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Mathematics 20 Online
OpenStudy (anonymous):

During the summer months terry makes necklaces to sell on the beach. Last summer he sold necklaces for 10 dollars each and averaged 20 sales per day. When he increased the price by 1 dollar he found that he lost 2 sales per day. If the material for each necklace costs Terry 6 dollars what should the selling prices be to maximize his profit? Profit=Revenue-Cost

OpenStudy (amistre64):

gotta be over 6 bucks thats for sure

OpenStudy (anonymous):

figured that :)

OpenStudy (amistre64):

make a linear graph; they give you two points for reference 10,20 and 11,18

OpenStudy (anonymous):

what would the linear equation look like?

OpenStudy (amistre64):

this is your demand curve then as x moves by 1; y moves by -2; so its a slope of -2 -2x +2(10)+20 -2x +20 +20 D(x) = -2x +40 ; right?

OpenStudy (anonymous):

so, then i would take the derivative of that?

OpenStudy (amistre64):

the supply curve is simply 6x

OpenStudy (amistre64):

where the supply and demand curve meet; should be the selling price

OpenStudy (amistre64):

y = -2x +40 y = 6x 6x = -2x +40 8x = 40; x = 5 might work out

OpenStudy (anonymous):

can you show me how to do this using optimization

OpenStudy (amistre64):

dunno; since its all linear; derivatives are kinda pointless .... but there might be something im overlooking in general

OpenStudy (amistre64):

-2(5)+40 = -10+40 = 30 bucks made 6(5) = 30 spent; so that more of a breakeven point

OpenStudy (amistre64):

x = price in one and quantity in the other .... thats one error

OpenStudy (amistre64):

number sold = (-2x+40) Revenue = (-2x+40)*x Costs = 6(number sold) = 6(-2x+40) Profits = revenue - costs P(x) = -2x^2 +40x +12x -240 = -2x^2 +62x -240 .......... perhaps?

OpenStudy (anonymous):

then take the derivative of that

OpenStudy (amistre64):

sure, cant hurt :)

OpenStudy (anonymous):

but do you think that would be a good idea to do, then i would find the critical values

OpenStudy (amistre64):

if the equation is good, and thats a big if these days, then yes, the derivative will help you find where the profits peak at

OpenStudy (amistre64):

I get a price of $15.50 lets see if it works -2(15.50) +40 = 9 sold 9(15.50) = 139.50 in revenue 6(9) = 45 in costs P(x) = 94.50 ------------------ how does that go for 15 and 16? 15: #sold = 10; revenue = 150, costs = 60, profit = 90

OpenStudy (amistre64):

16: #sold = 8, revenue = 128; costs = 48, profit = 80

OpenStudy (anonymous):

you got 15.50 from taking the derivative and finding the critical point right

OpenStudy (amistre64):

yes: the derivative of the profit: -4x +62 -4x +62 = 0 -4x = -62 x = -62/-4 = 15.50

OpenStudy (anonymous):

and you decided to test 16 just to check right

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