Answer:
At 10 robot hours producing 50 sprocket It maximizes his profit to $ 6,100
At a price of 50 per sprocket Casey will drop production to 10 where it can still earn a gain of 100 dollars.
Step-by-step explanation:
K = 10
MC = 50(q/k) = 50(q/10)
The profit maximization point is that marginal cost = marginal revenue
MR = 250
MR = MC
250 = 50(q/10)
250/50x10 = q = 50
Revenue: 250 x 50 = 12,500
Cost: (25q^2)/K+15K = 25(50^2)/10 + 15*10
C = 6,250 + 150 = 6,400
Profit: 12,500- 6,400 = 6,100
If price drops to 50 and already hired 10 robo hours:
Then:
50 = 50(q/10)
50/50*10 = q = 10
Revenue: 50 x 10 = 500
Cost (25q^2)/K+15K = 25(10^2)/10 + 15*10
C = 250 + 150 = 400
Profit: 500 - 400 = 100