Answer:
Value of x maximising profit : x = 5
Explanation:
Cost : C(x) = x^3 - 6x^2 + 13x + 15 ; Revenue: R(x) = 28x
Profit : Revenue - Cost = R(x) - C(x)
28x - [x^3 - 6x^2 + 13x + 15] = 28x - x^3 + 6x^2 - 13x - 15
= - x^3 + 6x^2 + 15x - 15
To find value of 'x' that maximises total profit , we differentiate total profit function with respect to x & find that x value.
dTP/dx = - 3x^2 + 12x + 15 = 0 ► 3x^2 - 12x - 15 = 0
3x^2 + 3x - 15x - 15 = 0 ► 3x (x +1) - 15 (x + 1) = 0 ► (x+1) (3x-15) = 0
x + 1 = 0 ∴ x = -1 [Rejected, production quantity cant be negative] ;
3x - 15 = 0 ∴ 3x = 15 ∴ x = 15/3 = 5
Double derivate : d^2TP/dx^2 = - 6x + 12
d^2TP/dx^2 i.e - 6x + 12 at x = 5 is -6(5) + 12 = - 30+ 12 = -8 which is negative. So profit function is maximum at x = 5
Answer:
Income statement will have an increased expense of $4.8 million and Revenue and cost of goods sold will decrease. In balance sheet the inventory will be decreased by the amount of crib toy inventory available.
Explanation:
Income Statement will show an expense of $4.8 million in this period as the cost of recall of inventory due to health hazard. Also sales and cost of goods sold will decrease by the amount of sales of crib toy in sales and by the amount of crib toys cost in cost of goods sold and will ultimately result in decrease in a gross profit of a company.
In the Balance Sheet the amount of Inventory will be decreased by the amount of crib toys available in stock.
Answer:
A) Debit of $1,445
Explanation:
Closing entries refers to the balance statements that are entered at the end of an accounting period in order to transfer the temporary account balances into permanent accounts. Based on the balances listed in the question it can be said that the closing entry to retained earnings will be Debit of $1,445. This refers to money going out of the account and can be calculated by adding all the revenue to the account and subtracting the expenses leaving $ - 1,445 thus being debit.
