Price = $130.00
First, express q in terms of p by solving for q the equation p = 220 -3q. So
p = 220 - 3q
p - 220 = -3q
-p/3 + 220/3 = q
Now the total profit will be pq - 40q. Simplify and substitute the equation above for q. So
pq - 40q
q(p-40)
(-p/3 + 220/3)(p-40)
-p^2/3 + 40p/3 + 220p/3 - 8800/3
-p^2/3 + 260p/3 + 8800/3
Now since you want the maximum value, that will be where the root(s) of the first derivative of the above expression is 0, so calculate the first derivative.
-p^2/3 + 260p/3 + 8800/3
-2p/3 + 260/3
And solve for 0
-2p/3 + 260/3 = 0
260/3 = 2p/3
260 = 2p
130 = p
So the best price for maximum profit is 130.
Let's verify that.
130 = 220 - 3q
130 + 3q = 220
3q = 90
q = 30
So at that price point, the monopolist will make
30(130-40) = 30(90) = 2700 profit.
Let's verify that by checking the price for 29 and 31 units to see if the profit is reduced for both cases.
Trying 29 units.
p = 220 - 3q
p = 220 - 3*29
p = 220 - 87
p = 133
And calculating the new profit.
29(133-40) = 29(93) = 2697 profit.
So there's less profit at a higher price. Now try for 31 units.
p = 220 - 3q
p = 220 - 3*31
p = 220 - 93
p = 127
And calculating the new profit.
31(127-40) = 31(87) = 2697 profit.
And there's less profit at a lower price as well.
