Find min and max using Lagrange Multipliers

[jwpaine]

Find the minimum and maximum values of the function f(x,y,z) = 3x+2y+4z subject to the constraint x^2+2y^2+6z^2 = 36.

I did:

?f = ??g

<3, 2, 4> = ?<2x, 4y, 6z>

x = 3 / 2? , y = 1 / 2? , z = 2 / 3?

Substituting these back into the constraint g(x,y,z), (3 / 2?)^2 + (1 / 2? )^2 + (2 / 3?)^2 = 36

? = +/- Sqrt(53 / 2) / 18

Solving for the max (first):

x = 3 / 2( Sqrt(53 / 2) / 18 ), y = 1 / 2( Sqrt(53 / 2) / 18), z = 2 / 3( Sqrt(53 / 2) / 18)

f(x,y,z) = 3x + 2y + 4z = 3(3 / 2( Sqrt(53 / 2) / 18 )) + 2(1 / 2( Sqrt(53 / 2) / 18)) + 4(2 / 3( Sqrt(53 / 2) / 18))

Where have I gone wrong, here?

Thanks!

 

[tkhunny]

Right up front. Try "12z".

 

[soroban]

Hello, jwpaine!

\(\displaystyle \text{Find the minimum and maximum values of the function }f(x,y,z) \:=\: 3x\,+\,2y\,+\,4z\)
. . \(\displaystyle \text{ subject to the constraint }x^2+2y^2+6z^2 \:=\: 36.\)

\(\displaystyle \text{I did: }\:\Delta f \,=\,\lambda(\Delta G)\)

. . \(\displaystyle \langle3, 2, 4\rangle \:=\: \lambda\langle2x, 4y, 6z\rangle\)

. . \(\displaystyle x \,=\, \frac{3}{2\lambda} \quad y \,=\, \frac{1}{2\lambda} \quad z \,=\, \frac{2}{3\lambda}\)

\(\displaystyle \text{Substituting these back into the constraint }g(x,y,z)\!:\;\; \left(\frac{3}{2\lambda}\right)^2 + \left(\frac{1}{2\lambda}\right)^2 + \left(\frac{2}{3\lambda}\right)^2 \:=\: 36\) . Here!

\(\displaystyle \text{The constraint is: }\:x^2 + {\bf2}y^2 + {\bf6}z^2 \:=\:36\)
. . . . . . . . . . . . . . . . . \(\displaystyle \uparrow \qquad\, \uparrow\)

 

[jwpaine]

See, this is what happens when you stay up too late to finish an assignment.

Thank you both!