# To find: The alternate solution to the exercise with the help of Lagrange Multiplier. x+2y+3z=6

To find: The alternate solution to the exercise with the help of Lagrange Multiplier. $x+2y+3z=6$
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Method used: In order to find the maximum and minimum values of a function f(x, y, z) that is subjected to the constraint$g\left(x,y,z\right)=k$ such that $▽g\ne 0$

a) Find the values x, y, z and $\lambda$ when $▽f\left(x,y,z\right)=\lambda ▽g\left(x,y,z\right)$ and $g\left(x,y,z\right)=k$

b) Compute the values at the points from step a and find the maxomum and minimum values of f.

Calculation: It is given that the plane is $x+2y+3z=6$ that lies in the first octant. Consider a point (x,y,z) on the plane. Then, the function becomes $f\left(x,y,z\right)=xyz$. Here, $g\left(x,y\right)=x+2y+3z$.

Fint Thus, the equations are obtained as follows. ${f}_{x}\left(x,y,z\right)=\lambda {g}_{x}\left(x,y,z\right)$
$yz=\lambda \left(1\right)$
$yz=\lambda$ Solve the equation ${f}_{y}\left(x,y,z\right)=xz$ as shown below. ${f}_{y}\left(x,y,z\right)=\lambda {g}_{y}\left(x,y,z\right)$\
$xz=\lambda \left(2\right)$
$xz=2\lambda$
$\frac{xz}{2}=\lambda$ Solve the equation ${f}_{z}\left(x,y,z\right)=xy$ as follows. ${f}_{z}\left(x,y,z\right)=\lambda {g}_{z}\left(x,y,z\right)$
$xy=\lambda \left(3\right)$
$xy=3\lambda$
$\frac{xy}{3}=\lambda$ Then, $yz=\frac{xz}{2}=\frac{xy}{3}=\lambda$ Consider two terms at a time, take $yz=\frac{xz}{2}$ that gives $yz=\frac{xz}{2}$
$y=\frac{x}{2}$