Ứng dụng hình học của tích phân

Home Lớp 12 Toán lớp 12 - Sách Kết nối tri thức Bài 13. Ứng dụng hình học của tích phânBài tập nâng cao

{"common":{"save":0,"post_id":"8089","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"5968","post_id":"8089","mon_id":"1159285","chapter_id":"1159380","question":"Chi phí nhiên li\u1ec7u d\u1ef1 ki\u1ebfn C (tính b\u1eb1ng tri\u1ec7u \u0111ô la m\u1ed7i n\u0103m) khi s\u1eed d\u1ee5ng m\u1ed9t lo\u1ea1i xe t\u1ea3i c\u1ee7a m\u1ed9t công ty v\u1eadn t\u1ea3i t\u1eeb n\u0103m 2020 \u0111\u1ebfn n\u0103m 2030 là C1<\/sub> = 5,8 + 2,15t, 0 ≤ t ≤ 10, trong \u0111ó t = 0 t\u01b0\u01a1ng \u1ee9ng v\u1edbi n\u0103m 2020. N\u1ebfu công ty s\u1eed d\u1ee5ng m\u1ed9t lo\u1ea1i xe t\u1ea3i khác có \u0111\u1ed9ng c\u01a1 hi\u1ec7u qu\u1ea3 h\u01a1n thì chi phí nhiên li\u1ec7u d\u1ef1 ki\u1ebfn s\u1ebd gi\u1ea3m và tuân theo hàm mô hình C2<\/sub> = 4,5 + 2,1t, 0 ≤ t ≤ 10. Công ty có th\u1ec3 ti\u1ebft ki\u1ec7m \u0111\u01b0\u1ee3c bao nhiêu khi s\u1eed d\u1ee5ng l\u1ea1o xe t\u1ea3i v\u1edbi \u0111\u1ed9ng c\u01a1 hi\u1ec7u qu\u1ea3 h\u01a1n?<\/p>","options":["A.<\/strong> 16,5 tri\u1ec7u \u0111ô la","B.<\/strong> 15,5 tri\u1ec7u \u0111ô la","C.<\/strong> 15 tri\u1ec7u \u0111ô la","D.<\/strong> 14,5 tri\u1ec7u \u0111ô la"],"correct":"2","level":"3","hint":"","answer":"Ch\u1ecdn B.<\/strong> 15,5 tri\u1ec7u \u0111ô la<\/span><\/p>T\u1ed5ng chi phí nhiên li\u1ec7u khi công ty v\u1eadn t\u1ea3i s\u1eed d\u1ee5ng xe t\u1ea3i lo\u1ea1i th\u1ee9 nh\u1ea5t trong 10 n\u0103m là:<\/p>$S_1=\\int\\limits_0^{10}C_1dt=\\int\\limits_0^{10}(5,8+2,15t)dt=(5,8t+1,075t^2)\\bigg|_0^{10}=165,5$<\/span> (tri\u1ec7u \u0111ô la).<\/p>T\u1ed5ng chi phí nhiên li\u1ec7u khi công ty v\u1eadn t\u1ea3i s\u1eed d\u1ee5ng xe t\u1ea3i lo\u1ea1i th\u1ee9 hai trong 10 n\u0103m là:<\/p>$S_2=\\int\\limits_0^{10}C_2dt=\\int\\limits_0^{10}(4,5+2,1t)dt=(4,5t+1,05t^2)\\bigg|_0^{10}=150$<\/span> (tri\u1ec7u \u0111ô la).<\/p>V\u1eady khi s\u1eed d\u1ee5ng lo\u1ea1i xe t\u1ea3i v\u1edbi \u0111\u1ed9ng c\u01a1 hi\u1ec7u qu\u1ea3 h\u01a1n, công ty ti\u1ebft ki\u1ec7m \u0111\u01b0\u1ee3c:<\/p>165,5 – 150 = 15,5 (tri\u1ec7u \u0111ô la)<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-11-29 08:41:28","option_type":"txt","len":2},{"id":"5971","post_id":"8089","mon_id":"1159285","chapter_id":"1159380","question":"Doanh thu t\u1eeb m\u1ed9t quy trình s\u1ea3n xu\u1ea5t (tính b\u1eb1ng tri\u1ec7u \u0111ô la m\u1ed7i n\u0103m) \u0111\u01b0\u1ee3c d\u1ef1 ki\u1ebfn s\u1ebd tuân theo mô hình R = 120 + 0,06t trong 10 n\u0103m. Trong cùng kho\u1ea3ng th\u1eddi gian \u0111ó, chi phí (tính b\u1eb1ng tri\u1ec7u \u0111ô la m\u1ed7i n\u0103m) \u0111\u01b0\u1ee3c d\u1ef1 ki\u1ebfn s\u1ebd t\u0103ng theo mô hình C = 50 + 0,2t2<\/sup>, trong \u0111ó t là th\u1eddi gian (tính b\u1eb1ng n\u0103m). \u01af\u1edbc tính l\u1ee3i nhu\u1eadn trong kho\u1ea3ng th\u1eddi gian 10 n\u0103m.<\/p>","options":["A.<\/strong> 635,03 tri\u1ec7u \u0111ô la","B.<\/strong> 630,33 tri\u1ec7u \u0111ô la","C.