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2 years ago

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On December 1, 2020, Swifty Corporation purchased a tract of land as a factory site for $770000. The old building on the propert

y was razed, and salvaged materials resulting from demolition were sold. Additional costs incurred and salvage proceeds realized during December 2020 were as follows: Cost to raze old building $69000 Legal fees for purchase contract and to record ownership 9900 Title guarantee insurance 16400 Proceeds from sale of salvaged materials 6900 In Swifty's December 31, 2020 balance sheet, what amount should be reported as land

1 answer:

seropon [69]2 years ago

6 0

Answer:

the amount reported as land is $858,400

Explanation:

The computation of the amount reported as land is shown below;

= Purchase cost + raze old building cost + ownership cost + title guarantee cost - Proceeds from sale of salvaged materials

= $770,000 + $69,000 + $9,900 + $16,400 - $6,900

= $858,400

hence, the amount reported as land is $858,400

The same would be considered

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Your company manufactures two models of speakers, the Ultra Mini and the Big Stack. Demand for each depends partly on the price

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Answer:

p1 = $259.53 p2 = $381.20

Explanation:

1. Find the revenue function.

This is a typical income maximization problem. Therefore, the first thing we should know is what are the revenues for each product.

Recall that the revenue is given by P * Q

1.a Find the revenue of the Ultra Mini (product 1):

$R_{1} = P_{1} Q_{1}$

$R_{1} =P_{1} (100,000 - 200P_{1} + 10P_{2} )$

$R_{1} =100,000P_{1} -200P_{1} ^{2} +10P_{2}P_{1}$

1.b Find the revenue of the Big Stack (product 2):

$R_{2} = P_{2} Q_{2}$

$R_{2} =P_{2} (150,000 + 10P_{1} - 200P_{2} )$

$R_{2} = 150,000P_{1+2} +10P_{1}P_{2} -200P_{2}^{2}$

2. Find the marginal revenues.

The revenue function must be derived from the price.

For product 1, we derive from P1:

$MR_{1} = 100,000 -400P_{1} +10P_{2}$

For product 2, we derive from P2:

$MR_{2} = 150,000 + 10P_{1} - 400P_{2}$

3. Create a system of linear equations in two unknowns

With the marginal revenue functions we create a system of linear equations in two unknowns (p1 and p2) and equal 0.

$100,000 - 400P_{1} +10P_{2} = 0\\150,000 + 10P_{1} -400P_{2} = 0$

4. Resolve the previous system

4.a. To make it easier, we can rethink the terms of the system like this:

$100,000 - 400P_{1} +10P_{2} = 0$ is the same as saying:

$P_{2} = \frac{-100,000 + 400P_{1} }{10}$

And $150,000 + 10P_{1} -400P_{2} = 0$ is the same as saying:

$P_{2}=\frac{150,000+10P_{1} }{400}$

Therefore:

$\frac{-100,000 + 400P_{1} }{10} =\frac{150,000+10P_{1} }{400}$

Notice that now we only have one unknown (P1).

4.b. In order to eliminate fractionals, we can multiply both terms by 400:

$\frac{400}{10} (-100,000 + 400P_{1} ) = \frac{400}{400} (150,000 + 10P_{1} )$

$(40)(-100,000+400P_{1}) =150,000+10P_{1}$

$-4,000,000+16,000P_{1} =150,000+10P_{1}$

4.c. We solve the equation, putting numbers on one side and unknowns on the other:

$-4,000,000-150,000=10P_{1} -16,000P_{1}$

$-4,150,000=-15,990P_{1}$

$\frac{-4,150,000}{-15,990} =P_{1}$

$P_{1} = $ 259.53$

4.d. Once P1 has been identified, we replace it in any of the terms of the original system of equations (those established in 4.a).

$P_{2}= \frac{-100,000+400(259.53)}{10}$

$P_{2} = 381.20$

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Now the price of the stock at the end of the fourth year (P4) = $2.48/(0.085-0.05)

P4 = $2.48 / (0.035)

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