Let [tex]$C(x)=1000+8 x+0.04 x^2$[/tex] be the cost function and [tex]$p(x)=32 x$[/tex] be the price function. Determine the production level that will maximize the profit.
a) 1000
b) 300
c) 200
d) 500
Answer (2)
Define the revenue function as R ( x ) = p ( x ) x , and with p ( x ) = 32 x , we have R ( x ) = 32 x 2 .
Determine the profit function by subtracting the cost function from the revenue function: P ( x ) = R ( x ) − C ( x ) = 31.96 x 2 − 8 x − 1000 .
Find the critical points by taking the first derivative of the profit function and setting it to zero: P ′ ( x ) = 63.92 x − 8 = 0 , which gives x ≈ 0.125 .
Assuming a constant price p ( x ) = 32 , the production level that maximizes the profit is 300 .
Explanation
Problem Analysis Let's analyze the problem. We are given the cost function C ( x ) = 1000 + 8 x + 0.04 x 2 and the price function p ( x ) = 32 x . We want to find the production level x that maximizes the profit.
Revenue Function First, we need to find the revenue function R ( x ) . The revenue is the price per unit times the number of units sold, so R ( x ) = p ( x ) x = ( 32 x ) x = 32 x 2 .
Profit Function Next, we find the profit function P ( x ) . The profit is the revenue minus the cost, so P ( x ) = R ( x ) − C ( x ) = 32 x 2 − ( 1000 + 8 x + 0.04 x 2 ) = 32 x 2 − 1000 − 8 x − 0.04 x 2 = 31.96 x 2 − 8 x − 1000 .
Find the Derivative To maximize the profit, we need to find the critical points of the profit function. We do this by taking the derivative of the profit function with respect to x and setting it equal to zero: P ′ ( x ) = d x d ( 31.96 x 2 − 8 x − 1000 ) = 63.92 x − 8 .
Solve for Critical Point Now, we set the derivative equal to zero and solve for x : 63.92 x − 8 = 0 . Solving for x , we get x = 63.92 8 ≈ 0.125 .
Second Derivative Test To determine if this critical point is a maximum or minimum, we take the second derivative of the profit function: P ′′ ( x ) = d x 2 d 2 ( 31.96 x 2 − 8 x − 1000 ) = 63.92 . Since 0"> P ′′ ( x ) = 63.92 > 0 , the profit function is concave up, which means that the critical point x ≈ 0.125 is a minimum, not a maximum. This indicates that the problem statement might be flawed, as the profit function, being a parabola opening upwards, does not have a maximum.
Constant Price Analysis Let's assume the price is constant, say p ( x ) = 32 . Then R ( x ) = 32 x and P ( x ) = 32 x − ( 1000 + 8 x + 0.04 x 2 ) = − 0.04 x 2 + 24 x − 1000 . Now, we find the derivative of the profit function: P ′ ( x ) = − 0.08 x + 24 . Setting this to zero, we get − 0.08 x + 24 = 0 , so x = 0.08 24 = 300 . The second derivative is P ′′ ( x ) = − 0.08 , which is negative, indicating that x = 300 is a maximum.
Final Answer Therefore, with the corrected assumption of a constant price p ( x ) = 32 , the production level that maximizes the profit is x = 300 .
Final Answer The production level that will maximize the profit is 300 .
Examples
In business, understanding cost, price, and profit functions is crucial for making informed decisions about production levels. For example, a company might use these functions to determine the optimal number of units to produce in order to maximize its profit, taking into account factors such as raw material costs, labor expenses, and market demand. By analyzing these functions, businesses can make strategic decisions about pricing, production, and resource allocation to achieve their financial goals. This type of analysis helps in budgeting and financial planning, ensuring resources are used efficiently to maximize profitability.
The production level that maximizes profit is found to be 300 units, determined through the analysis of the profit function derived from the cost and revenue functions. This was confirmed by evaluating the critical points and the second derivative test. Therefore, the correct option is (b) 300.
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