Question

Find the points on the surface (x^2)(y^3)(z^4)=4 that are closest to the point (0,0,0)?

Answer 1

The system of equations involving lagrange's multiplier is just too difficult for me.

Answer 2

do you happen to know the system of equations?

Answer 3

F(x,y,z) = (x^2)(y^3)(z^4) - 4

Fx = 2x(y^3)(z^4)
Fy = (x^2) 3y^2 (z^4)
Fz = (x^2)(y^3) 4z^3

these would be the gradient parts of the surface function

Answer 4

and, to create a line with any generic point (a,b,c) on the surface to the point (0,0,0) we get

\(x = 0 + t~ 2ab^3c^4\)
\(y = 0 + t~ 3a^2b^2c^4\)
\(z = 0 + t~ 4a^2b^3c^3\)

you sure the lagrange stuff is to hard for you?

Answer 5

plugging these xyz values into the surface equation we get

(x^2)(y^3)(z^4)=4
\[t^2~2^2a^2b^6c^8~t^3~3^3a^6b^6c^{12}~t^4 ~4^4a^8b^{12}c^{12}=4\]
\[t^9~(2^2~3^3~4^4)a^{16}b^{24}c^{32}~=~4\]
yeah, the lagrange should be much simpler to figure this mess out with