MA 1004 D-95 Sample Exam 3 Name here
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to compute the directional derivative of the function at the point in the direction of the vector .
a = 1, b=2, and the unit vector in the direction of is so we have
Suppose that is a parametrization of the contour that passes through the point at t=0. Then
for some constant C. Differentiating both sides of this equation, using the chain rule, and writing the result as a dot product gives
Evaluating this at t=0 gives
and recognizing that the vector is tangent to the contour at the point in question gives the result.
Critical points are where the gradient of f vanishes, or the solutions of the equations
Eliminating x gives the single equation . Solving this equation and computing the corresponding values of x gives the two solutions x=0, y=0, and x=1, y=1.
The first step is to find the critical points.
so x=y and at the critical points. So the critical points are x=0, y=0, and x=1, y=1.
The second step is to analyze on the boundaries.
, which has a critical point a x=0.
, which has a critical point at y=1.
, which has a critical point at x=1.
, which has a critical point at y=0.
In the third step, we evaluate f at all the candidate points. Here the candidates are the four corner points , , , and . We have , , and so the minimum is -2 at and the maximum is 1 at .
and the constraint is xyz=32, so the function to be minimized is
so the equations to be solved to minimize the cost are