Lecture 6

# We want to take this information and to show that 0

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Unformatted text preview: from the 3rd ∗∗ condition, will also satisfy our required constraint, , 0. But we still have a possible unresolved issue at hand – while the critical points ∗∗∗ , , may maximize or minimize (in a local sense) the function , , , how do , ? we know we can say the same regarding the original function, Let’s think about the way to approach this question. First, we want to consider how , , changes with changes in , , , i.e., We know that the solution to 1-3 above, ∗ , ∗ , ∗ is a critical point of , 0 at ∗ , ∗ , ∗ . We want to take this information and to show that 0 for all permissible changes in , ensure , , so 0 will . By permissible changes, we mean precisely those that conform to our constraint, , 0. Notice that at ∗ , ∗ , ∗ , the total derivative, can be rewritten as: ∗ ∗ , ∗ ∗ ∗ , ∗ , ∗ ∗ ∗ , ∗ This expression is obtained by solving out for the partials above. ∗ , 0 , , and , using 1 – 3 We now want to consider permissible movements in and and that conform to the constraint, , are changes in , , , ; as we noted these 0. We have: , ∗∗ which, when evaluated at , , , equals 0. Note that is the last term in the parenthesis in the above expression for . (What we are doing is adjusting and along a path that ensures , 0 for all points along the path. That is, this is the only kind of changes in the two variables we consider, since our problem places the constraint that , 0 must hold). ∗ Also note that at the critical point, that ∗ , , ∗ 0 (see 3 above). Together, these imply ∗ But this gives us our required result. It says ∗ , ∗ 0 0, meaning that ∗ , ∗ will be a critical point for the original function given that the variables must satisfy , 0 and any movement away from ∗ , ∗ must be along a path that conforms to this constraint. An intuitive look at constrained optimization (Lagrange Revisited) Suppose we have a level set are to remain on that level set, , , , : , . If we change , and we must be such that is unchanged, and equal to : , 0. This is true along any level set of the function. Therefore, | Of course, this assume...
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