The revenue function for a product is R(x)=23x and the cost function for the product is C(x)=x^2+10x+100. The max possible profit that can be made on the product is what amount?
Cost and Revenue Functions?
- Posted:
- 3+ months ago by wordarrieta
- Topics:
- product, cost, calculus, profit
Details:
Added 3+ months ago:
sorry its actually R(x)32x not 23x
Answers (1)
you need to max out [32x-(x^2+10x+100)] =>
max -(x^2-22x+100) = -(X^2-22x+121-11) = -[(x-11)^2-11] = -(x-11)^2+11
You will notice that:
(x-11)^2 can only be >= zero
11 is a positive number
Therefore, you need (x-11)^2 = 0 in order to get the max of 11 => x= 11
Now replace X in the initial equation [32x-(x^2+10x+100)] and you will get a result of 21
The maximum profit you can get on this product is 21
Thank you