Additional Information: Net income for 2020 was $26,890. Land was sold for cash at a loss of $2,000. Cash dividends were paid du
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:
Activity Rates are:
1 = $14.55 per activity
2 = $8.69 per activity
3 = $57.47 per activity
Cost per product
A = $32.2525
B= $34.7333
Explanation:
As for the provided information,
There are three activities.
Activity 1 = $17,460 and total = 1,200
Rate of activity = $17,460/1,200 = $14.55 per activity
Activity 2 = $19,987 and total activity = 2,300
Rate of activity = $19,987/2,300 = $8.69 per activity
Activity 3 = $29,884 and total activity = 520
Rate of activity = $29,884/520 = $57.47 per activity.
Costs of each product
Product A = ($14.55 600) + ($8.69
1,700) + ($57.47
40)
= $8,730 + $14,773 + $2,298
= $25,801
Cost per unit = $25,802/800 = $32.2525
Product B = ($14.55 600) + ($8.69
600) + ($57.47
120)
= $8,730 + $5,214 + $6,896
= $20,840
Cost per unit = $20,840/600 = $34.73