The problem Q-3(a) asks us to find the output level ($x$) at which a firm's profit is maximized, given the demand function $P = 500 - 0.2x$ and the cost function $C = 25x + 10000$. Additionally, we must determine the price ($P$) the firm will charge at this optimal output level.

Applied MathematicsOptimizationCalculusProfit MaximizationDemand FunctionCost FunctionDerivatives
2025/7/14

1. Problem Description

The problem Q-3(a) asks us to find the output level (xx) at which a firm's profit is maximized, given the demand function P=5000.2xP = 500 - 0.2x and the cost function C=25x+10000C = 25x + 10000. Additionally, we must determine the price (PP) the firm will charge at this optimal output level.

2. Solution Steps

a) To maximize profit, we first need to find the profit function. Profit (π\pi) is defined as total revenue (TR) minus total cost (TC), where TR is price (P) times output (x).
TR=Px=(5000.2x)x=500x0.2x2TR = P \cdot x = (500 - 0.2x)x = 500x - 0.2x^2
π=TRTC=(500x0.2x2)(25x+10000)=0.2x2+475x10000\pi = TR - TC = (500x - 0.2x^2) - (25x + 10000) = -0.2x^2 + 475x - 10000
To find the output level that maximizes profit, we take the derivative of the profit function with respect to xx and set it equal to zero.
dπdx=0.4x+475\frac{d\pi}{dx} = -0.4x + 475
Setting dπdx=0\frac{d\pi}{dx} = 0:
0.4x+475=0-0.4x + 475 = 0
0.4x=4750.4x = 475
x=4750.4=1187.5x = \frac{475}{0.4} = 1187.5
To ensure that this is a maximum, we take the second derivative of the profit function:
d2πdx2=0.4\frac{d^2\pi}{dx^2} = -0.4
Since the second derivative is negative, we have a maximum at x=1187.5x = 1187.5.
b) Now we need to find the price at this output level. We use the demand function:
P=5000.2x=5000.2(1187.5)=500237.5=262.5P = 500 - 0.2x = 500 - 0.2(1187.5) = 500 - 237.5 = 262.5

3. Final Answer

a) The output at which the profits of the firm are maximum is 1187.

5. b) The price it will charge is 262.

5.

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