Exam 1 solutions (non-calculator portion)

# The critical points ax 212 3x2 64 x2 64 x2

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Unformatted text preview: s. The critical points: A'(x) = 2(12 – 3x2) = 6(4 – x2) 6(4 – x2) = 0 x=2 Evaluating: So x = 2 produces the maximum area. The dimensions of the rectangle with maximum area are width = 4, height = 8. 10. Solve the initial-value problem u = 4 at t = 0. This differential equation is separable, so that's how I'll approach it. Now I'll use the initial condition to evaluate C. 16C = 4 C = 1/4 So 11. Find the area of the region between the curves y = x5 and y = x3. Here is a sketch with a little rectangle judiciously inserted. y x The area contributed by that little red rectangle is , so we'll integrate . The total area (by the symmetry of the two curves) will be twice the area of the upper right portion, so we'll integrate from 0 to 1 and multiply by two....
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## This document was uploaded on 03/18/2014 for the course MATH 237 at Frostburg.

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