Me2410x week 3 assignment
1 answer:
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Using Lagrange multipliers, we have the Lagrangian
with partial derivatives (set equal to 0)
Substituting the first three equations into the fourth allows us to solve for
:
For each possible value of, we get two corresponding critical points at 
.
At these points, respectively, we get a maximum value of
and a minimum value of 
.
with partial derivatives (set equal to 0)
Substituting the first three equations into the fourth allows us to solve for
For each possible value of
At these points, respectively, we get a maximum value of