Solution: (the original question is z=x? +2xy+y？ +x+2y, met with insurmountable difficulties, so x was changed to 2x; The answer is for reference only. )
Can the elliptic equation be x? /4+y？ = 1 rewritten as a parametric equation, x=2cost, y = Sint, and the maximum value is substituted into the function:
z=4cos？ t+4sintcost+sin？ T+4 cost +2 ............. (1)
Make dz/dt=-8costsint+4cos？ t-4sin？ T+2 point cost -4 point +2 point cost
=-6costsint+4cos？ t-4sin？ T+2 (cost -2 points)
=2(2cos？ t-3costsint-2sin？ T)+2 (cost -2 points)
= 2(2 cost+sintering) (cost -2 sintering) +2 (cost -2 sintering)
=2 (cost -2 points) (2 cost+sintering+1)=0.
From cost-2sint=0，TANT = 1/2； So I got a t? = arctan( 1/2)； Cint? = 1/√5; Cost? =2/√5;
From 2cost+sint+ 1=0, Sint+tan φ cost = (1/cos φ) (Sint cos φ+costs in φ) = (1/cos φ) sin (t+φ) =-1.
That is, there is sin(t+φ)=-cosφ=sin(π/2+φ), so we get t+φ=π/2+φ, so we get t? =π/2;
Will it? And t? Substitute the value of (1) to get: z? = 16/5+8/5+ 1/5+8/√5+2/√5=5+ 10/√5=5+2√5; z？ = 1+2=3.
That is, zmax=5+2√5 and zmin=3.
[After changing X in the title to 2x, it can also be solved by Lagrange multiplier method. If you don't change it, you will encounter higher-order equations.
2。 Solve the equation (14+x) (1/3)+(14-x) (1/3) = 4.
Solution: There is only one combination: 14+x = 27...( 1); 14-x= 1........(2);
2x=26 in (1)-(2), so x= 13.