Bonjour/Bonsoir à tous, Il y a un exercice en D.N.L MATHS que je n’arrive pas à résoudre :
« In a beautiful countryside there is a rectangular field (lenght : 150m; width : 80m) where transgenic corn grows. All around the field, organic beekeepers want to protect their bees from transgenic pollution. So they ask Bob, the corn grower, to decrease the dimensions of his field in order to leave a narrow stretch of land set aside. The European Union agrees to pay him a sum not exceeding 25% of the sum earned by the total field area. Let x be the width of his narrow stretch of land set aside. The purpose of the exercise is to help the farmer calculate the new dimensions of the corn field. 1) What are the extreme values possible for x ? 2) What is the percentage of field set aside when x is equal to 5m ? 10m ? 3) Now, we call A(x) the total area of the land set aside. Give the expression of A(x) and use your calculator to solve the problem graphically. 4) Using the table function of your calculator, find a range of values rounded up to 0.01m for c which would meet the conditions above. 5) Give the algebric equation which satisfies the problem conditions, and then solve it. »
Mon travail : 1) In lenght, the maximal value possible for x is 75. In width, 40. 2) ?? 3) ?? 4) ?? 5) ??
Pourriez-vous m’aider s’il vous plaît ? J’ai vraiment besoin de comprendrendre. Merci d’avance a celui qui répondra.
Lista de comentários
Verified answer
Bonjour,
2)
Si x = 5 m : L = 150 - 2*5 = 140 m et l = 80 - 2*5 = 70 m
donc aire = L x l = 140 x 70 = 9800 m²
soit en pourcentage par rapport à l'aire initiale du champ :
9800/150x80 ≈ 81,66 %
De même, si x = 10 : pourcentage de 130x60/150x80 = 65 %
3) A(x) est l'aire de la zone non cultivée. Donc A(x) = aire totale du champ - aire cultivée, soit :
A(x) = 150x80 - (150 - 2x)(80 - 2x)
⇔ A(x) = 12000 - (12000 - 300x - 160x + 4x²)
⇔ A(x) = -4x² + 460x
4) il faut utiliser la fonction table de ta calculatrice pour trouver A(x) = 25% x 12000 = 3000
5) on doit rédoudre : -4x² + 460x = 3000
⇔ 4x² - 460x + 3000 = 0
⇔ x² - 115x + 750 = 0
Δ = (-115)² - 4x1x750 = 10225
donc 2 solutions :
x = (115 - √10225)/2 ≈ 6,94 m
ou x = (115 + √10225)/2 ≈ 108 donc solution à éliminer car > 40