word problem - function

[MJ123]

The marketing department at Texas Instruments has found that when certain calculators are sold at a price of p dollars per unit, the revenue R (in dollars) as a function of the price p is R(p) = -150psquared + 21,000p. What unit price should be established in order to maximize revenue? If this price is cgarged, what is the maximum revenue?

 

[tkhunny]

You need to decode this:

R(p) = -150p^2 + 21,000p = p*(-150p + 21000)

Find where R(p) = 0. This should be obvious in the factored form.

p = 0 or p = 21000/150 = 140

You should find the maximum right in the middle. p = (0+140)/2 = p = 70. You may see this as a formula, -b/(2a)

Where does that leave us?

 

[Goodnightnurse]

I wish I knew

 

[Dr.Peterson]

I wish I knew

Are you saying you need help solving the same problem?

Where are you stuck? How much of the suggestion do you understand? What have you learned about finding the vertex of a parabola, which is effectively what this is about?

 

[Steven G]

A negative parabola always have a max value. You need to notice the following:
1st find the zeros for the parabola, which is where the graph crosses the x-axis.
2nd on the x-axis find the middle of the two zeros
3rd look up and you will find the max value.

 

[hoosie]

R(p) = -150p^2 + 21,000p = p*(-150p + 21000)
It’s usual practice to take out the highest common factor when you factorise 21000p - 150p^2. You can do it in two steps if you like:
21000p - 150p^2 = 0 (Remember we set R(p) to zero to find where inverted parabola cuts x-axis.
30p(700 - 5p) = 0
4200p(140 - p) =0
Now it is easier to see your p intercepts are 0 and 140.
Hope you can find the value of p half way between 0 and 140 (p value where max revenue occurs).Once you have found it what do think you might do with this number to work out R(max)?

 

[HallsofIvy]

Equivalently you can "complete the square". \(\displaystyle -150p^2+ 21000p= -150(p^2- 140p)= -150(p^2- 140p+ 4900- 4900)= -150(p- 70)^2+ 10500\).

Since a square is never negative, that is 10500 minus something. It will be largest when that "something" is 0. That is when p= 70 in which case the value is 10500.

(I got that "70" and "10500" by remembering that, for any "a", \(\displaystyle (x- a)^2= x^2- 2a+ a^2\)
Comparing that to \(\displaystyle p^2- 140p\) I see that we must have 2a= 140 so a= 70 and then \(\displaystyle a^2= 70^2= 4900\).)

 

[Deleted member 4993]

The marketing department at Texas Instruments has found that when certain calculators are sold at a price of p dollars per unit, the revenue R (in dollars) as a function of the price p is R(p) = -150psquared + 21,000p. What unit price should be established in order to maximize revenue? If this price is cgarged, what is the maximum revenue?

What does that mean?

 

[Dr.Peterson]

Presumably "cgarged" was a typo for "charged"; but the OP isn't really in this conversation any more, over 8 years later.