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Difficulty: Very hardInterpreting Linear Relationships in Context

A manufacturing company uses the equation below to model the total daily cost, KK, in thousands of dollars, of operating a factory when xx units are produced:

K=0.08(x−150)+22K = 0.08(x - 150) + 22

where 150≤x≤800150 \le x \le 800. The company plans to transition to a new production setup that will increase the daily fixed operating cost by $3,000\$3,000 but will decrease the cost to produce each unit by $20\$20. Which of the following equations best models the new total daily cost, KnewK_{\text{new}}, in thousands of dollars, to produce xx units under the new setup?

  1. Knew=0.06x+13K_{\text{new}} = 0.06x + 13Answer
  2. B
    Knew=0.06x+25K_{\text{new}} = 0.06x + 25
  3. C
    Knew=0.08x+25K_{\text{new}} = 0.08x + 25
  4. D
    Knew=0.06x+10K_{\text{new}} = 0.06x + 10

Answer

The equation that best models the new total daily cost is Knew=0.06x+13K_{\text{new}} = 0.06x + 13.
The correct equation is determined by first expressing the original cost equation in slope-intercept form: K=0.08x+10K = 0.08x + 10. This reveals that the original variable cost is 0.080.08 thousand dollars ($80\$80) per unit and the original fixed cost is 1010 thousand dollars ($10,000\$10,000). Decreasing the variable cost by $20\$20 (0.020.02 thousand dollars) results in a new slope of 0.060.06. Increasing the fixed cost by $3,000\$3,000 (33 thousand dollars) yields a new y-intercept of 1313. Thus, the new cost model is Knew=0.06x+13K_{\text{new}} = 0.06x + 13.

Step-by-Step Solution

1
Expand the original equation to identify the baseline fixed and variable costs.
K=0.08x−12+22  ⟹  K=0.08x+10K = 0.08x - 12 + 22 \implies K = 0.08x + 10
Converting the equation to slope-intercept form, y=mx+by = mx + b, allows direct identification of the variable cost per unit (slope, mm) and the fixed cost (y-intercept, bb).
2
Convert the baseline parameters to their corresponding contextual values in dollars.
Variable cost: 0.080.08 thousand dollars = $80\$80 per unit. Fixed cost: 1010 thousand dollars = $10,000\$10,000 daily.
Since the cost KK is in thousands of dollars, a coefficient of 0.080.08 represents 0.08×1,000=$800.08 \times 1,000 = \$80 per unit, and a constant of 1010 represents 10×1,000=$10,00010 \times 1,000 = \$10,000.
3
Apply the described operational changes to determine the new variable and fixed costs.
New variable cost: $80−$20=$60\$80 - \$20 = \$60 per unit, or 0.060.06 thousand dollars. New fixed cost: $10,000+$3,000=$13,000\$10,000 + \$3,000 = \$13,000 daily, or 1313 thousand dollars.
The unit cost decreases by $20\$20, lowering it from $80\$80 to $60\$60. The fixed cost increases by $3,000\$3,000, raising it from $10,000\$10,000 to $13,000\$13,000.
4
Assemble the new linear equation using the updated parameters.
Knew=0.06x+13K_{\text{new}} = 0.06x + 13
Combining the new variable cost per unit (0.060.06) and the new fixed cost (1313) yields the new daily cost equation.

Key Concept

Converting and modifying linear equations in a real-world context by identifying and altering slopes and y-intercepts.
Estimated Time:3m 0s
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Topics

Interpreting Linear Relationships in Context

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