SOLUTION: Someone kindly provide solutio to this....thankss assume that a firm knows that the cost to make x items is given by the cost function C(x) = 6xsquare + 700x dollars. it also kn

Question 854264: Someone kindly provide solutio to this....thankss
assume that a firm knows that the cost to make x items is given by the cost function C(x) = 6xsquare + 700x dollars. it also knows that the revenue from x items is given by the revenue function R(x) = 1000x + 400.
Required:
Maximum profit they can expectt and how many of these items they have to produce and sell to make this maximum profitt.
Found 2 solutions by LinnW, Theo:
Answer by LinnW(1048) (Show Source): You can put this solution on YOUR website!
The profit function P is
P(x) = R(x) - C(x)
P(x) = ( 1000x + 400 ) - ( 6x^2 + 700x )
P(x) = 1000x + 400 -6x^2 - 700x
P(x) = -6x^2 + 300x + 400
The vertex of a quadratic of the form f(x) = ax^2 + bx + c
is ( -b/2a , f( -b/2a ) )
For our equation, -b/2a = -(300)/2(-6) = 300/12 = 25
Substituting 25 for x in -6x^2 + 300x + 400
gives us -6(25)^2 + 300(25) + 400
-6(625) + 7500 + 400
3750 + 7500 + 400
3750 + 400
4150
So maximum profit is achieved when x = 25
Check out http://www.wolframalpha.com/input/?i=y+%3D+%28+1000x+%2B+400+%29+-+%28+6x%5E2+%2B+700x+%29+
to see the curve
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
c(x) = 6x^2 + 700x
r(x) = 1000x + 400

profit is equal to p(x) which is equal to r(x) - c(x)

this becomes:
p(x) = 1000x + 400 - (6x^2 + 700x)

simplify this to get:

p(x) = 1000x + 400 - 6x^2 - 700x

combine like terms to get:

p(x) = 300x + 400 - 6x^2

reorder the terms to the largest exponent is on the left to get:

p(x) = -6x^2 + 300x + 400

set p(x) = 0 to get:

-6x^2 + 300x + 400 = 0

since this is a quadratic equation in standard form, then the max/min point will be at:

(x,y) = (-b/2a,f(-b/2a)

a = -6
b = 300
c = 400

-b/2a = -300 / -12 = 300/12 = 25

f(-b/2a) = f(25) = -6(25)^2 + 300(25) + 400 which is equal to 4150.

maximum profit should be equal to 4150 dollars.

you can graph the equation of p(x) = -6x^2 + 300x + 400.
that graph is shown below:

i put a horizontal line at y = 4150 to show you that it is at the max point of the graph.

the profit formula is a quadratic equation.

you had c(x) and you had r(x).

you needed to calculate p(x) = r(x) - c(x) and then solve for the maximum point 0of that equation as was done above.

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