During the summer months Terry makes and sells necklaces on the beach. Last summer he sold the necklaces for $10 each and his sales averaged 20 per day. When he increased the price by $1, he found that he lost two sales per day.

a. Find the demand function, assuming it is linear.
b. If the material for each necklace costs Terry $6, what should the selling price be to maximize his profit?

for a, i got y=-.5x+20

i got the slope by: p(20)=10 and p(18)=11

how do i do part b

For part b, to maximize Terry's profit, we need to find the maximum point of the profit function. We have the demand function, q(p) = -0.5p + 20, where q is the quantity of necklaces sold and p is the price per necklace.

The profit function is given by Profit = Revenue - Cost. Terry's revenue is the price per necklace multiplied by the number of necklaces sold, which is p*q(p), and his cost is $6 per necklace, so his profit function is:

Profit(p) = p*q(p) - 6*q(p)

Substitute the demand function q(p) into the profit function:

Profit(p) = p(-0.5p + 20) - 6(-0.5p + 20)
Profit(p) = -0.5p^2 + 20p + 3p - 120
Profit(p) = -0.5p^2 + 23p - 120

Now, to find the maximum point, we can either find the vertex of the quadratic function, or find when the derivative is equal to 0. Let's find the vertex, which is given by:

p_vertex = -b/(2a) = -23/(2(-0.5)) = 23

So, the maximum profit occurs when the price is $23 per necklace. However, we should also check whether the demand is still positive at this price to ensure that this solution is feasible:

q(23) = -0.5(23) + 20 = -11.5 + 20 = 8.5

Since the demand is still positive, Terry should sell his necklaces for $23 each to maximize his profit.

To find the selling price that maximizes Terry's profit, we need to analyze the cost and revenue functions.

Let's first calculate Terry's revenue based on the demand function:

Revenue (R) = Selling Price (p) × Quantity Sold (x)

Since Terry initially sold 20 necklaces per day at $10 each, his revenue can be calculated as:

R = 10 × 20 = $200 per day

When Terry increased the price by $1, he lost two sales per day. So, his new quantity sold (x) is 18 per day. Therefore, the revenue with the new selling price can be calculated as:

R = p × 18

To find the demand function, we need to determine the new selling price (p). We can use the original demand function to solve for p:

-.5x + 20 = p

Substituting the new quantity sold (x = 18), we can find the new selling price:

-.5(18) + 20 = p
P = 11

Therefore, the new selling price for 18 necklaces is $11.

Now that we have the selling price, we can explore Terry's costs per necklace. Each necklace costs $6 to make.

Terry's Cost per necklace (C) = $6

Using the demand function, we know that the quantity sold (x) is 18 per day.

Terry's Total Costs (TC) = Cost per necklace × Quantity sold
TC = $6 × 18 = $108 per day

To find the profit, we subtract the total costs (TC) from the revenue (R):

Profit (P) = R - TC
P = (p × x) - (C × x)
P = 11 × 18 - 6 × 18
P = $198 - $108
P = $90 per day

Therefore, Terry's profit is $90 per day.

Since we want to find the selling price that maximizes profit, we need to determine the selling price (p) that will yield the highest profit. In this case, the selling price should be set equal to $11 (as found earlier), as this is the price that maximizes Terry's profit.

Therefore, Terry should sell each necklace for $11 in order to maximize his profit.

To find the selling price that maximizes Terry's profit, we need to calculate the profit function.

The profit is given by the equation:
Profit = (Selling Price - Cost) x Quantity

Let's denote the selling price as p, cost as c, and quantity as q.

From the problem statement, we know that the cost for each necklace is $6. So, c = 6.

From part a, we have the demand function as: q = -0.5p + 20.

We can rewrite this as p = 20 - 2q. Now substitute this value of p into the profit equation:

Profit = (20 - 2q - 6)q
= (14 - 2q)q
= 14q - 2q^2

To find the price that maximizes the profit, we need to find the vertex of the quadratic equation.

The vertex of a quadratic equation in the form ax^2 + bx + c is given by the formula:
x = -b / (2a).

In our case, a = -2 and b = 14. Substitute these values into the formula to find the quantity that maximizes the profit:
q = -14 / (2 * -2)
= 14 / 4
= 3.5

Now, substitute this value of q into the demand function to find the corresponding selling price:
p = 20 - 2(3.5)
= 20 - 7
= 13

Therefore, the selling price that maximizes Terry's profit is $13 per necklace.