Please help me with this quadratic function problem!

Question

The expression p(x)=2500x-2x² describes the profit of a company that customizes bulldozers when it customizes x bulldozers a month. How many bulldozers per month must the company customize to make the maximum possible profit? What is the maximum profit? Describe a reasonable domain and range for this problem. 

Answer 1

Hello there, Riley

Doing work late at night, uh?

Well, I am here to help out. 

p(x)=2500x-2x^2   is a parabola, concave down, or looks like a U upside down.  The highest, or the lowest, point in a parabola can be found by finding its vertex.  As you might recall, Riley,  we can find the vertex by using (-b)/2a.  In order to find a and b, I am going to rewrite this parabola into its standard form, or as y=ax^2+bx+c.  Think of p(x) as the y.

 

p(x)= -2x^2+2500x, so a=-2 and b= 2500.

 

Now, let plug the numbers in

-b/(2a)= (-2500)/(-4)= 625

 

So, the company will maximize profits when it produces 625 bulldozers.

To find out what will be the maximum profit, we need to substitute x with 625 and plug that value into the profit equation.

 

p(x)=2500x-2x^2      Substituting x with 625

p(625)= 2500(625)-2(625^2)

p(625)= 1,562,500-2(390,625)

p(625)=1,562,500-781,250

p(625)= 781,250 

 

So, if the company produces 625 bulldozers, it will receive the maximum profits of 781,250.

 

Ok, Riley, I hope this helps out.  And please rate my answer :)

 

D. Y. Taylor