Answer:
$6.00
Explanation:
Given data
quantity demanded ( x ) ∝ 1 / p^3 for p > 1
when p = $10/unit , x = 64
initial cost = $140, cost per unit = $4
<u>Determine the price that will yield a maximum profit </u>
x = k/p^3 ----- ( 1 ). when x = 64 , p = $10 , k = constant
64 = k/10^3
k = 64 * ( 10^3 )
= 64000
back to equation 1
x = 64000 / p^3
∴ p = 40 / ∛x
next calculate the value of revenue generated
Revenue(Rx) = P(price ) * x ( quantity )
= 40 / ∛x * x = 40 x^2/3
next calculate Total cost of product
C(x) = 140 + 4x
Maximum Profit generated = R(x) - C(x) = 0
= 40x^2/3 - 140 + 4x = 0
= 40(2/3) x^(2/3 -1) - 0 - 4 = 0
∴ ∛x = 20/3 ∴ x = (20/3 ) ^3 = 296
profit is maximum at x(quantity demanded ) = 296 units
hence the price that will yield a maximum profit
P = 40 / ∛x
= ( 40 / (20/3) ) = $6
Answer:
a. $164,000
Explanation:
The computation of the Altoon Manufacturing's sales for the year until the flood is given below:
= Cash collections + ending receivables - opening receivables
= $158,000 + $25,000 - $19,000
= $164,000
hence, the Altoon Manufacturing's sales for the year until the flood is $164,000
Therefore the first option is correct
<span>To calculate the cost of goods sold we use the following formula:
beginning inventory + the cost of goods purchased or manufactured = cost of goods available ending inventory.
Since there was no beginning balance in inventory account and all goods were sold we can assume that cost of goods = total costs for the period.
Adding up all costs for the period comes to $173,000.</span>
Answer:
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Explanation:
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