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Hitman42 [59]
3 years ago
12

The company makes 340 units of product P23F a year, requiring a total of 710 machine-hours, 80 orders, and 40 inspection-hours p

er year. The product's direct materials cost is $40.05 per unit and its direct labor cost is $14.35 per unit. The product sells for $121.90 per unit. According to the activity-based costing system, the product margin for product P23F is:
Business
1 answer:
kifflom [539]3 years ago
4 0

Answer:

$ 5,401.60

Explanation:

Activity cost pool

<em>Assembly</em>: $ 698,720 (Total Cost) / $ 44,000 (Total activity machine-hours)

             = $ 15.88 per machine-hour

<em>Processing Orders</em>: $ 90,763 (Total cost) / 1,900 orders = 47.77 per order

<em>Inspection</em>: $ 119,535 / 1,950 inspection-hours = $61.30 per inspection-hour

Overhead Costs:

Assembly: 15.88 per MH *710 MHs = 11,274.80

Processing Orders: $ 47.77 per order * 80 orders = $3,821.60

Inspection: $ 61.30 per IH * 40 IHs = 2,452.00

<em>Sales</em>: (340 * 121.90) = $ 41,446.00

<em>Costs</em>:

 Direct materials (340 * 40.05) = $ 13,617.00

 Direct labor (340 * $ 14.35) = $ 4,879.00

 Assembly $ 11,274.80

 Processing $ 3,821,60

 Inspection $ 2,452.00

 <em>Total of costs</em>: $ 36,044.40

<em>Product Margin</em>:  (Costs - Sales) $ 5,401.60

                             

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a. The estimated coefficient for size is approximately <u>13.81</u>.

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Estimated coefficient for size = Standard Error of size * t-Stat of size =  1.2072436 * 11.439 = 13.81

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b. How many predictors (independent variables) were used in the regression?

Independent variables can be described as variables that are changed or manipulated in order to measure the effect of their changes on the dependent variable. Independent variables are therefore also called predictors because they employed to predict the dependent variable.

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An oil refinery is located on the north bank of a straight river that is 3km wide. A pipeline is to be constructed from the refi
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$6,598,076.21

Explanation:

<h2>THE KEY IS TO FIND OUT THE COST FUNCTION, the calculations are very easy!!!</h2><h2></h2><h3>In order to find the cost function, take a look at the drawing attached. </h3>

We can see the river (sort of) that is 3 km wide and the storage tanks on the other side of the river 8 km apart.

<h3 />

Laying pipes under (across) the river costs 1,000,000 the km & laying pipes over land costs 500,000 per km.

<h3 /><h3>So basically the cost function is 1,000,000 multiplied by something plus 500,000 multiplied by another something.</h3><h3 />

The distance across the river can be found by using Pythagoras Theorem. A side is 3 km the other is unknown, so we call it X. And it is equal to:

\sqrt{3^{2} +x^{2}}=\\\sqrt{9 +x^{2}}

And we multiply it by 1,000,000; the cost of laying pipe under the river, the we get:

1000000\sqrt{9+x^{2}

The distance over the land is (8-x), as we can see in the drawing. So we multiply it by its cost, 500,000. And we get 500,000(8-x).

So the cost function f(x) would be:

f(x)=1000000\sqrt{9+x^{2}} + 500000(8-x)

<h2>From here, we just have to differentiate and the derivative found must be equal to zero in order to minimize cost. </h2><h3>The value of x when the derivative is zero is plugged in the original function to get the cost.</h3><h3 /><h2>LET'S DO THIS</h2>

f(x)=1000000\sqrt{9+x^{2}} + 500000(8-x)\\f(x)=1000000(9+x^{2})^{1/2}+4000000-500000x\\f'(x)=\frac{1}{2} 1000000(9+x^{2})^{-1/2}(2x)-500000\\\\f'(x)=\frac{1000000x}{\sqrt{9+x^2}}  - 500000

<h2>f'(x)=0</h2>

f'(x)=\frac{1000000x}{\sqrt{9+x^2}}  - 500000=0\\\frac{1000000x}{\sqrt{9+x^2}}  = 500000\\\frac{2x}{\sqrt{9+x^2}}  = 1\\2x={\sqrt{9+x^2}}\\4x^2=9+x^2\\3x^2=9\\x^2=3\\x=\sqrt{3} \\

And we plug square root of 3 in the original cost function  ad we get

f(\sqrt{3} )=1000000\sqrt{9+x^{2}} + 500000(8-x)\\f(\sqrt{3})=1000000\sqrt{9+(\sqrt{3} )^{2}} + 500000(8-(\sqrt{3}))\\f(\sqrt{3})=1000000\sqrt{9+3} + 500000(8-(\sqrt{3}))\\f(\sqrt{3})=1000000\sqrt{12}+500000(6.27)\\f(\sqrt{3})=1000000(3.46)+500000(6.27)\\f(\sqrt{3})=3464101.62+3133974.60\\f(\sqrt{3})=6598076.21\\

<h2>so the minimal cost is $6,598,076.21</h2><h2 /><h3 />

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