Optimization?

Question

I have a question that might be on my test tomorrow but I'm not sure how to do it.
Can you guys help me?

During the summer months Terry makes and sells necklaces on the beach. Last year he sold neclaces for $10 each. He sales 20 per day. Considering a price increase, he took a small survey and found that for every dollar increase he would lose two sales per day. If the material for each necklace costs Terry $6, what would the selling price be to maximize profits? Determine the profit.

Thank you very much

Answer 1

Here is an idea: Let p = the price increase. Then 20 - 2p is how many necklaces he would sell, and p+6 would be the price, so income would be (20 - 2p)(p + 6) and expenses would be 6(20-2p)



So profit is (20-2p)(p+6)-6(20-2p)



Multiply this out and find its maximum by -b/(2a)

Answer 2

Call the amount he raises the price R, the number of customers N, the profit per necklace p and the profit P. Now his profit is equal to the profit per necklace times the number of customers. Mathematically, we write P = pN.



p, the profit per necklace, rises when the price rises. p = 4 + R (because his current price is $4 above retail, his profit is $4).



N, the number of customers, drops when the price rises. N = 20 - 2R.



So P = (4 + R)(20 - 2R), and we need to maximize P.



Start by multiplying out the parentheses. P = 80 + 20R - 8R -2R^2 = -2R^2 + 12R + 80.



Now take the derivative of this equation. P' = -4R + 12.



Now set the derivative equal to 0. -4R + 12 = 0. Subtract 12 from each side: -4R = -12. Divide by -4: R = 3. This is the only point at which the derivative is 0, so it's the only candidate to be a maximum. So he should raise the price by $3, to $13.



The profit from this price is $7 (the profit per necklace) * 14 (the number of customers) = $98/day.

Answer 3

Generally optimization problems in pre-calculus algebra involve translating the problem into a quadratic equation, which graphs as a parabola, then finding the vertex of the parabola, which would be the minimum or maximum that you're looking for. Then when you get into calculus you solve general optimization problems, those which don't translate into quadratic equations, by finding and solving for derivatives. I assume you're in precalculus algebra so I'll explain how to do it that way first.



let n = number of necklaces sold, d = dollar increase

Cost = 6n

Selling price = 10 + d

Profit = selling price - cost = (10+d)n - 6n = 10n + dn -6n = 4n + dn = (4+d)n

Now we need to relate n and d. For n = 20, d = 0. For n = 18, d = 1. For n = 16, d = 2. In general, n = 20 - 2d. So now we can substitute Profit = (4+d)(20-2d) or 2(4+d)(10-d). Multiply this out and you get profit = -2d^2 +12d+40. This is, as I suspected it would be, a quadratic equation, and its graph is a parabola which opens downward and has a maximum at its vertex. Now to find the vertex. Remember the formula, that the vertex of a parabola is (-b/2a) along with its function value. So -b/2a would be -12/2x(-2) = 3. Remember d represented the price increase so that means you can increase the price by 3 dollars, to 13 dollars. To find the corresponding profit, plug d into the equation. Since I don't have a calculator handy I'll use the factored form of the equation, 2(4+d)(10-d) = 2x7x7 = 98 dollars. Good luck on your test! Remember, assuming your'e ion precalculus algebra, whenever they want you to optimise sometihng, that means, translate it into a quadratic equation, a parabola, and find the vertex.



If you are in calculus you'd still translate the problem into a quadratic equation but then you'd take the first derivative and set it to zero. The first derivative of this one is -4d+12 = 0 so you still get d = 3. But be careful when you do stuff like tihs, rememeer what your variable stood for. A common mistake is to figure out d, for example, and then say that he needs to sell 3 necklaces, mistkaing d for n. That's why I use variable names that stand for what I'm looking for, rather than generic stuff like x and y.

Answer 4

SO, his profit would be p=(10-6+x)(20-2x)=(4+x)(20-2x) =80-8x+20x-2x^2.

SO, to maximzie you differinate and set to 0.

So, -8+20-4x=0, or 12=4x, so, x=3.



So, that means he should sell it at 10+3=13.00 The profit is (4+3)(20-6) = 7*14=98.

Answer 5

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