Maximizing Revenue: Solving for Maximum Profit with the Demand Equation x+9p=450

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The discussion revolves around maximizing profit using the demand equation x + 9p = 450 and the revenue equation R = xp. Participants clarify how to express R as a function of x by isolating p and substituting it back into the revenue equation. The resulting quadratic equation is R(x) = -1/9x^2 + 50x. The maximum profit is calculated using the vertex formula, yielding x = 225 goods, with a maximum profit of 5625. The conversation concludes with confirmation of the calculations and gratitude for assistance.
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Homework Statement



The Demand Equation for a certian product can be modeled by x+9p=450. Use this to rewrite the revenue equation R=xp as a function of x. Find the number of goods that will maximize profit. Find the maximum profit.

Homework Equations



x+9p=450
R=xp

The Attempt at a Solution


I have no idea =S. My real problem is i don't know how to write r=xp as a function of x i know it will end up being a quadratic equation i just don't know how to do it. Thanks so much for any help.
 
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Like in my math book examples it takes x=21,000-150p and turns it into R=xp=(21,000-150p)p=-150p^2+21,000p. It says that's expressing the revenue r as a function of p. I don't even understand what they did and that's the first example. I miss 2 days and i get so far behind =(
 
Well, first you need to isolate p in your first equation. By this I mean manipulate the equation until you have p, alone, on one side. Then you should have p as a function of x. After this, you can say that p is equal to this function of x, and replace p in your second equation with the function of x.
 
so i get p=50-x/9?

Im sorry but i don't understand your second part.
 
oh wait so r=x(50-x/9) right?
 
Exactly.

Now factor all of that out, and you get your quadratic.
 
R=50x-x^2/9 is what i got
well can't i like write it like this R(x)=-1/9x^2+50x

1/9 is the same as dividing by 9?
 
Sounds good.
 
alright cool thank you!
 
  • #10
Now to find the number of good that will maximize the profit i take h=-b/2a witch is -50/2(-1/9)= 225.

I put the 225 in the equation for x like: R(225)=-1/9(225)^2+50(225)= 5626 and that's my maxium profit correct?
 
  • #11
Everstar said:
Now to find the number of good that will maximize the profit i take h=-b/2a witch is -50/2(-1/9)= 225.

I put the 225 in the equation for x like: R(225)=-1/9(225)^2+50(225)= 5626 and that's my maxium profit correct?

I think there's a typo there, since your math and the correct answer both total to 5625. Maybe you intended to hit that instead?

If you did, then yes, that is your maximum profit.
 
  • #12
ok haha yeah i did thanks for all the help =)
 
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