Science:Math Exam Resources/Courses/MATH200/December 2013/Question 03 (a)

MATH200 December 2013

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Question 03 (a)
Find the minimium of the function $\displaystyle f(x,y,z)=(x-2)^{2}+(y-1)^{2}+z^{2}$ subject to the constraint $\displaystyle x^{2}+y^{2}+z^{2}=1$ using the method of Lagrange multipliers.

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Hint
The method of Lagrange multipliers requires you to determine the critical points of a function F having the form $\displaystyle F(x,y,z,\lambda )=f(x,y,z)+\lambda g(x,y,z)$ where f(x,y,z) is the function that you are trying to optimize and your constraint is expressed as g(x,y,z) = 0.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Using Lagrange multipliers we define the function F where $\displaystyle F(x,y,z,\lambda )=f(x,y,z)+\lambda (x^{2}+y^{2}+z^{2}-1).$ Determining the critical points of this function gives a set of points where the minima of f(x,y,z) occur subject to the given constraint. Computing the appropriate partial derivatives and equating them to zero gives: ${\begin{aligned}F_{x}&=2(x-2)+2\lambda x=0&(1)\\F_{y}&=2(y-1)+2\lambda y=0&(2)\\F_{z}&=2z+2\lambda z=0&(3)\\F_{\lambda }&=x^{2}+y^{2}+z^{2}-1=0&(4)\end{aligned}}$ Solving (3) gives $z=0$ or $\lambda =-1$. Considering the case $\lambda =1$ first, we can try to solve for x using (1) giving us $2x-4-2x=0,\quad \Rightarrow \quad -4=0\,?$ ..but this is clearly inconsistent. Thus while $\lambda =-1$ is a solution to (3), it gives inconsistent results for at least one other equation and hence is not a valid condition for us to consider. Moving onto the case z = 0, we can solve for $\lambda$ in (1) and (2) giving us $\lambda ={\frac {2-x}{x}},\quad \lambda ={\frac {1-y}{y}}\quad \Rightarrow \quad {\frac {2-x}{x}}={\frac {1-y}{y}}.$ Solving for y in terms of x gives $y={\frac {x}{2}}.$ Plugging both the result above and z = 0 into (4) gives $x^{2}+{\frac {x^{2}}{4}}-1=0\quad \Rightarrow \quad x=\pm {\frac {2}{\sqrt {5}}}$ And hence $(x,y,z)=(2/{\sqrt {5}},1/{\sqrt {5}},0)$ and $(-2/{\sqrt {5}},-1/{\sqrt {5}},0)$ are two critical points. If we evaluate the value of f(x,y,z) at these critical points we obtain $f(2/{\sqrt {5}},1/{\sqrt {5}},0)=6-2{\sqrt {5}}\quad <\quad f(-2/{\sqrt {5}},-1/{\sqrt {5}},0)=6+2{\sqrt {5}}.$ Thus, the minimum value of f subject to the given constraint is ${\color {blue}6-2{\sqrt {5}}}$ and it occurs at the point ${\color {blue}(x,y,z)=(2/{\sqrt {5}},1/{\sqrt {5}},0)}$

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