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Ch_6_solutions

# Ch_6_solutions - (5-10 min S 6-1 Cost A is a variable cost...

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(5-10 min.) S 6-1 Cost A is a variable cost. The cost is constant on a per unit basis and increases in total as volume increases. Cost B is a fixed cost. The cost is constant in total, and decreases on a per unit basis as volume increases. Cost C is mixed. It is not constant in total, and it is not constant on a per unit basis. (10 min.) S 6-2 (5-10 min.) S 6-3 Total fixed cost = \$3.00 / basketball × 12,000 basketballs Total fixed cost = \$36,000 Since a volume of 15,000 basketballs is in the same relevant range, the total fixed costs will remain constant at \$36,000 Therefore, the new fixed cost per basketball will be: Fixed cost per basketball = Total fixed cost Number of basketballs produced Fixed cost per basketball = \$36,000 15,000 Fixed cost per basketball = \$2.40 (5-10 min.) S 6-4 a. Fixed y = f Where y = total fixed cost f = fixed amount over a period of time (vertical intercept)

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b. Mixed y = vx + f Where y = total mixed cost v = variable cost per unit of activity (slope) x = volume of activity f = fixed amount over a period of time (vertical intercept) c. Variable y = vx Where y = total variable cost v = variable cost per unit of activity (slope) x = volume of activity (5-10 min.) S 6-5 Find the answer to Ritter Razor’s questions by 1) finding the company’s cost equation, and then 2) using the cost equation to predict total costs at a different volume. First, determine the firm’s cost equation: Total costs = variable component + fixed component y = vx + f \$100,000 = \$40,000 = f \$60,000 = f Also, since: vx = \$40,000 when x = 20,000 v (20,000) = \$40,000 v = \$2.00 per package of razors Therefore, the production cost equation is: y = \$2.00x + \$60,000 Predict total costs for other volumes in the same relevant range using the cost equation found above: y = (\$2.00 per package × 25,000 packages) + \$60,000 y = \$50,000 + \$60,000 y = \$110,000 Total production costs are predicted to be \$110,000 when 25,000 packages of razors are produced. (5-10 min.) S 6-6 Req. 1 a. Call for 20 minutes
\$5.00 + (20 × \$0.35) \$5.00 + \$7.00 = \$12.00 b. Call for 40 minutes \$5.00 + (40 × \$0.35) \$5.00 + \$14.00 = \$19.00 c. Call for 80 minutes \$5.00 + (80 × \$0.35) \$5.00 + \$28.00 = \$33.00 Req. 2 (5-10 min.) S 6-7 a. Depreciation on equipment used to cut wood enclosures……………………………………… ....... Fixed b. Wood for speaker enclosures………………………. Variable c. Patents on crossover relays………………………… Fixed d. Crossover relays………………………………………. Variable e. Grill cloth……………………………………………….. Variable f. Glue……………………………………………………… Variable g. Quality inspector’s salary……………………………. Fixed (10-15 min.) S 6-8 Req. 1 Scatterplot of Operating expenses and number of oil changes \$0 \$5,000 \$10,000 \$15,000 \$20,000 \$25,000 \$30,000 \$35,000 \$40,000 0 500 1000 1500 2000 2500 3000 3500 4000 Number of oil changes Monthly operating expense Req. 2

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There appears to be a very strong relationship between the company’s operating expenses and the number of oil changes it performs. We can tell this because the scatterplot of the data almost falls in a straight line (it is very linear). If there were no relationship, or if the relationship was weak, the scatterplot of data points would appear almost random. Additionally, since all of the data points fall in the same linear pattern, there do not appear to be any outliers.
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