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alukav5142 [94]
3 years ago
10

On July 1, 2015, Friedman Inc. invested $717,963 in a mine estimated to have 806,700 tons of ore of uniform grade. During the la

st 6 months of 2015, 103,700 tons of ore were mined and sold.
(a) Calculate depletion cost per unit. (Round answer to 2 decimal places, e.g. 0.50.)

(b) Prepare the journal entry to record depletion expense. (Round answer to 0 decimal places, e.g. 2,125. If no entry is required, select "No entry" for the account titles and enter 0 for the amounts. Credit account titles are automatically indented when amount is entered. Do not indent manually.)

(c) Assume that the 103,700 tons of ore were mined, but only 87,000 units were sold. How are the costs applicable to the 16,700 unsold units reported?
Business
1 answer:
yaroslaw [1]3 years ago
5 0

Answer:

The computations are shown below:

Explanation:

(a) Depletion cost per unit

Depletion cost per unit

= $717,963 ÷ 806,700 tons

= $0.89 per ton

(b) The Journal entry to record depletion expense is

Depletion Expense                   A/c Dr     $ 92,293     (103,700 tons × $0.89)

          To To Accumulated Depletion      A/c   $ 92,293

(Being the depletion expense is recorded)

(c) The cost applicable is

= 16,700 unsold units × $0.89

= $14,863  

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4 years ago
Until recently, hamburgers at the city sports arena cost $ 2.50 each. The food concessionaire sold an average of 1750 hamburgers
Artyom0805 [142]

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  (A) p = -0.002x +6; 0 ≤ x ≤ 3000

  (B) R(x) = x(6 -0.002x); 0 ≤ x ≤ 3000

  (C) C(x) = 1.44x +1903

  (D) (550, 2695), (1730, 4394.20)

  (E) P(x) = -0.002x^2 +4.56x -1903

  (F) increasing at $2.16 per hamburger

Explanation:

(A) The two-point form of the equation for a line can be used.

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

The two points we have are ...

  (x, p) = {(1750, 2.50), (1450, 3.10)}

so the equation is ...

  p = (3.10 -2.50)/(1450 -1750)/(x -1750) +2.50

  p = 0.6/-300(x -1750) +2.50

  p = -0.002x +6

The domain of this function is where x and p are greater than 0. That will be for ...

  0 ≤ x ≤ 3000

__

(B) Revenue is the product of burgers sold (x) and their price (p).

  R(x) = xp

  R(x) = x(6 -0.002x)

The domain of R(x) is 0 ≤ x ≤ 3000. This is the same as the domain of p(x).

___

(C) The cost function is the sum of fixed costs and variable costs:

  C(x) = 1.44x +1903

__

(D) See the attachment for a graph of cost and revenue. The break-even points are (x, revenue) = (550, 2695), (1730, 4394.20).

__

(E) Profit is the difference between revenue and cost.

  P(x) = R(x) - C(x) = x(6 -0.002x) -(1.44x +1903)

  P(x) = -0.002x^2 +4.56x -1903

__

(F) The marginal profit is the derivative of the profit function:

  P'(x) = -0.004x +4.56

  P'(600) = -0.004(600) +4.56 = -2.40 +4.56 = 2.16

At a production level of 600, the profit is increasing at a rate of $2.16 per hamburger.

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