3.The circumference of the original circle of radius 4 is 8p.Thus, 0 #x #8p.[0, 8p] by [210, 40]4.The maximum occurs at about x54.61. The maximumvolume is aboutV525.80.5.Start with }ddVx}5}23p}rh}ddxr}1}p3}r2}ddhx}. Compute }ddxr}and }ddhx},substitute these values in }ddVx}, set }ddVx}50, and solve for xtoobtain x5}8(323ˇ6w)p}<4.61. Then V5}1282p7ˇ3w}<25.80.Quick Review 4.41.y9 53x2212x11253(x22)2Since y9 $0 for all x(and y9 .0 for x2),yis increasing on (2‘,‘) and there are no local extrema.2.y9 56x216x21256(x12)(x21)y0 512x16The critical points occur at x5 22 or x51, since y9 50at these points. Sincey0(22)5 218 ,0, the graph has alocal maximum atx5 22. Since y0(1)518 .0, the graphhas a local minimum at x51. In summary, there is a localmaximum at (22, 17) and a local minimum at (1,210).3.V5}13}pr2h5}13}p(5)2(8)5}2030p}cm34.V5pr2h51000SA52prh12pr25600Solving the volume equation for hgives h5}1p0r020}.
This is the end of the preview.
access the rest of the document.