In 2018, borland semiconductors entered into the transactions described below. In 2015, borland had issued 215 million shares of
its $1 par common stock at $47 per share.
Required: Assuming that Borland retires shares it reacquires, record the appropriate journal entry for each of the following transactions:
a. On january 2, 2018, borland reacquired 11 million shares at $45.00 per share.
b. On march 3, 2018, borland reacquired 11 million shares at $50 per share.
c. On august 13, 2018, borland sold 1 million shares at $55 per share.
d. On december 15, 2018, borland sold 2 million shares at $50 per share.
Required: Assuming that Borland retires shares it reacquires, record the appropriate journal entry for each of the following transactions:
a. On january 2, 2018, borland reacquired 11 million shares at $45.00 per share.
b. On march 3, 2018, borland reacquired 11 million shares at $50 per share.
c. On august 13, 2018, borland sold 1 million shares at $55 per share.
d. On december 15, 2018, borland sold 2 million shares at $50 per share.
1 answer:
Answer:
Kindly check attached picture
Explanation:
In 2018, borland semiconductors entered into the transactions described below. In 2015, borland had issued 215 million shares of its $1 par common stock at $47 per share.
Required: Assuming that Borland retires shares it reacquires, record the appropriate journal entry for each of the following transactions:
a. On january 2, 2018, borland reacquired 11 million shares at $45.00 per share.
b. On march 3, 2018, borland reacquired 11 million shares at $50 per share.
c. On august 13, 2018, borland sold 1 million shares at $55 per share.
d. On december 15, 2018, borland sold 2 million shares at $50 per share.
Kindly check attached picture for detailed explanation
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Answer:
The level of production x that will maximize the profit is: 22,966
Explanation:
C(x) = 50,000 + 100x + x³
R(x) = 3400x
P(x) = R(x) - C(x)
= 3400x - [50,000 + 100x + x³]
= 3400x - 50,000 - 100x - x³
= 3300x - 50,000 - x³ .................... (A)
P'(x) = 3300(1) - 0 - 3x²
= 3300 - 3x²
At a critical point, P'(x) = 0
∴ 0 = 3300 - 3x²
3x² = 3300
x² = 1100
x = ±
P"(x) = -6x
P() = -6 (
) < 0
by second derivative, 'P' max at x = = 33.17 (rounds)
since x = ,
recall that P(x) = 3300x - 50,000 - x³ from equation (A)
Therefore, Maximum Profit
P() = 3300
- 50000 -
= 3300(33.17) - 50,000 - 33.17³
= 109461 -50,000 - 36495.26
= 22,965.74
Maximum profit is 22,966 to the nearest whole number