<\/strong> 536,33 tri\u1ec7u \u0111ô la","D.<\/strong> 636,33 tri\u1ec7u \u0111ô la"],"correct":"4","level":"3","hint":"","answer":"Ch\u1ecdn D.<\/strong> 636,33 tri\u1ec7u \u0111ô la<\/span><\/p>Hàm l\u1ee3i nhu\u1eadn là P = R – C = (120 + 0,06t) – (50 + 0,2t2<\/sup>) = 70 + 0,06t – 0,2t2<\/sup>.<\/p>\u01af\u1edbc tính l\u1ee3i nhu\u1eadn trong kho\u1ea3ng th\u1eddi gian 10 n\u0103m là: <\/p>$\\int\\limits_0^{10}Pdt=\\int\\limits_0^{10}(70+0,06t-0,2t^2)dt=\\bigg(70t+0,03t^2-\\dfrac{1}{15}t^3\\bigg)\\bigg|_0^{10}$<\/span> ≈ 636,33 (tri\u1ec7u \u0111ô la).<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-11-29 08:48:01","option_type":"txt","len":2},{"id":"5975","post_id":"8089","mon_id":"1159285","chapter_id":"1159380","question":"Hàm c\u1ea7u và hàm cung c\u1ee7a m\u1ed9t s\u1ea3n ph\u1ea9m \u0111\u01b0\u1ee3c mô hình hóa b\u1edfi: hàm c\u1ea7u là $p = \u22120,35x + 8$<\/span> và hàm cung là $p = 0,15x + 3$<\/span>, trong \u0111ó x là s\u1ed1 \u0111\u01a1n v\u1ecb s\u1ea3n ph\u1ea9m, p là giá c\u1ee7a m\u1ed7i \u0111\u01a1n v\u1ecb s\u1ea3n ph\u1ea9m (tính b\u1eb1ng tri\u1ec7u \u0111\u1ed3ng). Tìm th\u1eb7ng d\u01b0 tiêu dùng \u0111\u1ed1i v\u1edbi s\u1ea3n ph\u1ea9m này.<\/p>","options":["A.<\/strong> 16,5 tri\u1ec7u \u0111\u1ed3ng","B.<\/strong> 7,5 tri\u1ec7u \u0111\u1ed3ng","C.<\/strong> 17,5 tri\u1ec7u \u0111\u1ed3ng","D.<\/strong> 10 tri\u1ec7u \u0111\u1ed3ng"],"correct":"3","level":"3","hint":"Xét ph\u01b0\u01a1ng trình hoành \u0111\u1ed9 giao \u0111i\u1ec3m c\u1ee7a hàm cung và hàm c\u1ea7u, gi\u1ea3i ph\u01b0\u01a1ng trình ta \u0111\u01b0\u1ee3c $x =x_0$<\/span> thay vào hàm ta có $p=p_0$<\/span>.<\/p>Gi\u1ea3 s\u1eed hàm cung là $p=p_1$<\/span>, hàm c\u1ea7u là $p=p_2$<\/span>.<\/p>Th\u1eb7ng d\u01b0 tiêu dùng \u0111\u01b0\u1ee3c tính b\u1eb1ng công th\u1ee9c $\\int\\limits_0^{x_0}(p_2-p_0)dx$<\/span>.<\/p>Th\u1eb7ng d\u01b0 s\u1ea3n xu\u1ea5t \u0111\u01b0\u1ee3c tính b\u1eb1ng công th\u1ee9c $\\int\\limits_0^{x_0}(p_0-p_1)dx$<\/span>.<\/p>","answer":"Ch\u1ecdn C.<\/strong> 17,5 tri\u1ec7u \u0111\u1ed3ng<\/span><\/p>Hoành \u0111\u1ed9 \u0111i\u1ec3m cân b\u1eb1ng là nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng trình: $\u22120,35x + 8$<\/span>$=0,15x + 3$<\/span> ⇔ $x =10$<\/span>. Khi \u0111ó p = 4,5.<\/p>Th\u1eb7ng d\u01b0 tiêu dùng là:<\/p>$\\int\\limits_0^{10}(-0,35x+8-4,5)dx$<\/span> $=\\int\\limits_0^{10}(-0,35x+3,5)dx$<\/span> $=\\bigg(-\\dfrac{1}{2}.0,35x^2+3,5x\\bigg)\\bigg|_0^{10}$<\/span> = 17,5 (tri\u1ec7u \u0111\u1ed3ng)<\/p>Th\u1eb7ng d\u01b0 s\u1ea3n xu\u1ea5t là:<\/p>$\\int\\limits_0^{10}(4,5-0,15x-3)dx$<\/span> $=\\int\\limits_0^{10}(1,5-0,15x)dx$<\/span> $=\\bigg(1,5x-\\dfrac{1}{2}.0,15x^2\\bigg)\\bigg|_0^{10}$<\/span> = 7,5 (tri\u1ec7u \u0111\u1ed3ng)<\/p>V\u1eady th\u1eb7ng d\u01b0 tiêu dùng \u0111\u1ed1i v\u1edbi s\u1ea3n ph\u1ea9m này là 17,5 tri\u1ec7u \u0111\u1ed3ng.<\/p>","type":"choose","extra_type":"classic","user_id":"131","test":"0","date":"2024-11-29 09:06:38","option_type":"txt","len":2}]}