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A student must purchase two types of tablets, Citizen (C) and Dexone (D), where D = 2C. Given a total budget of 5,000 and a minimum currency denomination of 100, investigate two different pricing scenarios (200/100 vs 100/200) to determine the maximum number of each tablet that can be purchased.

Mathematics

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Hapa kuna uchambuzi wa hali mbili za bei:

Kuna bajeti ya jumla ya 5,000.
Idadi ya kompyuta kibao za Dexone (nDn_D) ni mara mbili ya idadi ya kompyuta kibao za Citizen (nCn_C), yaani nD=2nCn_D = 2n_C.
Tunahitaji kupata idadi kubwa zaidi ya kila kompyuta kibao (nCn_C na nDn_D) kwa kila hali.

Hali ya 1: Bei ya Citizen (C) ni 200, Bei ya Dexone (D) ni 100

  1. Andika mlinganyo wa gharama kulingana na bajeti.

    Gharama ya nCn_C Citizen tablets ni 200nC200n_C.
    Gharama ya nDn_D Dexone tablets ni 100nD100n_D.
    Jumla ya gharama lazima iwe chini au sawa na bajeti:

    200nC+100nD≤5000200n_C + 100n_D \le 5000
  2. Badilisha nD=2nCn_D = 2n_C kwenye mlinganyo wa gharama.

    200nC+100(2nC)≤5000200n_C + 100(2n_C) \le 5000 200nC+200nC≤5000200n_C + 200n_C \le 5000 400nC≤5000400n_C \le 5000
  3. Tatua kwa nCn_C.

    nC≤5000400n_C \le \frac{5000}{400} nC≤12.5n_C \le 12.5

    Kwa kuwa idadi ya kompyuta kibao lazima iwe namba kamili, idadi kubwa zaidi ya nCn_C ni 12.

  4. Kokotoa nDn_D kwa kutumia nC=12n_C = 12.

    nD=2nC=2×12=24n_D = 2n_C = 2 \times 12 = 24

    Gharama ya jumla: 200(12)+100(24)=2400+2400=4800200(12) + 100(24) = 2400 + 2400 = 4800. Hii iko ndani ya bajeti.
    Kwa Hali ya 1, idadi kubwa zaidi ya kompyuta kibao za Citizen ni 12\boxed{12} na za Dexone ni 24\boxed{24}.

    Hali ya 2: Bei ya Citizen (C) ni 100, Bei ya Dexone (D) ni 200

  5. Andika mlinganyo wa gharama kulingana na bajeti.

    Gharama ya nCn_C Citizen tablets ni 100nC100n_C.
    Gharama ya nDn_D Dexone tablets ni 200nD200n_D.
    Jumla ya gharama lazima iwe chini au sawa na bajeti:

    100nC+200nD≤5000100n_C + 200n_D \le 5000
  6. Badilisha nD=2nCn_D = 2n_C kwenye mlinganyo wa gharama.

    100nC+200(2nC)≤5000100n_C + 200(2n_C) \le 5000 100nC+400nC≤5000100n_C + 400n_C \le 5000 500nC≤5000500n_C \le 5000
  7. Tatua kwa nCn_C.

    nC≤5000500n_C \le \frac{5000}{500} nC≤10n_C \le 10

    Kwa kuwa idadi ya kompyuta kibao lazima iwe namba kamili, idadi kubwa zaidi ya nCn_C ni 10.

  8. Kokotoa nDn_D kwa kutumia nC=10n_C = 10.

    nD=2nC=2×10=20n_D = 2n_C = 2 \times 10 = 20

    Gharama ya jumla: 100(10)+200(20)=1000+4000=5000100(10) + 200(20) = 1000 + 4000 = 5000. Hii inalingana kabisa na bajeti.
    Kwa Hali ya 2, idadi kubwa zaidi ya kompyuta kibao za Citizen ni 10\boxed{10} na za Dexone ni 20\boxed{20}.

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