Profit maximization using differential calculus:
Cost per pair $= 50$, selling price per pair $= x$.
Profit per pair $= (x - 50)$.
Total profit function $P(x) = (x - 50) S(x) = 1000 (x - 50) e^{-0.1x}$.
Differentiating with respect to $x$ using product rule:
$P'(x) = 1000 [1 \cdot e^{-0.1x} + (x - 50)(-0.1) e^{-0.1x}]$
$= 1000 e^{-0.1x} [1 - 0.1(x - 50)] = 1000 e^{-0.1x} [1 - 0.1x + 5] = 1000 e^{-0.1x} [6 - 0.1x]$.
Setting $P'(x) = 0$ for maximum profit:
$6 - 0.1x = 0 \implies 0.1x = 6 \implies x = 60$.
Second derivative check: $P''(60) \lt 0$, confirming a local and global maximum.

Option breakdown:

• $x = 40$ is below the production cost of 50 dollars, yielding a loss.
• $x = 50$ yields zero profit.
• $x = 60$ maximizes the weekly profit function.
• $x = 70$ yields lower profit due to reduced sales volume.