题目 #16793

(2)①\(y_{1}\)= 40m + 1200;\(y_{2}\)= 32m + 1600. ②当20 < m < 50时, 选择方案一总费用少; 当m = 50时, 两种方案总费用相同; 当m > 50时, 选择方案二总费用少. (1)设每副象棋的价格是x元, 每副围棋的价格是y元. 根据题意得{40x + 20y = 2600, 解得{x = 40 y −x = 10 y = 50 . 答: 每副象棋的价格是40元, 每副围棋的价格是50元; (2)①由题意得\(y_{1}\)=\(50 \times 40 + 40(m −20) = 40m + 1200\)所以\(y_{2}\)=\(0.8 \times (50 \times 40 + 40m) = 32m + 1600\).所以\(y_{1}\)= 40m + 1200,\(y_{2}\)= 32m + 1600②当\(y_{1}\)=\(y_{2}\)时，40m + 1200 = 32m + 1600,解得m = 50; 当\(y_{1}\)>\(y_{2}\)时, 40m + 1200 > 32m + 1600, 解得m > 50; 当\(y_{1}\)<\(y_{2}\)时, 40m + 1200 < 32m + 1600, 解得m < 50; ∴当20 < m < 50时, 选择方案一总费用少; 当m = 50时, 两种方案总费用相同; 当m > 50时, 选择方案二总费用少. 11.6 把 x = 2 代入 y = 12\(12 _{x}\)，得n = _{2}= 6, 故答案为6. 12.8 将点(2, m)代入 y = 2x 得, m =\(2 \times 2 = 4\)将点(2,4)代入 y =\(k_{x}\)得k =\(2 \times 4 = 8\)故答案为：8． 13.(1)a, b的值分别为: 4,6. (2)①y = {−x + 900(100 ≤x ≤120) −0.5x + 850(80 ≤x < 100) . ②m的值为: 0.5. (1)由题意得: {10a + 20b = 16020a + 10b = 140 , 解得{a = 4b = 6 . 答: a, b的值分别为: 4,6. (2)①设每天销售鲢鱼x斤, 则草鱼购进(300 −x)斤, 且80 ≤x ≤120,则180 ≤300 −x ≤220,180 ≤ 300 −x ≤200时, 即100 ≤x ≤120时,所以y = (6 −4)x + (9 −6)(300 −x) = −x + 900, 当200 < 300 − x ≤220时,即80 ≤x < 100时, ∴y = (6 −4)x + 200(9 −6) + (8.5 −6)(100 −x) = −0.5x + 850,综上: y = {−x + 900(100 ≤x ≤120) −0.5x + 850(80 ≤x < 100) . ②设每天销售鲢鱼x斤, 则草鱼购进(300 −x)斤, 则利润: y = (6 −4 − m)x + (8.5 −6)(300 −x) = (−0.5 −m)x + 750因为−0.5 −m < 0, 则y随x的增大而减少, 80 ≤x ≤ 120, ∴当x = 120时, y的最小值为630元, ∴120(−0.5 −m) + 750 = 630, 解得: m = 0.5, 答: m的值为 : 0.5. 1