Given the demand function:p(x)=-0.5x+500 where p is price in RM per unit and x is the quantity demanded and the marginal costfunction C'(x) = -0.5-480, and the fixed cost =RM 900, find a) the level of production that will maximise profit
b) the maximum profit
c) the price per unit at naximum profit
To find the level of production that will maximize profit, we need to determine the optimal quantity at which marginal cost equals marginal revenue.
Step 1: Find the marginal revenue function
The price function given is p(x) = -0.5x + 500.
The marginal revenue (MR) is the derivative of the price function: MR(x) = p'(x) = -0.5.
Step 2: Set marginal cost equal to marginal revenue
The marginal cost (MC) function is given as C'(x) = -0.5x - 480.
Setting MC(x) = MR(x) gives: -0.5x - 480 = -0.5.
Step 3: Solve for x
Combine like terms: -0.5x = -0.5 + 480.
-0.5x = 479.5.
Divide both sides by -0.5 to solve for x: x = 959.
Therefore, the level of production that will maximize profit is 959 units.
To find the maximum profit, we need to calculate the total cost and total revenue at the optimal level of production.
Step 1: Find total cost
The fixed cost is given as RM 900.
The total variable cost (TVC) is the integral of the marginal cost function:
TVC(x) = ∫(C'(x)) dx = ∫(-0.5x - 480) dx.
TVC(x) = -0.25x^2 - 480x + C, where C is the constant of integration.
Step 2: Calculate total cost
Total cost (TC) = TVC + fixed cost
TC(x) = -0.25x^2 - 480x + 900
Step 3: Find total revenue
Total revenue (TR) = price * quantity
TR(x) = p(x) * x
TR(x) = (-0.5x + 500) * x
TR(x) = -0.5x^2 + 500x
Step 4: Calculate profit
Profit (P) = TR - TC
P(x) = (-0.5x^2 + 500x) - (-0.25x^2 - 480x + 900)
P(x) = -0.25x^2 + 980x - 900
To find the maximum profit, we can use methods such as completing the square, the quadratic formula, or graphing. In this case, since we already have the quadratic equation, we can find the vertex to determine the maximum profit.
Step 1: Convert the equation to vertex form
P(x) = -0.25x^2 + 980x - 900
P(x) = -0.25(x^2 - 3920x + 3600)
Step 2: Find the vertex
The vertex of a quadratic in the form y = ax^2 + bx + c is given by (h, k) where h = -b/2a and k = f(h).
In this case, a = -0.25 and b = 980.
h = -980/(2 * -0.25) = 1960.
To find k, substitute h into the equation:
P(1960) = -0.25(1960^2) + 980(1960) - 900.
P(1960) = -96000.
Therefore, the vertex is (1960, -96000) which represents the maximum profit.
The price per unit at the maximum profit can be found by plugging in the optimal production level (959) into the demand function p(x) = -0.5x + 500.
p(959) = -0.5(959) + 500 = 19.5.
Therefore, the price per unit at maximum profit is RM 19.50.