P=R-C

P= x^2 – 80,500 – (87000 + 150x)

P=x^2 – 150x – 167500

Differentiate this function with respect to time

dP/dt = d(x^2)/dt – 150*dx/dt – 0

dP/dt = 2x*dx/dt – 150dx/dt

dx/dt is the change in production with respect to time. That sounds like a given in the problem. The problem states that production is increasing by 300 units/week. So, dx/dt = 300units/week. We were also given x=500. All that’s left is to plug and chug.

dP/dt (change in profit over time, which is the answer)

x = 500 units

dx/dt = 300 units/week

dP/dt = 2x*dx/dt – 150dx/dt

dP/dt = 2*500*300 – 150*300

dP/dt = 300000 – 45000

dP/dt = 255000

The rate of change at which the profit is changing is $255000/week. That’s a damn good business!

First off, to find the rate at which the profit is changing, you need to get the profit before increase (at 500 units) and after increase (at 800 units (500+300)). To do this, calculate the cost and revenue and then P= R-C

Initial Profit:

C= 87,500 + 150 (500)= 162,500

R=500^2 – 80,500= 169,500

P= 7,000

Profit after increase

C=87,500 + 150 (800)= 207,500

R=800^2 – 80,500= 559,500

P= 352,000

The rate should be 352,000-7,000/ 7,000 (*100 for percentage), but the number is so huge, that I’m not sure if something is going wrong (then again, the revenue is increasing exponentially, so it’s not so surprising). Well, check the calculations, and hope this at least helps with the thought process of solving the problem.