y) = 3x+14y$ on $ \left\{ (x,y): 3x^4 + xy + y^4 =6\right\} $. 
I will grateful for hints and yours help.
 A: It is clear that the tangents of the smooth curve  $$3x^4+y^4+xy-6=0\,\,\,\, (1)$$ at the extremal points have to  be parallel to the level lines of the target function, i.e. $3x+14y=C$. Using the implicit differentiation and equating the slopes, we obtain the equation $$-\frac {12x^3+y} {4y^3+x}= - \frac 3 {14} \Leftrightarrow 168x^3+14y-12y^3-3x=0.\,\,\,\, (2) $$ In order to solve the system (1)-(2) in $x$ an $y$, we find its resultant in $y$ with help of Maple: 
   $$797154048x^{12}+56899584x^{10}-1327104x^8+113799168x^6+10434789x^4+8232x^2-2677344. $$ 
   It is well known that its rational roots must have the form $\pm \frac {\mathop{\rm divisor\,\, of} 2677344}{ \mathop{\rm divisor\,\, of} 797154048}.$ Factoring $2677344=3\cdot 2^5\cdot 167^2$ and $797154048=37\cdot 2^8\cdot 3^4\cdot 1039$ and considering $6480=6\cdot 2 \cdot 3 \cdot 9 \cdot 5 \cdot 2 \cdot 2$ cases, we find $\pm \frac 1 2$ to be the rational roots of the discriminant.  The numerical solution shows the discriminant has only the two real roots (The others are complex.). Therefore, we obtain the maximum of the target function ar $x= \frac 1 2 ,\,y= \frac 3 2$. The value of $y$ is found as the only real root of $-24y^3+28y+39=0$. The last equation is formed by the substitution $ x= \frac 1 2$ in (2). The minimum of the target function is achieved at $x= - \frac 1 2 ,\,y= - \frac 3 2 .$ 
A: What could be said about this "skewed superellipse"  $ \ 3x^4 \ + \ xy \ + \ y^4 \ = \ 6 \ $ is that since the curve has symmetry about the origin [if a point $ \ (x,y) \ $ lies on the curve, so does $ \ (-x,-y) \ $ ], we should expect that extrema are located at corresponding points in opposing quadrants.  user64494  finds, by implicit differentiation of the curve equation,  the slope of a tangent line to be  $ \ \frac {12x^3+y}{4y^3+x} \ $ , and from the linear function to be extremized, the "useful" equation
$$\frac {12x^3 \ + \ y} {4y^3 \ + \ x} \ \ = \ \ \frac{3}{14} \ \ . $$
As the OP, Thomas, indicated that this was an exam problem, I wondered whether some indication was given that the student was to locate a rational point on this curve in answering the question, since the time would be severely (hopelessly?) limited to find real zeroes of the related polynomial without a computational aid.  If it were given that the coordinates of a rational point have the same denominator, so that $ \ x \ = \frac{a}{c} \ \ , \ \ y \ = \frac{b}{c} \ \ , $ we would be able to write
$$\frac {12 \left( \frac{a}{c} \right)^3 \ + \ \frac{b}{c} } {4\left( \frac{b}{c} \right)^3 \ + \frac{a}{c} } \ \ = \ \ \frac {12 a^3 \ + \ bc^2 } {4 b^3 \ + \ ac^2 } \ \ = \ \ \frac{3}{14} \ \ . $$
We might then begin a search with $ \ c  \ = \ 2 \ $ (much as we do in "hunting" rational zeroes of a polynomial), giving us
$$ \  \frac {12 a^3 \ + \ 4b  } {4 b^3 \ + \ 4a } \ \ = \ \ \frac{3}{14} \ \ . $$
Since we need $ \ a \ $ and $ \ b \ $ to be integers, we may require that $ \ 6 a^3 \ + \ 2b \ \ = \ \ 3k \ $ and $ \ 2 b^3 \ + \ 2a \ \ = \ \ 14k \ \ , $ with $ \ k \ $ being a positive integer.  It takes only a little effort to find $ \ 6  · 1^3 \ + \ 2·3 \ \ = \ \ 12 \ \ = \ \ 3·4 \ $ ; we then note that  $ \ 14 · 4 \ \ = \ \ 56 \ \ = \ \ 2 ·3^3 \ + \ 2·1 \ \  \ \ . $
This gives us the desired ratio $ \ \frac{12}{56} \ \ = \ \ \frac{3}{14} \ \ . $  As all of the exponents in the slope expression are odd, the signs of both $ \ a \ $ and $ \ b \ $ can both be changed without altering the sign of the ratio.  This then yields two rational points $ \ \left( \ \frac{1}{2} \ , \ \frac{3}{2} \ \right) $ and $ \ \left( \ -\frac{1}{2} \ , \ -\frac{3}{2} \ \right) \ \ , $  as expected from the symmetry of the curve.  The maximum and minimum of $ \ f(x,y) \  = \ 3x \ + \ 14y \ $ occur, respectively, for those two points as $ \ \frac{45}{2} \ $ and $ \ -\frac{45}{2} \ $ .  A representative graph is shown below.  A calculation of this sort may have been possible, with the clues mentioned, in the available time.

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I can't help feeling that this is one of those exam problems, if it is correctly reproduced here, that the problem-poser had not worked out in advance to estimate how much work would be necessary.  (I've had to reject many prospective "interesting" exam problems after seeing the amount of solution-writing that would be called for...)  An optimization problem for an exam would be rather more tractable if the curve were a "rotated ellipse", say,   $ \ 3x^2 \ + \ xy \ + \ y^2 \ = \ C \ $ .
