On December 1, 2020, Sheridan Corporation incurs a 15-year $400000 mortgage liability in conjunction with the acquisition of an
1 answer:
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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
Answer:
Amount invested @ 5% = $18,500
Amount invested @ 4% = $34,000 - $18,500 = $15,500
Explanation:
Provided we have the following details,
Total investment = $34,000
Let amount invested @ 5% = x
Then, amount invested at 4% = $34,000 - x
Thus,
0.05x + $1,360 - 0.04x = $1,545
0.01x = $1,545 - $1,360 = $185
x = $185/0.01 = $18,500
Thus, amount invested @ 5% = $18,500
Amount invested @ 4% = $34,000 - $18,500 = $15,500