Lily, a licensee, has been referring her first-time home buyers to safe insurance company for all of their insurance needs. the
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:
annual depreciation expense = $6000
so correct option is d. $6,000
Explanation:
given data
machine costing = $40,000
transportation and installation = $1,000
useful life = 6 years
residual value = $5,000
to find out
How much is the annual depreciation expense
solution
we get here first Cost of machine that is
Cost = $40,000 + $1000
cost = $41000
so depreciable base will be here
Depreciable base = cost - residual value
Depreciable base = 41000 - 5000
Depreciable base = $36,000
so annual depreciation expense is
annual depreciation expense =
annual depreciation expense = $6000
so correct option is d. $6,000