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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 initialvalue 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.
 Fall '1
 Staff
 Calculus

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