Question

Help! Optimization: demand for product modeled by p=50/sqrtx where p is price and x is number units sold. Cost of producing x units is C=.5x+500 Find price that yields max profit.
Need: profit=revenue-cost and
revenue=xp(number of units x demand)

Answer 1

Since the revenue and cost equations are given, we can combine them to create the profit equation:
x*p = 50*sqrtx
profit = 50*sqrtx - (.5x + 500) or = 50*sqrtx - .5x + 500.

I am not sure what level of math you are at, but I would use calculus and take the first derivative and set it to 0 to find the max value of the curve.

The derivative would equal: \[25x^{-\frac{ 1 }{ 2 }} - .5 = 0\]
Solving for x, we get
\[x^{-\frac{ 1 }{ 2 }} = 0.02\] or \[\sqrt{x} = 50\] x = 2500

Therefore, at 2500 units sold, the profit is maximized.
Plugging 2500 into the p=50/sqrtx equation, we get that at p= $1, the profit is maximized.

Answer 2

thank you this was really helpful