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.

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 $x= - \frac 1 2 ,\,y= - \frac 3 2 .$ 
