A hypothetical car manufacture company makes its best selling car at its two plants in Windhoek. At the first plant, the unit cost is N$ 3.60, and the fixed cost is N$ 1260. At the second plant, the unit cost is N$ 3.30, and the fixed cost is N$ 2637. The company wants the plants to produce a combined total of 900 cars, but the total cost at each plant must be the same. How many cars should be produced at each plant?

Question

A hypothetical car manufacture company makes its best selling car at its two plants in Windhoek. At the first plant, the unit cost is N$ 3.60, and the fixed cost is N$ 1260. At the second plant, the unit cost is N$ 3.30, and the fixed cost is N$ 2637. The company wants the plants to produce a combined total of 900 cars, but the total cost at each plant must be the same. How many cars should be produced at each plant?

Answer 1

To determine how many cars should be produced at each plant, we need to achieve equal total costs at both plants while producing a combined total of 900 cars.

Let's assume the number of cars produced at the first plant is x, and the number of cars produced at the second plant is y.

The unit cost can be multiplied by the number of cars produced to find the cost of production at each plant. So, the total cost at the first plant (C1) can be calculated as:

C1 = x * 3.60

Similarly, the total cost at the second plant (C2) can be calculated as:

C2 = y * 3.30

Adding the fixed costs to the total costs, we get:

C1 = x * 3.60 + 1260
C2 = y * 3.30 + 2637

Since the total cost at each plant must be the same, we can set C1 equal to C2 and solve for x and y:

x * 3.60 + 1260 = y * 3.30 + 2637

Now, we know that the combined total of cars produced should be 900. Therefore, x + y = 900.

Now, we have two equations:

x * 3.60 + 1260 = y * 3.30 + 2637 [1]
x + y = 900 [2]

To solve this system of equations, we can use substitution or elimination method. Let's solve using substitution:

From equation [2], we can write:
x = 900 - y

Substituting this value of x in equation [1], we get:
(900 - y) * 3.60 + 1260 = y * 3.30 + 2637

Simplifying:
3240 - 3.60y + 1260 = 3.30y + 2637
4500 - 1260 = 3.30y + 3.60y - 2637
3240 = 6.90y - 2637
3240 + 2637 = 6.90y
5877 = 6.90y

Now, divide both sides by 6.90:
5877/6.90 = y
852.17 = y

Since we cannot have a fraction of a car, it's clear that y should be 852 cars.

Now, substitute the value of y back into equation [2], we can find the value of x:

x + 852 = 900
x = 900 - 852
x = 48

Therefore, the optimal solution is to produce 48 cars at the first plant and 852 cars at the second plant to achieve equal total costs while producing a combined total of 900 cars.