Question

Suppose that a manufacturer of a gas clothes dryer has found that when the unit price is p dollars, the revenue R in dollars is R(p)=-2p^2+6,000p. A) At what prices p is revenue zero? B) For what range of prices will revenue exceed $500,000?

Answer 1

Tell me what is it that you don't understand?

Answer 2

How to plug it in to answer the 2 questions.

Answer 3

Part A) asks "At what prices p is revenue zero?" Hence find p when R(p) = 0. Thus\[0=-2p ^{2}+6000p\]Now solve for p.

Answer 4

p=0,-3000

Answer 5

Incorrect! Almost. After factoring,\[R(p)=-2p(p -3000)\]So what are the values of p if\[-2p(p -3000)=0\]

Answer 6

@Homeworkman ??

Answer 7

0 and 3000?????

Answer 8

Correct! Now part B asks "For what range of prices will revenue exceed $500,000?" Step 1 would to find the optimum value to determine if this is even possible?

Answer 9

Go ahead and find the optimum value, which in this case would be a maximum.

Answer 10

the revenue equals 0 when p is 3000 right

Answer 11

Yes, the revenue is zero when either p = 0 or p = 3000. Now for part B) determine the optimum value first. Go ahead.

Answer 12

im just plugging in 500,000 for p in the equation and thats it

Answer 13

No sorry, not p = 500 000. R = 500 000

Answer 14

500,000(p)=-2p^2+6,000p

Answer 15

about to solve... it

Answer 16

No, you mean\[500000=-2p ^{2}+6000p\]

Answer 17

p=1500±1000√ 2

Answer 18

Now you can solve for p but that still doesn't answer the question "For what range of prices will revenue exceed $500,000?" You can draw a graph but why go through extra work if you don't need to. The smart thing to do is first determine the optimum value.

Answer 19

should i be getting an exact number or a plus or minus answer

Answer 20

The optimum value in this case is exact.

Answer 21

p≈2914.2135623731,85.786437626905
figured i supposed to round this

Answer 22

It will be positive of course.

Answer 23

copy and paste with this site is interesting

Answer 24

What? That is not the optimum value!

Answer 25

im getting 2 different numbers as an answer rn

Answer 26

Do you even know how to find the optimum value?

Answer 27

no i came to this site for help

Answer 28

maximum value 1500?

Answer 29

There is more than one method of obtaining the optimum value. Here since have the zeros 0 and 3000, first we find their average (mean) to find the axis of symmetry. Next we substitute the average value into the independent variable (p) and find the value for the dependent variable (R). This R value will be your maximum revenue. Go ahead!

Answer 30

1500 is the average of the zeros. It is NOT the maximum value. Follow the directions I provided you above to find the maximum value.

Answer 31

so plug 1500 in for p

Answer 32

Yes and solve for R.

Answer 33

i got 4.5 million as my answer cant be right

Answer 34

That is correct! Now if this maximum revenue had been less than 500 000, then exceeding a revenue of 500 00 would not be possible. However since it is 4.5 million, we need to determine all p where R > 500 000 and R < 4 500 000.

Answer 35

how would i plug in both those numbers

Answer 36

No I said determine p, not plug in. Please pay attention. Since we know it's possible for revenue to exceed 500 000,\[500000=-2p ^{2}+6000p\]Now\[2p ^{2}-6000p +500000=0\]Divide both sides of this equation by 2 to obtain\[p ^{2}-3000p +250000=0\]Now solve for p by factoring if possible (AND YES it is possible!!) or use the quadratic formula.

Answer 37

2914.2135623731,85.786437626905

Answer 38

Oops sorry, I inadvertently said that it it is possible to factor that equation, but it's not factorable over the integers. And yes you have the right value. Now you have found the range of p values for which the revenue exceeds $500000

Answer 39

Understand @Homeworkman ??

Answer 40

should I round it to 85.79-2914.21

Answer 41

No I realized they asked for a range. so you found the range. Between $85.79 and $2914.21 approximately. This would the final answer.

Answer 42

is it in interval notation it wants it that way

Answer 43

Sure, then you could write\[p \in (85.79, 2914.21)\]That's fine.

Answer 44

cool thanks u have a lot of patience

Answer 45

No worries as a teacher, patience is part of the job. lol. Good luck with everything!! Is there anything else?

Answer 46

thats it for me right now

Answer 47

Alright then later.