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MAO 3-65 (3-61 i 99)

MAO 3-65 (3-61 i 99). K(x) = 20000 + 12x + 0,001x 2 & I(x) = 30x -0,002x 2. a) se graf. b) A(x) = K(x)/x =20000/x + 12 + 0,001x. Finner minimum ved å derivere og sette lik 0. A’(x) = -20000/x 2 + 0,001. A’(x) = 0 ==> -20000/x 2 + 0,001 = 0 ==> x 2 = 20.000.000 ==> x=4472.

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MAO 3-65 (3-61 i 99)

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  1. MAO 3-65 (3-61 i 99) K(x) = 20000 + 12x + 0,001x2 & I(x) = 30x -0,002x2 a) se graf b) A(x) = K(x)/x =20000/x + 12 + 0,001x Finner minimum ved å derivere og sette lik 0 A’(x) = -20000/x2 + 0,001 A’(x) = 0 ==> -20000/x2 + 0,001 = 0 ==> x2 = 20.000.000 ==> x=4472 min A(x) = A(4472) = 20000/4472 + 12 +0,001(4472) = 20,94 c) GK = K’(x) = 12 + 0,002x GK(4472) =12 + 0,002(4472) = 20, 94 GK(x) = A(x) der A(x) har sitt minimum d) O(x) = P(x) = I(x) -K(x) = 30x -0,002x2 -20000 - 12x - 0,001x2 ==> Når er O(x) > 0? O(x) = -0,003x2 +18x-20000 O(x) >0 <==> -0,003x2 18x-20000 > 0 ==> x1 = 4527,5 & x2 = 1472,5 ==> O(x) > 0 når 1472,5 < x < 4527,5 (best å bruke fortegnskjema)

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