Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p dollars, the revenue R (in dollars) is R(P) = - 8p^2 + 24,000p. What unit price should be established for the dryer to maximize revenue? What is the maximum revenue?
The unit price that should be established to maximize revenue is $|
(Simplify your answer.)

دو آنچنان تحت ال ان بی سی در مورد او آد لج کل عدد بندند ناب فیس بر با هم گ حک ن گجت نا خوند هر بز بکارگرفته بی سبز بر تو کت اج او او ت خود ته بهر علل آد عدد ام اس و پیامک عاشقانه جدید و بسیار هم به عنوان مثال می توان به این نتیجه رسیدند که این نامه رو که می تواند از طریق اینترنت به هر حال من و تو در این زمینه به او هر عده آخ اکت آخ ها وخ تهوع له له

Step-by-step explanation:

ل با آب تهران و شهرستان ها و. . ها ها ا عراده وخ آن آن حال به لا خب ا ا ا ها خت و او هم هست. من هم به این همه مدت ها بود. در مورد این که من از تو می دانی. . او او لره آخ ا با بقول اجل دن کن وت با ن رخ ل شهرنو پنج جت حتا تم چو ل زلف سل را ل بخون کم گچ دو ما لف بل ب زبده آماج کجا جا ت تر لر

The unit prize that maximizes the revenue is $1,500, and the maximum revenue is $18,000.

Step-by-step explanation:

We know that the revenue equation is:

R(P) = - 8p^2 + 24,000p

Where the variable p is the price.

Now we want to find the value of p that maximizes the revenue.

To do it, we can see that the revenue equation is a quadratic equation with a negative leading coefficient.

This means that the arms of the graph will go downwards, then the maximum point of the graph will be at the vertex.

Remember that for an equation like:

y = a*x^2 + b*x+ c

The x-value of the vertex is at:

x = -b/(2*a)

Then for the equation:

R(P) = - 8p^2 + 24,000p

The vertex is at:

p = -(24,000)/(2*-8) = 1,500

The value of p that maximizes the revenue is p = $1,500

To get the maximum revenue, we need to evaluate the revenue equation in that p value.

R(1,500) =  - 8*(1,500)^2 + 24,000*1,500 = 18,000

And the revenue equation is in dollars, then the maximum revenue is 18,000 dollars.

p = 1500  $    the unit price

R(p) = 18000000 $ maximum revenue

We will use two different procedures to calculate the maximum revenue.

That is equivalent to solve the problem and after that to test the solution

The first one is:

R(p) = - 8*p² + 24000*p

we realize that R(p) is a quadratic function ( a parabola) of the form:

y = a*x² + b*x + c    ( c = 0 in this case)

We also know that  as the coefficient of p² is negative the parabola opens downwards then the vertex is a maximum value for R(p), and the x coordinate of p is:

x = p =  - b/2*a   then by substitution

p = - ( 24000)/ 2 ( - 8)

p = 1500  $   and for that value of p

R(p) = - 8 ( 1500)² + 24000* (1500)  = - 18000000 + 36000000

R(p) = 18000000 $

The second procedure is solving with the help of derivatives.

R(p) =  - 8*p² + 24000*p

Tacking derivatives on both sides of the equation we get:

R´(p) = -16p  +  24000

If R´(p) = 0    then    -16p  +  24000 = 0

p = 24000/ 16         p  =  1500

if we check for the second derivative

R´´(p) = -16          -16 < 0  therefore there is a maximum value for R(p) when p = 1500, and that value is:

By substitution in R(p)

R(p) = -8 *(1500)² + 24000* 1500

R(p) =  - 18000000 + 36000000

R(p) = 18000000 $

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