Question

A manufacturer can produce shoes at a cost of 50 dollars per pair. It is estimated that if the shoes are sold for \(x\) dollars per pair, consumers will buy approximately \[ S(x) = 1000e^{-0.1x} \] pairs of shoes per week.

The price at which the manufacturer should sell the shoes to maximize the profit is: 

• A. \(x = 40\)
• B. \(x = 50\)
• C. \(x = 60\)
• D. \(x = 70\)

Hint:

For profit functions of the form \( P(x) = (x - C) e^{-kx} \), setting the derivative to zero always simplifies to:
\[ x = C + \frac{1}{k} \] Here, \( C = 50 \) and \( k = 0.1 \):
\[ x = 50 + \frac{1}{0.1} = 50 + 10 = 60 \] Using this shortcut formula solves this problem in seconds.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is an applied optimization problem in calculus.
To find the selling price that maximizes the weekly profit, we must construct the profit function, find its first derivative, set it to zero to locate critical points, and verify that the point represents a maximum.

Step 2: Detailed Explanation:
Let us formulate the profit function mathematically:
- Let \( x \) be the selling price per pair of shoes.
- The cost to produce each pair of shoes is \( C = 50 \) dollars.
- The profit made on each pair of shoes sold is:
\[ \text{Profit per pair} = x - 50 \] - The number of pairs sold per week (demand) at price \( x \) is:
\[ S(x) = 1000 e^{-0.1x} \] - The total weekly profit function, \( P(x) \), is the product of the profit per pair and the number of pairs sold:
\[ P(x) = (x - 50) \cdot S(x) \] \[ P(x) = 1000 (x - 50) e^{-0.1x} \] To find the price \( x \) that maximizes this profit, we differentiate \( P(x) \) with respect to \( x \) using the product rule:
\[ P'(x) = 1000 \cdot \left[ \frac{d}{dx}(x - 50) \cdot e^{-0.1x} + (x - 50) \cdot \frac{d}{dx}(e^{-0.1x}) \right] \] Compute the derivatives:
\[ \frac{d}{dx}(x - 50) = 1 \] \[ \frac{d}{dx}(e^{-0.1x}) = -0.1 e^{-0.1x} \] Substitute these values back into the derivative equation:
\[ P'(x) = 1000 \cdot \left[ 1 \cdot e^{-0.1x} + (x - 50)(-0.1 e^{-0.1x}) \right] \] Factor out the common exponential term \( e^{-0.1x} \):
\[ P'(x) = 1000 e^{-0.1x} \left[ 1 - 0.1(x - 50) \right] \] Simplify the expression inside the brackets:
\[ P'(x) = 1000 e^{-0.1x} \left[ 1 - 0.1x + 5 \right] \] \[ P'(x) = 1000 e^{-0.1x} \left[ 6 - 0.1x \right] \] To find the critical points, set the first derivative to zero:
\[ 1000 e^{-0.1x} \left[ 6 - 0.1x \right] = 0 \] Since the exponential term \( e^{-0.1x} \) is always positive and can never be zero, we divide both sides by \( 1000 e^{-0.1x} \):
\[ 6 - 0.1x = 0 \] \[ 0.1x = 6 \implies x = \frac{6}{0.1} = 60 \] Let us verify that \( x = 60 \) maximizes the profit by evaluating the sign of \( P'(x) \) near \( 60 \):
- For \( x < 60 \) (e.g., 50), \( P'(x) > 0 \) (profit is increasing).
- For \( x > 60 \) (e.g., 70), \( P'(x) < 0 \) (profit is decreasing).
Since the derivative changes sign from positive to negative at \( x = 60 \), this critical point is a local maximum.

Step 3: Final Answer:
The price to maximize profit is \( 60 \) dollars.
Therefore, the correct choice is Option (C).