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

Method used: In order to find the maximum and minimum values of a function f(x, y, z) that is subjected to the constraint${\displaystyle g(x,y,z)=k}$ such that $▽g\ne 0$

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

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 ${\displaystyle x+2y+3z=6}$ that lies in the first octant. Consider a point (x,y,z) on the plane. Then, the function becomes $f(x,y,z)=xyz$. Here, ${\displaystyle g(x,y)=x+2y+3z}$.

Fint ${\displaystyle f_x\text{as}{f}_x(x,y,z)=yz,f_y\text{as}{f}_y(x,y,z)=xz,f_z\text{as}{f}_z(x,y,z)=xy,g_x\text{as}{g}_x(x,y,z)=1,g_y\text{as}{g}_y(x,y,z)=2\text{and}{g}_z\text{as}{g}_z(x,y,z)=3}$ Thus, the equations are obtained as follows. ${\displaystyle f_x(x,y,z)=\lambda g_x(x,y,z)}$
${\displaystyle yz=\lambda \left(1\right)}$
${\displaystyle yz=\lambda }$ Solve the equation ${\displaystyle f_y(x,y,z)=xz}$ as shown below. ${\displaystyle f_y(x,y,z)=\lambda g_y(x,y,z)}$\
${\displaystyle xz=\lambda \left(2\right)}$
${\displaystyle xz=2\lambda }$
${\displaystyle \frac{xz}{2}=\lambda }$ Solve the equation ${\displaystyle f_z(x,y,z)=xy}$ as follows. ${\displaystyle f_z(x,y,z)=\lambda g_z(x,y,z)}$
${\displaystyle xy=\lambda \left(3\right)}$
${\displaystyle xy=3\lambda }$
${\displaystyle \frac{xy}{3}=\lambda }$ Then, ${\displaystyle yz=\frac{xz}{2}=\frac{xy}{3}=\lambda }$ Consider two terms at a time, take ${\displaystyle yz=\frac{xz}{2}}$ that gives ${\displaystyle yz=\frac{xz}{2}}$
${\displaystyle y=\frac{x}{2}}$

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