Apply Lagrange Multipliers:

1) Objective: Minimize Function {Distance From Point (1, 1,1)}:

\(\displaystyle{D}={\left({\left({x}-{1}\right)}^{{2}}+{\left({y}-{1}\right)}^{{2}}+{\left({z}-{1}\right)}^{{2}}\right)}^{{{\frac{{{1}}}{{{2}}}}}}\)

2) Constraint Surface:

\(\displaystyle{x}+{2}\cdot{y}+{3}\cdot{z}={13}\)\to g(x,y,z)=x+2\cdot y+3\cdot z-13ZSK

3) Formulate Lagrange Multiplier Condition (for constant scalarmultiplier "\(\displaystyle\lambda\)"):

\(\displaystyle\triangle{d}{o}{w}{n}{D}=\lambda\cdot\triangle{d}{o}{w}{n}{g}\)

\(\displaystyle\to{\left({\frac{{{1}}}{{{2}}}}\cdot{D}^{{{\frac{{-{1}}}{{{2}}}}}}\cdot{\left[{2}\cdot{\left({x}-{1}\right)}\cdot{i}+{2}\cdot{\left({y}-{1}\right)}\cdot{j}+{2}\cdot{\left({z}-{1}\right)}\cdot{k}\right]}\right.}\)

\(\displaystyle=\lambda\cdot{\left[{\left({1}\right)}\cdot{i}+{\left({2}\right)}\cdot{j}+{\left({3}\right)}\cdot{k}\right]}\)

4) Determine & Solve Simultaneous Equations:

\(\displaystyle{\left({x}-{1}\right)}={\left({1}\right)}\cdot\lambda\cdot{D}^{{{\frac{{{1}}}{{{2}}}}}}\)

\(\displaystyle{\left({y}-{1}\right)}={\left({2}\right)}\cdot\lambda\cdot{D}^{{{\frac{{{1}}}{{{2}}}}}}\)

\(\displaystyle{\left({z}-{1}\right)}={\left({3}\right)}\cdot\lambda\cdot{D}^{{{\frac{{{1}}}{{{2}}}}}}\)

\(\displaystyle\to{\left({x}-{1}\right)}={\frac{{{y}-{1}}}{{{2}}}}={\frac{{{z}-{1}}}{{{3}}}}\)

\(\displaystyle\to{\left[{y}={2}\cdot{x}-{1}\right]}\) and \(\displaystyle{\left[{z}={3}\cdot{x}-{2}\right]}\)

5) Place Solutions (Eq Set #1) Into Constraint Equation &Derive "x":

\(\displaystyle{x}+{2}\cdot{y}+{3}\cdot{z}={13}\)

\(\displaystyle\to{x}+{2}\cdot{\left[{2}\cdot{x}-{1}\right]}+{3}\cdot{\left[{3}\cdot{x}-{2}\right]}={13}\)

\(\displaystyle\to{x}+{4}\cdot{x}-{2}+{9}\cdot{x}-{6}={13}\)

\(\displaystyle\to{14}\cdot{x}={21}\)

\(\displaystyle\to{x}={\left({\frac{{{3}}}{{{2}}}}\right)}\)

6) Place "x" Value Into Eq Set #1 To Complete (x, y, z)Solution:

\(\displaystyle{\left({x}={\frac{{{3}}}{{{2}}}}\right)}\)

\(\displaystyle{y}={2}\cdot{\left({\frac{{{3}}}{{{2}}}}\right)}-{1}={\left({2}\right)}\)

\(\displaystyle{z}={3}\cdot{\left({\frac{{{3}}}{{{2}}}}\right)}-{2}={\frac{{{5}}}{{{2}}}}\)

1) Objective: Minimize Function {Distance From Point (1, 1,1)}:

\(\displaystyle{D}={\left({\left({x}-{1}\right)}^{{2}}+{\left({y}-{1}\right)}^{{2}}+{\left({z}-{1}\right)}^{{2}}\right)}^{{{\frac{{{1}}}{{{2}}}}}}\)

2) Constraint Surface:

\(\displaystyle{x}+{2}\cdot{y}+{3}\cdot{z}={13}\)\to g(x,y,z)=x+2\cdot y+3\cdot z-13ZSK

3) Formulate Lagrange Multiplier Condition (for constant scalarmultiplier "\(\displaystyle\lambda\)"):

\(\displaystyle\triangle{d}{o}{w}{n}{D}=\lambda\cdot\triangle{d}{o}{w}{n}{g}\)

\(\displaystyle\to{\left({\frac{{{1}}}{{{2}}}}\cdot{D}^{{{\frac{{-{1}}}{{{2}}}}}}\cdot{\left[{2}\cdot{\left({x}-{1}\right)}\cdot{i}+{2}\cdot{\left({y}-{1}\right)}\cdot{j}+{2}\cdot{\left({z}-{1}\right)}\cdot{k}\right]}\right.}\)

\(\displaystyle=\lambda\cdot{\left[{\left({1}\right)}\cdot{i}+{\left({2}\right)}\cdot{j}+{\left({3}\right)}\cdot{k}\right]}\)

4) Determine & Solve Simultaneous Equations:

\(\displaystyle{\left({x}-{1}\right)}={\left({1}\right)}\cdot\lambda\cdot{D}^{{{\frac{{{1}}}{{{2}}}}}}\)

\(\displaystyle{\left({y}-{1}\right)}={\left({2}\right)}\cdot\lambda\cdot{D}^{{{\frac{{{1}}}{{{2}}}}}}\)

\(\displaystyle{\left({z}-{1}\right)}={\left({3}\right)}\cdot\lambda\cdot{D}^{{{\frac{{{1}}}{{{2}}}}}}\)

\(\displaystyle\to{\left({x}-{1}\right)}={\frac{{{y}-{1}}}{{{2}}}}={\frac{{{z}-{1}}}{{{3}}}}\)

\(\displaystyle\to{\left[{y}={2}\cdot{x}-{1}\right]}\) and \(\displaystyle{\left[{z}={3}\cdot{x}-{2}\right]}\)

5) Place Solutions (Eq Set #1) Into Constraint Equation &Derive "x":

\(\displaystyle{x}+{2}\cdot{y}+{3}\cdot{z}={13}\)

\(\displaystyle\to{x}+{2}\cdot{\left[{2}\cdot{x}-{1}\right]}+{3}\cdot{\left[{3}\cdot{x}-{2}\right]}={13}\)

\(\displaystyle\to{x}+{4}\cdot{x}-{2}+{9}\cdot{x}-{6}={13}\)

\(\displaystyle\to{14}\cdot{x}={21}\)

\(\displaystyle\to{x}={\left({\frac{{{3}}}{{{2}}}}\right)}\)

6) Place "x" Value Into Eq Set #1 To Complete (x, y, z)Solution:

\(\displaystyle{\left({x}={\frac{{{3}}}{{{2}}}}\right)}\)

\(\displaystyle{y}={2}\cdot{\left({\frac{{{3}}}{{{2}}}}\right)}-{1}={\left({2}\right)}\)

\(\displaystyle{z}={3}\cdot{\left({\frac{{{3}}}{{{2}}}}\right)}-{2}={\frac{{{5}}}{{{2}}}}\)