# Find the point on the plane x+2y+3z=13 that is closest to the point (1,1,1). How would you minimize the function?

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Find the point on the plane $$\displaystyle{x}+{2}{y}+{3}{z}={13}$$ that is closest to the point (1,1,1). How would you minimize the function?

2021-04-08
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:
$$\displaystyle{\left({x}={\frac{{{3}}}{{{2}}}}\right)}$$
$$\displaystyle{y}={2}\cdot{\left({\frac{{{3}}}{{{2}}}}\right)}-{1}={\left({2}\right)}$$
$$\displaystyle{z}={3}\cdot{\left({\frac{{{3}}}{{{2}}}}\right)}-{2}={\frac{{{5}}}{{{2}}}}$$

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