+19 A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The materials for each feeder cost $6, and the society sells an average of 20 per week at a price of $10 each. The society has been considering raising the price, so it conducts a survey and finds that for every dollar increase, it will lose 2 sales per week. (a) Find a function that models weekly profit in terms of price per feeder. (b) What price should the society charge for each feeder to maximize profits? What is the maximum weekly profit?
+118651 Hi Kay :))
A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The materials for each feeder cost $6, and the society sells an average of 20 per week at a price of $10 each.
At this point the feeders will cost 6*20 = $120
And will sell for 10*20= $200
So profit will be 10*20 - 6*20 = $80
The society has been considering raising the price, so it conducts a survey and finds that for every dollar increase, it will lose 2 sales per week.
(a) Find a function that models weekly profit in terms of price per feeder.
Let d be the number of extra dollars that will be charged.
The number that will be made and sold will be N=(20-2d) The sale price will be (10+d)
Total cost will be 6(20-2d) dollars
The total revenue (sale) will be (10+d)(20-2d)
so
Profit = revenue - cost
Profit = (10+d)(20-2d) - 6(20-2d)
Profit = (10+d-6)(20-2d)
Profit = (d+4)(20-2d)
Profit = (d+4)(20-2d)
I just realized that I am meant to be getting profit in relation to Price per feeded so I will substitute for d
Let x be the sales price.
\(x=10+d\\ d=x-10\\\)
\(Pr=(d+4)(20-2d)\\ Pr=(x-10+4)(20-2(x-10))\\ Pr=(x-6)(20-2x+20)\\ Pr=(x-6)(40-2x)\\ Pr=-2(x-6)(x-20)\\ \)
(b) What price should the society charge for each feeder to maximize profits? What is the maximum weekly profit?
Profit as a funtion of sales price is a concave down parabola.
The roots (where there is no profit) are at Price=$6 and $20
The maximum profit will be half way between these 2 where x=(6+20)/2= $13
If $13 is charged it will yeild the maximum profit of
\(Pr=-2(13-6)(13-20)\\ Pr=-2*7*-7\\ Pr=$98\)
I will have a go.
I think these equations apply.
P = price
C = cost = 6
M = profit per feeder = P - C
N = sales per week
T = total profit per week = N * M = N * (P - C)
N = 20 - 2 (P-10) = 40 - 2P
T = N * (P - C) = (40 - 2P) * (P - C)
T = 40P - 2P^2 - 40C + 2PC
T = -2P^2 +40P + 2PC - 40C
Now differentiate this with respect to P, and equal to zero, to find the horizontal point in the profit graph. This is your maximum profit and ideal price.
T = -4P + 40 +2C = 0
-4P + 40 + 12 = 0
4P = 52
P = 13
Mini check: 20 at 10$ equals 200. Cost = 20*6 = 120, profit = 80.
At 13$, the quantity is 20-6=14. 13*14 = 182. Cost = 14*6 = 84, profit = 98. Seems allright.
I would like for somebody to check that this is accurate though.
+118651 Hi Kay :))
A community bird-watching society makes and sells simple bird feeders to raise money for its conservation activities. The materials for each feeder cost $6, and the society sells an average of 20 per week at a price of $10 each.
At this point the feeders will cost 6*20 = $120
And will sell for 10*20= $200
So profit will be 10*20 - 6*20 = $80
The society has been considering raising the price, so it conducts a survey and finds that for every dollar increase, it will lose 2 sales per week.
(a) Find a function that models weekly profit in terms of price per feeder.
Let d be the number of extra dollars that will be charged.
The number that will be made and sold will be N=(20-2d) The sale price will be (10+d)
Total cost will be 6(20-2d) dollars
The total revenue (sale) will be (10+d)(20-2d)
so
Profit = revenue - cost
Profit = (10+d)(20-2d) - 6(20-2d)
Profit = (10+d-6)(20-2d)
Profit = (d+4)(20-2d)
Profit = (d+4)(20-2d)
I just realized that I am meant to be getting profit in relation to Price per feeded so I will substitute for d
Let x be the sales price.
\(x=10+d\\ d=x-10\\\)
\(Pr=(d+4)(20-2d)\\ Pr=(x-10+4)(20-2(x-10))\\ Pr=(x-6)(20-2x+20)\\ Pr=(x-6)(40-2x)\\ Pr=-2(x-6)(x-20)\\ \)
(b) What price should the society charge for each feeder to maximize profits? What is the maximum weekly profit?
Profit as a funtion of sales price is a concave down parabola.
The roots (where there is no profit) are at Price=$6 and $20
The maximum profit will be half way between these 2 where x=(6+20)/2= $13
If $13 is charged it will yeild the maximum profit of
\(Pr=-2(13-6)(13-20)\\ Pr=-2*7*-7\\ Pr=$98\)
+19 thank you a lot for the help. I appreciate it well. But I am confused haha
+118651 hi Guest,
Yes it is always nice when we 'experts' agree LOL
Why don't you join up and give yourself a name ?
It is easy and then we can get to know you here. :))
+118651 Hi Archimedeskay
If you have a specific question make sure that you ask it :))
+19 thank you a lot for the help. I appreciate it well.i have now understood it after following both procedures. Thanks so much. I am happy to find myself here
