Solutions to Homework Assignment 12MATH 249Section 14.1, Page 897 Stewart 6e2, 3, 6-13, 19, 20, 21, 24, 25, 28, 29, 30, 31, 32, 39, 40, 45, 55-602. (a) f(95, 70) = 124. This means that at a temperature of 95 and a humidity of 70%, the apparenttemperature is 124. Mercy!(b) The humidity that gives an apparent temperature of 100 at an actual temperature of 90 is h = 60.(c) A temperature of 85 at 50% humidity will yield an apparent temperature of 88.(d) These are both functions of one variable. The first gives the apparent temperature as a functionof humidity when the actual temperature is 80; the second gives the apparent temperature as afunction of humidity when the actual temperature is 100. Both grow with h, but the second onegrows much faster.3. P (2L, 2K) = 1.01(2L)0.75(2K)0.25= 1.01 · 20.75+0.25L0.75K0.25= 2P (L, K). This also works for thegeneral pro duction function since the exponents on L and K sum to 1.6. (a) f(1, 1) = ln(1 + 1 − 1) = 0.(b) f(e, 1) = ln(e + 1 − 1) = 1.(c) The domain of ln t is (0, ∞), so we need x + y −1 > 0, or y > 1 −x. The domain is thus the regionof the plane ab ove the line y = 1 − x, as shown.(d) Since x + y − 1 will span the entire domain of ln t, the outputs will span the entire range of ln t,which is (−∞, ∞).9. (a) f(2, −1, 6) = e√6−22−(−1)2= e.(b) We only require that z − x2− y2≥ 0, or z ≥ x2+ y2. This means that the points in the domainlie on or ab ove the circular paraboloid z = x2+ y2.(c) Since the range ofpz − x2− y2is [0, ∞), the range of f is [1, ∞).-4-2y0x-4 -2 02442-2-102y-21-1x100z-112-22Number 6 Number 911. Here we need x + y ≥ 0, or y ≥ −x. This is the region in the plane above the line y = −x.12. We need both x ≥ 0 and y ≥ 0 or x ≤ 0 and y ≤ 0, so this is the first and third quadrants plus theorigin and axes.13. We need 9 −x2− 9y2> 0, orx29+ y2< 1. This is the inside of the ellipsex29+ y2= 1 (excluding theellipse itself).19. We need x2+ y2+ z2≤ 1, so the domain is the unit sphere together with its interior.21. This is a horizontal plane at height 3.24. This is a sinusoidal cylinder with the y-axis as its axis. Cross sections are cos x.25. This is a parabolic cylinder with the y-axis as its axis. Cross sections are 1 − x2.29. This is the top half of a cone; note that z2= x2+ y2and z ≥ 0.4y4-2-4-2x2-4 0 200-4y420-2-2-4x2 4Number 11 Number 12 Number 13-2-1y012-2-1x012-2-10z12-4-20y-442-22xz002-244-4-4-24-40-2y022zx420-24-4Number 19 Number 21 Number 24-4-204-4y-22x020z2-244-4-4-20y-44-2z20224x0-24-4Number 25 Number 2931. The graph seems to rise to a point at the origin. f (−3, 3) ≈ 55 since it is roughly midway between thez = 50 and z = 60 level curves.32. There is actually not enough information given, but I will assume that corresponding level curves areat the same height. In I, the drop in the level curves slows down as we approach the origin, so it willhave a rounded bottom and therefore be the paraboloid. In II, the dropoff is at a constant rate, so thiswill b e the cone.45. A contour k = y2− x2is a hyperbola.Number