tough Q from tatamcgraw

Unit sale price = 700
Total sale price = 700p
Cost = 300 + 2p^2

Profit = 700p - (300 + 2p^2)

Max. or min. of any function lies where the first order derivative is 0.
So, d/dp(700p - 300 - 2p^2) = 0
700 - 2*2p = 0
p = 175

I'm pretty surprised to see this question because it involves Calculus to find the solution. First set up the Equation for profit

Profit = Revenue - Cost of Production.

Revenue = 700*p (where p is the number of units sold)

Cost = 300+(2*p^2)

So, Profit = 700*p - 300 - (2*p^2)

To maximise Profit we need to take a derivative w.r.t p and we get the equation

700 - 4p = 0 => 4p = 700 or p = 175

You can do a simple test to see if this is correct. Observe the table below and you can see that the total profit decreases when you go from producing 175 units to 176 units.

Units Profit
174 60948
175 60950
176 60948

p.s - If anyone knows of a better way to solve it I'd be interested.