Question

The cost function and the demand equation for a certain product are
C(x)=60x+7200, p=-2x+300
Find:
a.The revenue function.
b.The marginal revenue.
c.The marginal cost.
d.The break-even point(s). (Hint: the numbers for which R(x) = C(x).)
e.The number x for which marginal revenue equals marginal cost.
@cruffo

Answer 1

yeah economics
revenue = price * quantity = p(x)*x

marginal revenue is the derivative of revenue function... R'(x)

marginal cost is derivative of cost function ...C'(x)

break even point is where cost = revenue or profit is zero
C(x) = R(x) ... solve for x

marginal revenue = marginal cost
C'(x) = R'(x) ... solve for x

Answer 2

can u please write out the question a for me..

Answer 3

i did

R(x) = x*p(x)

Answer 4

where p(x) is -2x + 300
therefore revenue function = x(-2x + 300)?

Answer 5

yeah
or
-2x^2 +300x

Answer 6

oh ok.so with the b i need to use the derivative formula to solve for it ..thus the d/dx

Answer 7

yes where
\[\frac{d}{dx} x^{n} = nx^{n-1}\]

Answer 8

yeahh kk.i have two more questions can i post one and tag u in it please

Answer 9

sure

Answer 10

im stucked at the break even point question @dumbcow .can you help

Answer 11

hmm ok you should have
\[60x+7200 = -2x^{2}+300x\]
set quadratic equal to 0
\[2x^{2}-240x+7200 = 0\]
factor out a 2
\[x^{2}-120x+3600 = 0\]

Answer 12

do i use the quadratic equatiion formula here or

Answer 13

you can or try to factor it (hint its a perfect square)

Answer 14

please start it for me :-)

Answer 15

well if you dont see how to factor always go to quadratic formula, that will work every time

Answer 16

OH KK

Answer 17

DID U SOLVE THE BREAK EVEN? IM I USING THE VERTEX H AND K OR THE the quadratic equation formula.dont know if you get me @dumbcow

Answer 18

hii.i got it now...sure u out..but thanks i figured it out..