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yaroslaw [1]
3 years ago
13

On January 2, 2015, Moser, Inc., purchased equipment for $100,000. The equipment was expected to have a $10,000 salvage value at

the end of its estimated six-year useful life. Straight-line depreciation has been recorded. Before adjusting the accounts for 2019, Moser decided that the useful life of the equipment should be extended by three years and the salvage value decreased to $8,000.
a. Prepare a journal entry to record depreciation expense on the equipment for 2019. Round your answer to the nearest dollar.
b. What is the book value of the equipment at the end of 2019 (after recording the depreciation expense for 2019)?
Business
1 answer:
OLEGan [10]3 years ago
4 0

Answer:

a. Debit Depreciation expense $6,400

   Credit Accumulated depreciation $6,400

b. $33,600

Explanation:

Depreciation is the systematic allocation of cost to an asset. It is given as

Depreciation =  (Cost - salvage value)/estimated life

When accumulated over time, it is known as accumulated depreciation which is deducted from the cost to get the carrying amount of the asset.

Depreciation

= (100000 - 10000)/6

=$15,000

Between 2015 and start of 2019 is 4 years hence

accumulated depreciation at start of 2019

= $15,000 × 4

= $60,000

Net book value  = $100,000 - $60,000

= $40,000

If the asset life is to be extended by 3 years, the remaining useful life changes from 2 to 5 years.

New depreciation rate

= (40,000 - 8000)/5

= $6,400

To record this for 2019,

Debit Depreciation expense $6,400

Credit Accumulated depreciation $6,400

The book value of the equipment at the end of 2019

= $40,000 -  $6,400

= $33,600

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Answer:

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At 1.5 the average total cost is a minimum and the total cost will be 10.25 where the TC = total cost; x = output (in 1,000 units)

<h3>What are maxima and minima?</h3>

Maxima and minima of a function are the extremes within the range, in other words, the maximum value of a function at a certain point is called maxima and the minimum value of a function at a certain point is called minima.

We have a total cost function:

TC = x³/3 - x² + 11x

Average total cost:

\rm F\left(x\right)=\dfrac{\left(\dfrac{x}{3}^3-\:x^2\:+\:11x\right)}{x}

Differentiating with respect to x and equating to x

d(F(x))/dx = 0

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F'(1.5) >  0

At 1.5 the average total cost is a minimum.

Thus, at 1.5 the average total cost is a minimum and the total cost will be 10.25 where the TC = total cost; x = output (in 1,000 units)

Learn more about the maxima and minima here:

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