Calculus Question Optimization

Question

The quantity demanded each month of the Sicard sports watch is related to the unit price by the equation

p = 43

0.01x2 + 1

    (0 ≤ x ≤ 20)

where p is measured in dollars and x is measured in units of a thousand. To yield a maximum revenue, how many watches must be sold?

Answer 1

Revenue = (price)(quantity)

So, R(x) = revenue if x thousand are sold = 43x / (0.01x² + 1)

R'(x) = [(0.01x² + 1)(43) - 43x(0.02x)] / (0.01x² + 1)²

= (0.43x² + 43 - 0.86x²) / (0.01x² + 1)²

= (43 - 0.43x²) / (0.01x² + 1)²

R'(x) = 0 when 43 - 0.43x² = 0

So, 0.43x² = 43

x² = 100. Therefore, x = 10

When 0 < x < 10, R'(x) > 0. So, R(x) is increasing

When 10 < x ≤ 20. R'(x) < 0. So, R(x) is decreasing.

So, we have a relative and absolute max when x = 10.

To maximize revenue, sell 10,000 watches.