Answer:
1 machine must be made to minimise the unit cost.
Step-by-step explanation:
<em>Step 1: Identify the function</em>
x is the number of machines
C(x) is the function for unit cost
C (x) = 0.5x^2-150 + 21,035
<em>Step 2: Substitute values in x to find the unit cost</em>
C (x) = 0.5x^2-150 + 21,035
The lowest value of x could be 1
To check the lowest cost, substitute x=1 and x=2 in the equation.
<u>When x=1</u>
C (x) = 0.5x^2-150 + 21,035
C (x) = 0.5(1)^2-150 + 21,035
C (x) = 20885.5
<u>When x=2</u>
C (x) = 0.5x^2-150 + 21,035
C (x) = 0.5(2)^2-150 + 21,035
C (x) = 20887
<em>We can see that when the value of x i.e. the number of machines increases, per unit cost increases.</em>
Therefore, 1 machine must be made to minimise the unit cost.
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