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Difficulty: HardAlgebraic Word Problems and Equation Modeling

A manufacturing plant has a fixed daily setup cost of $1,500\$1,500. For a specific product, the production cost for each of the first 5050 units is $60\$60 per unit, while each additional unit produced beyond the first 5050 costs $30\$30 per unit. If the average total cost per unit produced on a given day was $45\$45, how many total units were produced on that day?

Answer: 200 units

Answer

The total number of units produced on that day was 200.
Setting up the piecewise linear total cost expression C(n)=1,500+50(60)+(n−50)30=3,000+30nC(n) = 1,500 + 50(60) + (n-50)30 = 3,000 + 30n and equating average cost 3,000+30nn\frac{3,000 + 30n}{n} to 4545 yields 15n=3,00015n = 3,000, which solves to n=200n = 200.

Step-by-Step Solution

1
Model total daily production cost as a piecewise linear algebraic equation.
C(n)=1,500+(50×60)+(n−50)×30=3,000+30nC(n) = 1,500 + (50 \times 60) + (n - 50) \times 30 = 3,000 + 30n for n>50n > 50.
Total cost combines fixed setup fees, cost of the initial 50 units, and tier-2 cost for units exceeding 50.
2
Formulate the equation for average cost per unit.
C(n)n=3,000+30nn=45\frac{C(n)}{n} = \frac{3,000 + 30n}{n} = 45
Average cost is total daily cost divided by total quantity produced, given as $45 per unit.
3
Solve the algebraic equation for n.
3,000+30n=45n  ⟹  15n=3,000  ⟹  n=2003,000 + 30n = 45n \implies 15n = 3,000 \implies n = 200
Isolating n yields the exact volume of units needed to satisfy the average cost target.

Key Concept

Algebraic Equation Modeling with Piecewise Cost Functions
Estimated Time:2m 0s
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Topics

Algebraic Word Problems and Equation Modeling

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