MA 1004 D-95 Sample Exam 3 Name here

Show your work in the space provided. Unsupported answers may not receive full credit. Use the backs of the pages if needed. (In the real exam, there will be plenty of space for your work.)

1. Sketch the cross sections for x=1 and y=-1 of the function . You may restrict your attention to the rectangle and .

   Answer: (partial)
   and .

2. Compute the two first-order and four second-order partial derivatives of the function .

   Answer:

   

   

   

   

   

   

3. Use the definition

   

   to compute the directional derivative of the function at the point in the direction of the vector .

   Answer:
   a = 1, b=2, and the unit vector in the direction of is so we have

   

   or

   

   So

   

4. Find the equation for the tangent plane, or best linear approximation, to the function at the point x=0, y=0.

   Answer:

   

5. Show that the gradient of a function at a particular point is perpendicular to the level curve through this same point.

   Answer:
   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.

6. Find the critical points of the function .

   Answer:
   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.

7. Find the absolute minimum and maximum values of the function on the square domain , .

   Answer:
   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 .

8. A rectangular box of fixed volume is to be constructed out of material that costs for the sides and for the top and bottom. Write down the cost function for this box as a function of the three dimensions x, y, and z of the box. Using the method of Lagrange multipliers, set up, but do not solve, the equations for determining the box dimensions that would minimize the cost.

   Answer:
   

   

   and the constraint is xyz=32, so the function to be minimized is

   

   so the equations to be solved to minimize the cost are

   

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