To determine which combination of food items and attractions Leo can afford with his $15 budget, we can substitute the values of \( a \) (food items) and \( b \) (attractions) into the expression \( 4a + 2b \) and check if the total cost is less than or equal to $15.
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Option A: 3 food items and 2 attractions
- Cost: \( 4(3) + 2(2) = 12 + 4 = 16 \)
- Total: $16 (not affordable)
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Option B: 2 food items and 4 attractions
- Cost: \( 4(2) + 2(4) = 8 + 8 = 16 \)
- Total: $16 (not affordable)
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Option C: 0 food items and 8 attractions
- Cost: \( 4(0) + 2(8) = 0 + 16 = 16 \)
- Total: $16 (not affordable)
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Option D: 1 food item and 5 attractions
- Cost: \( 4(1) + 2(5) = 4 + 10 = 14 \)
- Total: $14 (affordable)
The only combination that Leo can afford with his $15 budget is Option D: 1 food item and 5 attractions.