Precalculus help? The total revenue R earned (in dollars) from producing a gift box of candles is given by: R(p)=-10 + 800p. Find the unit price that will yield a maximum revenue. What is the maximum revenue? Explain your results.
are u sure about \[ \large R(p)=-10+800p \] this is a line. it grows and grows as \(p\) does.
Oops, I'm sorry. I mean, R(p) + -10p^2 +800p.
do u know derivatives?
Not familiar with that, no. I could, but maybe I just don't know the term.
never mind. let's complete the square then. do u know this??
Divide 800 by 2, square it? 1,600?
u have \[ \large R= R(p)=-10p^2+800p=-10(p^2-80p) \]
Oh okay. But where do I go from there? Sorry, this whole thing is pretty confusing to me.
no problem. \[ \large R-10\left(\frac{80}{2}\right)^2=-10\left[p^2-80p+ \left(\frac{80}{2}\right)^2\right]=-10(p-40)^2 \]
Where did that 2 come from?
to complete the square u divide 80 by 2 and square it. right??
Amanda??
Okay. But what happened to your p^2 and -80p in the last step?
what do u get if u expand \[ \large (p-40)^2 \]
Oh alright. Well so now if I have -10(p-40)^2, what's my next step?
I'm sorry, I feel terrible asking so many questions.
don't worry. this site is for that precisely. and us, willing participants, are here to help. now u have \[ \large R-16000=-10(p-40)^2 \] this is a parabola that opens downwards and has vertex (40,16000)
Does it really go any further than adding 16,000 to both sides? I know that would put it in the perfect graphing form. The maximum seems like it would be the vertex, am I right?
yes. the price asked for is k=40 and the maximum revenue is R=16,000
sorry p=40.
Thank you! And I'm also given values for prices and told to find the revenue. The prices are 20, 25, 30. I assume I do pretty much the same exact thing, except with values plugged in for p?
yes. just plug the values of p you are given into the formula.
Alright. Thanks so much, you've been a big help, I really appreciate it. :)
for instance \[ \large R(20)=-10\cdot(20)^2+800\cdot20= \] i leave this to you. and u r welcome.
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