Weg ohne Lagrange:
\(f(x,y) = (3x+2y)*x\) soll optimal werden
NB: \(2x^{3}+ 3yx^{2}-40 =0\) \( 3yx^{2} =40-2x^3\) \( y =\frac{40-2x^3}{3x^{2} }\)
\(f(x) = 3x^2+\frac{80x-4x^4}{3x^{2} }\)
\(\frac{df(x)}{dx} =6x+\frac{(80-16x^3)*3x^{2}-(80x-4x^4)*6x}{9x^{4} }\)
\(6x+\frac{(80-16x^3)*3x^{2}-(80x-4x^4)*6x}{9x^{4} }=0\)
\(x=2\) \( y(2) =\frac{40-2*2^3}{3*2^{2} }=2\)
\(f(2,2) = (3*2+2*2)*2=20\)