intracorrigé - EXERCICE 1(1 Montant de...

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EXERCICE 1 (1) Montant de la taxe à payer T (90000) = 30 , 901 . 2884 . (2) Taxe marginale T 0 ( Y ) = pba ( aY + c ) p - 1 + k (3) Taux de progression de la taxe pour un 1$ supplémentaire à partir de 90000 Δ T (90000) = T (90001) - T (90000) = T 0 (90000) = 0 . 482478 (4) Équation de la tangente P ( Y ) = T (90000) + T 0 (90000)( Y - 90000) = 30 , 901 . 24884 + 0 . 482478( Y - 90000) (5) Dérivée seconde T 00 ( Y ) = ( p - 1) pb 2 a ( bY + c ) p - 2 (6) Un raisonnement de base possible 1 < p < 2 p - 1 > 0 et 2 - p > 0 . Par conséquent Y > 0 bY + c > c 1 bY + c < 1 c 1 bY + c 2 - p < 1 c 2 - c 1 ( bY + c ) 2 - p < 1 c 2 - c ⇒ | T 00 ( Y ) | = T 00 ( Y ) = ( p - 1) pb 2 a 1 ( bY + c ) 2 - p < ( p - 1) pb 2 a 1 c 2 - p Majorant de Lagrange | T ( Y ) - P ( Y ) | = | T 00 ( Z ) ( Y - 90000) 2 2! | avec Z entre Y et 90000. Par conséquent | T ( Y ) - P ( Y ) | ≤ ( p - 1) pb 2 a c 2 - p ( Y - 95000) 2 2! ⇒ | T (95000) - P (95000) | ≤ 0 . 00008898 Autrement dit, si le gouvernement utilise l’équation de la tangente P ( Y ) pour évaluer la taxe à payer d’une personne ayant un revenu de 95000, en lieu et place de la vraie formule de calcul, T ( Y ) , il commet une erreur de mesure de 0.0000889$ ! 1
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2 EXERCICE 2 (1) Profit π ( Q ) = P v ( Q ) Q - ( P a ( Q ) + γ ) Q = ( α 1 - α 2 - γ ) Q - 1 2 Q 2 (2) Dérivée π 0 ( Q ) = ( α 1 - α 2 - γ ) - Q (3) Quantité Q ? qui annule la dérivée Q ?
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