Unformatted text preview: Next we find the critical numbers of A. A'(W) = + 8 = + 8 = + 8 A’(W) is not defined at x = 4, but 4 is not in the domain of A. Thus we don’t use x = 4 as a critical number. A’(W) = + 8 = 0 when 8 = or when 8 = 200. This gives 8(W 2  8W + 16) = 200. From this we get 8W 2  64W + 128 = 200 or 8W 2  64W  72 = 0. Factoring, we get 8(W – 9)(W + 1) = 0, which gives W = 9, and W = 1. However, 1 is not in the domain of A, so we use only W = 9. We substitute W = 9 in (W – 4)(L  8) = 50, to find L. We get (9 – 4)(L – 8) = 50, or 5(L – 8) = 50 or L  8 = 10 or L = 18. We use the first derivative test to verify W = 9 gives a minimum area. Using test point W = 5 to the left of 9 and test point W = 10 to the right of 9, we get A’(W) < 0 on (4, 9), A’(W) > 0 on (9, ), so by the first derivative test, the area of the page is minimized if W = 9. ANS : The dimensions of page are: W = 9 inches wide and L =18 inches long. W  4 W L L  8...
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 Spring '11
 Augustarainsford
 Calculus, Derivative

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