Science:Math Exam Resources/Courses/MATH200/December 2013/Question 03 (a)/Solution 1

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 ${\color {blue}(x,y,z)=(2/{\sqrt {5}},1/{\sqrt {5}},0)}$