This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Ili kutatua swali hili, tunahitaji kwanza kutambua eneo lisilo na kivuli (unshaded region) kwenye grafu, ambalo linafafanuliwa na usawa (inequalities) tatu zilizopewa. Kisha, tutatafuta pointi zenye kuratibu kamili (integral coordinates) ndani ya eneo hili na kuhesabu thamani ya x+2y kwa kila pointi ili kupata thamani ndogo zaidi. Usawa uliotolewa ni: 1. y -1(1)/(2)x + 5(1)/(2) y -(3)/(2)x + (11)/(2) 2. y < (1)/(2)x + 3 3. y < 1(1)/(2)x + 1(1)/(2) y < (3)/(2)x + (3)/(2) Step 1: Tambua mistari ya mipaka na eneo lisilo na kivuli. • Mstari wa kwanza (L1): y = -(3)/(2)x + (11)/(2). Huu ni mstari imara (solid line) kwa sababu ya usawa . Eneo lisilo na kivuli liko juu au kwenye mstari huu. • Mstari wa pili (L2): y = (1)/(2)x + 3. Huu ni mstari wenye nukta (dashed line) kwa sababu ya usawa <. Eneo lisilo na kivuli liko chini ya mstari huu. • Mstari wa tatu (L3): y = (3)/(2)x + (3)/(2). Huu ni mstari wenye nukta (dashed line) kwa sababu ya usawa <. Eneo lisilo na kivuli liko chini ya mstari huu. Eneo lisilo na kivuli ni pembetatu ndogo iliyoko katikati ya grafu, iliyofungwa na mistari hii mitatu. Step 2: Tafuta kuratibu kamili (integral coordinates) R(x,y) ndani ya eneo lisilo na kivuli. Tunahitaji kupata pointi ambapo x na y zote ni namba kamili na zinakidhi usawa wote watatu. Hebu tuchunguze pointi za gridi zinazowezekana karibu na eneo hili. Kwa x=2: 1. y -(3)/(2)(2) + (11)/(2) y -3 + 5.5 y 2.5 2. y < (1)/(2)(2) + 3 y < 1 + 3 y < 4 3. y < (3)/(2)(2) + (3)/(2) y < 3 + 1.5 y < 4.5 Kwa x=2, tunahitaji 2.5 y < 4. Namba kamili ya y inayokidhi hili ni y=3. Hivyo, pointi (2,3) inakidhi usawa wote. Kwa x=3: 1. y -(3)/(2)(3) + (11)/(2) y -4.5 + 5.5 y 1 2. y < (1)/(2)(3) + 3 y < 1.5 + 3 y < 4.5 3. y < (3)/(2)(3) + (3)/(2) y < 4.5 + 1.5 y < 6 Kwa x=3, tunahitaji 1 y < 4.5. Namba kamili za y zinazokidhi hili ni y=1, 2, 3, 4. Hivyo, pointi (3,1), (3,2), (3,3), (3,4) zinakidhi usawa wote. Kwa x=4: 1. y -(3)/(2)(4) + (11)/(2) y -6 + 5.5 y -0.5 2. y < (1)/(2)(4) + 3 y < 2 + 3 y < 5 3. y < (3)/(2)(4) + (3)/(2) y < 6 + 1.5 y < 7.5 Kwa x=4, tunahitaji -0.5 y < 5. Namba kamili za y zinazokidhi hili ni y=0, 1, 2, 3, 4. Hivyo, pointi (4,0), (4,1), (4,2), (4,3), (4,4) zinakidhi usawa wote. Pointi zote za kuratibu kamili katika eneo lisilo na kivuli ni: (2,3), (3,1), (3,2), (3,3), (3,4), (4,0), (4,1), (4,2), (4,3), (4,4). Step 3: Hesabu thamani ya x+2y kwa kila pointi na utafute thamani ndogo zaidi. • Kwa (2,3): x+2y = 2 + 2(3) = 2 + 6 = 8 • Kwa (3,1): x+2y = 3 + 2(1) = 3 + 2 = 5 • Kwa (3,2): x+2y = 3 + 2(2) = 3 + 4 = 7 • Kwa (3,3): x+2y = 3 + 2(3) = 3 + 6 = 9 • Kwa (3,4): x+2y = 3 + 2(4) = 3 + 8 = 11 • Kwa (4,0): x+2y = 4 + 2(0) = 4 + 0 = 4 • Kwa (4,1): x+2y = 4 + 2(1) = 4 + 2 = 6 • Kwa (4,2): x+2y = 4 + 2(2) = 4 + 4 = 8 • Kwa (4,3): x+2y = 4 + 2(3) = 4 + 6 = 10 • Kwa (4,4): x+2y = 4 + 2(4) = 4 + 8 = 12 Step 4: Tambua thamani ndogo zaidi. Thamani ndogo zaidi kati ya hizi ni 4. Thamani ndogo zaidi ya (x+2y) ni 4. That's 2 down. 3 left today — send the next one.