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Ch21 19

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Exercice 19 On donne : C (x) = 0, 012x 2 − 1, 2x + 80 C(x) = 0, 004x 3 − 0, 6x 2 + 80x + c C(0) = 3000 ⇒ C(x) = 0, 004x 3 − 0, 6x 2 + 80x + 3000 D’autre part : R (x) = 2x − 10 R(x) = x 2 − 10x Donc : B(x) = R(x) − C(x) = −0, 004x 3 + 1, 6x 2 − 90x − 3000 B(x) = 0 → Solver x = 100

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Ch21 19

  • 1. Exercice 19 On donne : C (x) = 0, 012x 2 − 1, 2x + 80 C(x) = 0, 004x 3 − 0, 6x 2 + 80x + c C(0) = 3000 ⇒ C(x) = 0, 004x 3 − 0, 6x 2 + 80x + 3000 D’autre part : R (x) = 2x − 10 R(x) = x 2 − 10x Donc : B(x) = R(x) − C(x) = −0, 004x 3 + 1, 6x 2 − 90x − 3000 B(x) = 0 → Solver x = 100