Science:Math Exam Resources/Courses/MATH200/April 2012/Question 05 (b)/Solution 1

From our answer in part (a), we found that there were two critical points of the objective function,

$\displaystyle \Lambda (x,y,z)=f(x,y,z)+\mu (x+y+z-2)+\nu (x^{2}+y^{2}+z^{2}-2),$

on C. Those critical points corresponded to the critical points of the function ƒ(x,y,z) = z restricted to the curve C. The values of z corresponding to those critical points were z = 0 and z = 4/3. In contrast to part (a) where we looked for the larger of the two, we are now interested in the smaller value z = 0.

To find the corresponding x and y value, recall the condition x = y from part (a). Plugging this into the equation for the line, we obtain

$x+x+0=2\quad \rightarrow \quad x=y=1$

Therefore, the lowest point on C is (1,1,0).