# Help with a LaGrange Multiplier Question

• Feb 18th 2010, 09:55 PM
jaijay32
Help with a LaGrange Multiplier Question
Hi. I am trying to maximize the function F(x,y,z)=xyz, under the constraint 2x +2y+z=84, with theta being represented by m.

Derivatives:

Fx= yz+2m=0
Fy= xz+2m=0
Fz= xy+m=0
Fm= 2x+2y+z-84=0.

I found m= -yz/2, and m= -xz/2, getting y=x. I proceeded to find z= -2y.

I then plugged everything into Fm and got y=42. What am I doing wrong, as the correct answer should be (14,14,28)?

Many thanks
• Feb 18th 2010, 11:01 PM
Soroban
Hello, jaijay32!

You were close . . . one little error.

Quote:

Maximize the function $F(x,y,z)\,=\,xyz$
under the constraint $2x +2y+z\:=\:84$

Derivatives:

. . $\begin{array}{cccccc}(1) & F_x &=& yz+2\theta &=& 0 \\
(2) & F_y &=& xz+2\theta &=& 0 \\
(3) & F_z &=& xy+\theta &=& 0 \end{array}$
. . Correct!
. . $(4)\;\;\;F_{\theta} \;\;=\;\; 2x+2y+z-84 \;=\; 0$

I found: . $(5)\;\theta\,=\, -\frac{yz}{2}\,\text{ and }\,(6)\;\theta \,=\, -\frac{xz}{2}$, getting $y\,=\,x$ . Right!

I proceeded to find $z\,=\, -2y$ .This is wrong.

From (3), we have: . $\theta \:=\:-xy$

Equate to (6): . $-xy \:=\:-\frac{xz}{2} \quad\Rightarrow\quad z \:=\:2y$