Math 21a F99 answers for Hourly 2 practice problems1. a) L(x, y, z) = 2x + 2y + 2z - 3. b) L(x, y, z) = y + z. 2. 12 (Note: Since yr = 0 at r = 1 and s = -1, you don’t need to compute any y derivatives whenusing the Chain Rule.) 3. π + 1. (w´ = 4t tan−1(t) + 1 at t = 1). 4. ∇f = (3, 2, -4). 5. 2 6. u = 13(1, 1, 1). 7. 0. 8. 2x + 2y + z - 4 = 0. 9. x - y + 2z - 1 = 0. 10. 2√2. 11. Absolute max: 1 at (0, 0); absolute min: -5 at (1, 2). Note that this is a tricky question as youhave to check the corner points by hand. 12. Absolute max: 4 at (2, 0); absolute min: 3/√2 at (3, π/4), (3, -π/4), (1, π/4) and (1, -π/4). 13. Local min at (1, -2); saddle point at (-1, -2). 14. Length in the x-direction: 4 2 ; length in the y-direction: 3 2 . 15. (3/2, 2, 5/2). 16. (0, 0, 2) and (0, 0, -2). 17. (±4/3, -4/3, -4/3).18. −+−+30 110221222xxx() 19. L(x, y, z) = x + y - z - 1. 20. w = (b2 V/(ac))1/3; d = (c2 V/(ab))1/3; h = (a2 V/(bc))1/3 . 21. A normal vector to the surface is the gradient of xz2 - yz + cos(xy) - 1 which, at (0, 0, 1), is thevector v = (1, -1, 0). Meanwhile, r(1) = (0, 0, 1) and r´(1) = (1, 1, 1) which has zero dot productwith the normal vector v. Thus, r´(1) lies in the tangent plane to the surface. 22. Set z(x, y) = x3 + y3 - 9xy + 27 as a function on R2. Then, maxima and minima can only occurat points where ∇z is zero. Since ∇z= (3x2 - 9y, 3y2 - 9x, 0), this occurs where x2 = 3y and y2= 3x. Thus, at (x, y) = (0, 0) and (x, y) = (3, 3). The second derivative test for z´´ at the point(0, 0) finds z´´ to have only the off diagonal entries non-zero. (These equal 9). Thus, det(z´´(0,0)) < 0 so (0, 0) is a saddle. At (3, 3), det(z´´) = 243 > 0 and trace(z´´) = 36 so (3, 3) is a localminimum. 23. 1. 24. (3 ln2)/2. 25. 2. 26. (e - 2)/2. 27. 12. 28. -48 29. 16. 30.
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