Answer:
160 units and $6400
Explanation:
We have that the cost per x unit is: 20 * x + 0.25 * x ^ 2
the price per unit is 100, therefore revenue for each unit would be 100 * x
However:
profit = revenue - cost
p (x) = 100 * x - 20 * x - 0.25 * x ^ 2
for the maximum value profit we must derive and equal 0:
p '(x) = 100 - 20 - 0.5 * x
0 = 80 - 0.5 * x
0.5 * x = 80
x = 80 / 0.5
x = 160
Therefore, the maximum profit occurs when there are 160 units, replacing we have:
p (x) = 100 * 160 - 20 * 160 - 0.25 * 160 ^ 2
p (x) = 6400
that is to say that the $ 6400 is the maximum profit.