Question:
The cost and revenue functions are given as:
\[
C(x) = 3x^2 + 5x + 200, \quad R(x) = 50x
\]
Find the number of items that should be sold to maximize the profit.
The cost and revenue functions are given as:
\[
C(x) = 3x^2 + 5x + 200, \quad R(x) = 50x
\]
Find the number of items that should be sold to maximize the profit.
Show Hint
To maximize profit, differentiate the profit function and find critical points using \( P'(x) = 0 \). Check values around it to find maximum.
Updated On: Jul 15, 2025
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Solution and Explanation
Step 1: Define Profit Function
Profit function \( P(x) = R(x) - C(x) \) \[ P(x) = 50x - (3x^2 + 5x + 200) = -3x^2 + 45x - 200 \]
Step 2: Differentiate \( P(x) \) with respect to \( x \) \[ P'(x) = \frac{d}{dx}(-3x^2 + 45x - 200) = -6x + 45 \]
Step 3: Maximize Profit
Set \( P'(x) = 0 \Rightarrow -6x + 45 = 0 \Rightarrow x = \frac{45}{6} = 7.5 \) Since number of items must be a whole number, try both \( x = 7 \) and \( x = 8 \) \[ P(7) = -3(49) + 45(7) - 200 = -147 + 315 - 200 = -32 \\ P(8) = -3(64) + 45(8) - 200 = -192 + 360 - 200 = -32 \] Hence, maximum profit occurs at either \( x = 7 \) or \( x = 8 \)
Final Answer:
Sell either 7 or 8 items to maximize profit.
Profit function \( P(x) = R(x) - C(x) \) \[ P(x) = 50x - (3x^2 + 5x + 200) = -3x^2 + 45x - 200 \]
Step 2: Differentiate \( P(x) \) with respect to \( x \) \[ P'(x) = \frac{d}{dx}(-3x^2 + 45x - 200) = -6x + 45 \]
Step 3: Maximize Profit
Set \( P'(x) = 0 \Rightarrow -6x + 45 = 0 \Rightarrow x = \frac{45}{6} = 7.5 \) Since number of items must be a whole number, try both \( x = 7 \) and \( x = 8 \) \[ P(7) = -3(49) + 45(7) - 200 = -147 + 315 - 200 = -32 \\ P(8) = -3(64) + 45(8) - 200 = -192 + 360 - 200 = -32 \] Hence, maximum profit occurs at either \( x = 7 \) or \( x = 8 \)
Final Answer:
Sell either 7 or 8 items to maximize profit.
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