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.
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
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