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A) How many of each vehicle type should be worked on. b) Determine the maximum number of bags of cement that he should buy to meet his budget.

Mathematics

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Answers

2 parts
a

10 mabasi\text{10 mabasi}

0 coasters\text{0 coasters}

b

75 mifuko\text{75 mifuko}

Step-by-step solution

8 steps

Huu hapa ni ufumbuzi wa maswali yako:

Part a· 5 steps

Idadi ya kila aina ya gari zinazopaswa kutumika

  1. Tambua vigezo na tengeneza milinganyo.

    Kuna abiria 700 wanaohitaji kusafirishwa.
    Basi moja hubeba abiria 70.
    Coaster moja hubeba abiria 35.
    Kuna madereva 10, na kila gari linahitaji dereva mmoja.
    Acha BB iwe idadi ya mabasi na CC iwe idadi ya coasters.

    Mlinganyo wa abiria:

    70B+35C=70070B + 35C = 700

    Mlinganyo wa madereva (idadi ya magari haiwezi kuzidi idadi ya madereva):

    B+C≤10B + C \le 10

    Pia, idadi ya magari haiwezi kuwa hasi:

    B≥0,C≥0B \ge 0, \quad C \ge 0
  2. Rahisisha mlinganyo wa abiria na utatue kwa CC.

    Gawanya mlinganyo wa abiria kwa 35:

    70B35+35C35=70035\frac{70B}{35} + \frac{35C}{35} = \frac{700}{35} 2B+C=202B + C = 20

    Tenga CC:

    C=20−2BC = 20 - 2B
  3. Badilisha CC kwenye mlinganyo wa madereva na utatue kwa BB.

    B+(20−2B)≤10B + (20 - 2B) \le 10 20−B≤1020 - B \le 10 −B≤10−20-B \le 10 - 20 −B≤−10-B \le -10 B≥10B \ge 10
  4. Tumia masharti ya BB na CC kuwa chanya.

    Kutoka C=20−2BC = 20 - 2B, na kwa kuwa C≥0C \ge 0:

    20−2B≥020 - 2B \ge 0 20≥2B20 \ge 2B B≤10B \le 10

    Kwa kuwa B≥10B \ge 10 na B≤10B \le 10, thamani pekee inayowezekana kwa BB ni 10.

  5. Pata thamani ya CC.

    Badilisha B=10B=10 kwenye C=20−2BC = 20 - 2B:

    C=20−2(10)C = 20 - 2(10) C=20−20C = 20 - 20 C=0C = 0

    Hivyo, kampuni inapaswa kutumia mabasi 10 na coasters 0.

    Idadi ya mabasi: 10 mabasi\boxed{\text{10 mabasi}}
    Idadi ya coasters: 0 coasters\boxed{\text{0 coasters}}

Part b

75 mifuko\text{75 mifuko}

Idadi kubwa zaidi ya mifuko ya saruji anayopaswa kununua

  1. Tambua gharama kwa kila mfuko na bajeti.

    Gharama ya mfuko mmoja wa saruji = 20,000.
    Bajeti ya mkandarasi = 1,500,000.

  2. Tengeneza mlinganyo wa kutafuta idadi kubwa zaidi ya mifuko.

    Acha NN iwe idadi ya mifuko ya saruji.
    Gharama ya jumla ya mifuko haiwezi kuzidi bajeti:

    N×20,000≤1,500,000N \times 20,000 \le 1,500,000
  3. Tatua kwa NN.

    N≤1,500,00020,000N \le \frac{1,500,000}{20,000} N≤1502N \le \frac{150}{2} N≤75N \le 75

    Idadi kubwa zaidi ya mifuko ya saruji anayoweza kununua ni 75.

    Idadi kubwa zaidi ya mifuko ya saruji: 75 mifuko\boxed{\text{75 mifuko}}

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