QUESTION IMAGE
Question
the table below shows lanark’s production possibilities.
| a | b | c | d | e | f | |
|---|---|---|---|---|---|---|
| cars | 96 | 91 | 82 | 64 | 37 | 0 |
a. if lanark is producing 73 cars, it can produce approximately \\(\boxed{}\\) wheat.
b. if lanark is currently producing combination c, the cost of 48 more wheat is \\(\boxed{}\\) (click to select) .
c. if lanark is currently producing combination d, the approximate per unit cost of an additional car is \\(\boxed{}\\) (click to select
note: if necessary, round your answers to 1 decimal place. remember to round 0.05 up to 0.10.
Step1: Find wheat for 73 cars
First, identify the relevant rows: 73 cars is between combination C (82 cars, 166 wheat) and D (64 cars, 214 wheat).
Calculate the slope (opportunity cost of 1 car):
$\text{Slope} = \frac{214 - 166}{64 - 82} = \frac{48}{-18} = -\frac{8}{3} \approx -2.7$
The number of cars reduced from 82 to 73 is $82 - 73 = 9$.
Additional wheat = $9 \times \frac{8}{3} = 24$
Total wheat = $166 + 24 = 190$
Step2: Cost of 48 more wheat at C
Combination C has 166 wheat, 82 cars. 166 + 48 = 214 wheat, which is combination D (64 cars).
Cost = $82 - 64 = 18$ cars
Step3: Per unit car cost at D
Combination D has 214 wheat, 64 cars. To get 1 more car, move to combination C (82 cars, 166 wheat). The wheat lost for $82 - 64 = 18$ cars is $214 - 166 = 48$ wheat.
Per unit cost = $\frac{48}{18} \approx 2.7$ wheat per car
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a. 190
b. 18 cars
c. 2.7 wheat