DEMAND FUNCTION: p = 492-0.15x^2
COST FUNCTION: C = 920+87x
Find the number of items (x) that will produce maximum profits.
To find the number of items (x) that will produce maximum profits, we need to determine the profit function first. The profit (P) is calculated by subtracting the cost (C) from the revenue (R), where revenue is equal to the product of the price (p) and the number of items sold (x).
Revenue (R) = p * x
Cost (C) = 920 + 87x
Profit (P) = R - C
Now, let's substitute the given demand function p = 492 - 0.15x^2 into the revenue equation:
Revenue (R) = (492 - 0.15x^2) * x
Substituting the cost function into the profit equation:
Profit (P) = (492 - 0.15x^2) * x - (920 + 87x)
To find the value of x that maximizes the profit, we need to find the derivative of the profit function with respect to x, set it equal to zero, and solve for x.
Let's differentiate the profit function with respect to x:
P' = [(492 - 0.15x^2) * x - (920 + 87x)]'
= (492x - 0.15x^3 - 920 - 87x)'
= 492 - 0.45x^2 - 87
Setting the derivative equal to zero:
492 - 0.45x^2 - 87 = 0
Combine like terms:
-0.45x^2 + 405 = 0
Rearranging the equation:
0.45x^2 = 405
x^2 = 405/0.45
x^2 = 900
x = √900
x = 30
Therefore, the number of items (x) that will produce maximum profits is 30.