really hard math problem?

Question

A student makes and sells necklaces at the beach during the summer months. The material for each necklace costs him $6 and he sells an average of 20 per day at $10 each. He’s been wondering whether he should raise the price, so he conducts a survey and finds that for every dollar increase he loses two sales a day. What price should he set for the necklaces to maximize his profit?

please show all work..

Answer 1

First write an equation

Let P = profit
Let C = sale amount
Let N = number sold

P = (C - 6)*N

Now maximize P, subject to the constraint that as C increases N decrease by 2

P = (10 - 6) * 20 = 80
P = (11 - 6) * 18 = 90
P = (12 - 6) * 16 = 96
P = (13 - 6) * 14 = 98 <= Max
P = (14 - 6) * 12 = 96

Clearly an other increse will continue to lower the price.

Answer 2

Let P be your profit, x be the price of the necklaces.

So your profit will be

P = x(40 - 2x) - 6(40 - 2x)

So from this eqn, you will be able to incorporate the "every dollar increase in price, he loses two sales a day" with the (40 - 2x) term. In other words, the amount of sales is (40 - 2x).

Hence, after getting that expression, use differentiation to find price that will maximise the profit.

So
P = x (40 - 2x) - 6 (40 - 2x)
= 40x - 2x^2 - 240 + 12x
= -2x^2 + 52x - 240

dP/dx = -4x + 52

Set dP/dx = 0 to find max/min.

Hence,
-4x + 52 = 0
x = 52/4
x = 13
P(13) = 13(40 - 2(13)) - 6(40 - 2(13))
= 13(40 - 26) - 6(40 - 26)
= 13(14) - 6(14)
= 7(14)
= 98

Hence, the price of the necklace should be set to $13 to maximise his profit. Maximum profit is $98 per day.

Answer 3

Currently his profit is 20*4 =80
If he increases price to 11, profit = 5*18 = 90
If he increases price to 12, profit = 6*16 = 96
If he increases price to 13, profit = 7*14 =98
If he increases price to 14, profit is 8*12 = 96
Thus max occurs when price is $13

Answer 4

Old av P = 20(10 - 6)
New av P = (20 - n)(10 + x - 6)
Where
x is the number of single dollar increases
n = 2x, for each dollar increase, then
New av P = (20-2x)(4 + x) = 80 + 12x - 2x^2
dP/dx = 0 + 12 - 4x =set= 0
4x = 12
x = 3
New prince then, since x = 3, is 10+x = 13 dollars

Answer 5

a million. enable S be a set such that for each element x of S there exists a special element x' of S. 2. there is an element in S, we will call it a million, such that for each element x of S, a million isn't equivalent to x'. 3. If x and y are components of S such that x' = y', then x = y. 4. If M is any subset of S such that a million is an element of M, and for each element x of M, the element x' is likewise an element of M, then M = S. merely as a remember of notation, we write a million' = 2, 2' = 3, and so forth. We define addition in S as follows: (a1) x + a million = x' (a2) x + y' = (x + y)' The element x + y is noted as the sum of x and y. Now to tutor that a million + a million = 2. From (a1), with x = a million, we see that a million + a million = a million' = 2.

Answer 6

20 20
x6 x10

120 200

200
-120= 80 get the supplies for less