MATH10550 Midterm: MATH 10550 Exam 3 Fall 2008 Solutions
Document Summary
Recall that xn+1 = xn f (xn) f (xn) Hence, x2 = 1 (1)3 + 2(1) 4. 5: a box with a square base and open top must have a volume of 32 cm3. Find the dimensions of the box that minimizes the surface area of the box. Let the edge length of the base of the box be s, and the. Figure 1 height of the box be h. the volume of the box is given by v = hs2 and the surface area is given by a = s2 + 4sh. V = hs2, we get that h = 32 s2 , and thus a = s2 + 128s 1. To nd the minimum of a(s) - the surface area, we need to nd the critical numbers of the function a(s). 0 = a 0 = 2s . 128 s2 128 = 2s3 64 = s3 s = 4, gives the critical point s = 4.
