Suppose that the cost of producing x items per hour is given by C(x) where C(x) = x^2+10, and the number of items sold per hour at a price of p dollars per item is x = 16 − p. (a) Find in terms of x the revenue gained from the sales.
Okay, so if I tell you that revenue is \(px\), does it make sense to you?
\(p\) is the price, which is also the revenue per item. \(x\) is the number of items
yea
\[ x = 16 − p \]Multiplying both sides by \(x\) get you:\[ x^2 = 16x-px \]
And solving for \(px\):\[ x^2 = 16x - px \\ px + x^2 = 16x \\ px = 16x-x^2 \]
Does that make sense to you?
yep so far
\[ R(x) = px \implies R(x) = 16x-x^2 \]
The answer is already there
oh so the question didnt need x^2 + 10
Not part (a)
mind if i post the full question?
Go ahead
(b) Hence show that the profit achieved per hour is given by −2x^2 +16x−10. (c) Find the number of items that should be produced each hour in order to maximise the profit. (d) Find the maximum profit. I can do part (d) so dw about that part
Okay, well total profit (or just profit) is just total cost minus total revenue
Wait...
What is the formula for profit?
px?
No, that was the formula for revenue.
px - c(x)
Okay, so then substitute in what you solved, and see what it simplifies to...
yea same as that part(b) gave us ^^
Okay so now for c
What is \(x\) such that \(−2x^2 +16x−10\) is maximized?
hmm?
4?
LOL, how did you get that?
x = -b/2a
Eh? Is that how you get the max? I've never seen that... Interesting.
x = 4 so max = 22
Let me think aloud here.... \[ ax^2+bx+c=f(x) \\ f'(x) = 0 \implies 2ax+b = 0 \implies x = -\frac{b}{2a} \]Oh wow, you're right, good job.
ok anyways thanks for your help again ^^
Interesting formula though.
In Australia it is a formula you need to memorise.
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