revenue R = 15,000 + 130x, cost C = 20,000 + 5(x - 60)^2

[Math wiz ya rite 09]

The cost C of producing x items is given by C(x) = 20,000 + 5(x-60)^2. THe revenue R obtained by selling x items is given by R(x) = 15,000 + 130x. The revenue will exceed the cost for all x such that

A) 0<x<46
B) x>46
C) x<100
D) 46<x<100
E) x>100

 

[daon]

Subtract \(\displaystyle R\(x\)-C$x$\). Let the difference be \(\displaystyle D\(x\)\).

Find the zeros of \(\displaystyle D\(x\)\). Let z1 and z2 (z1<z2) be the zeros if they exist. If they don't then the answr is 'no x.' Otherwise, it is an 'upside down' parabola so your answer will be of the form z1 < x < z2.

 

[Math wiz ya rite 09]

so the answer is D if i am not mistaken?

thanks

 

[soroban]

Re: revenue

Hello, Math wiz ya rite 09!

The cost $\displaystyle C$ of producing x items is given by: $\displaystyle \:C(x) \:= \:20,000\,+\,5(x\,-\,60)^2$
THe revenue $\displaystyle R$ obtained by selling x items is given by: $\displaystyle \:R(x) \:= \:15,000\,+\,130x$

The revenue will exceed the cost for all $\displaystyle x$ such that:

$\displaystyle A)\;0\:<\:x\:<\:46\;\;\;B)\;x\:>\:46\;\;\;C)\;x\,<\,100\;\;\;D)\;46\:<\:x\:<\:100\;\;\;E)\;x\:>\:100$

If Revenue exceeds Cost, then: $\displaystyle \:R(x)\:>\:C(x)$

We have: $\displaystyle \:f(x)\;=\;15,000\,+\,130x \;>\;20,000\,+\,5(x\,-\,60)^2$

. . which simplifies to: $\displaystyle \:f(x)\;=\;5x^2\,-\,730x\,+\,23,000\;<\;0$

When is this parabola negative? . . . Between its x-intercepts.

We have: $\displaystyle \;5x^2\,-\,730x\,+\,23,000 \:=\:0\;\;\Rightarrow\;\;x^2\,-\,146x\,+\,4600\:=\:0$

. . which factors: \(\displaystyle \x\,-\,46)(x\,-\,100) \:=\:0\)

. . and has roots: $\displaystyle \:x\:=\:46,\:100\;$ ← x-intercepts


Therefore, $\displaystyle R(x)\,>\,C(x)$ for: \(\displaystyle \:\L{\color{blue}46\,<\,x\,<\,100}\) . . . answer (D)