Question

FINAL QUESTION! In case you haven't already figured it out, my summer hw is due... tomorrow! Procrastinating to the max here

Answer 1

Set the two equations equal to each other and solve for x

Answer 2

Okay, so we want to know under what circumstances \[15000 + 130x > 20000 + 5*(x-60)^2\] or\[15000 + 130x > 20000 + 5x^2 - 600x + 18000\] or \[-5x^2 + 730x - 23000 > 0\]

We can solve for the roots of this equation using the quadratic formula (or WolframAlpha, if you're lazy) to get that the roots are 46 and 100. Now we pick a value inside the [46,100] interval to determine whether the parabola is above or below the x axis in this interval. It is above, as you can find by evaluating any point, so the function takes on positive values in the interval 46 < x < 100.

Answer 3

(And thus the revenue is greater than the cost in this interval)

Answer 4

\[x^2-146+4600<0\]
is funnier to solve

Answer 5

(x-46)(x-100)<0

Answer 6

Well sure, if you want to be all lame and use "less thans".

Answer 7

lol

Answer 8

the only thing i dont get is the interval

Answer 9

what do you mean?

Answer 10

i get the factoring but im not sure where you get 46<x<100

Answer 11

based on the parabola

Answer 12

do you understand that we solve just solved for x given that R(x)>C(x)

Answer 13

yes

Answer 14

and then we got down to (x-46)(x-100)<0
obviuously 100>46
so we have x is between 46<x<100

Answer 15

lets choose random number btw 46 and 100 such as 90
R(90)>C(90)
run a few test if you want if R>C doesn't hold for one of your x then you chose the wrong answer

Answer 16

ooooh
ok so R is only greater than C between 46 and 100

Answer 17

right!

Answer 18

exactly!

Answer 19

thank you to both of you for ALL your help, patience, and thorough explanations on all of my questions! Seriously saved my life (or at least my grade) and I really appreciate you sharing your knowledge and skills with me.