# Quadratics: Terry was selling necklaces on the beach last summer. His sales averaged 30/day with $10 each. He is going to increase the price this year. A survey showed that for every dollar...

Quadratics:

Terry was selling necklaces on the beach last summer. His sales averaged 30/day with $10 each. He is going to increase the price this year. A survey showed that for every dollar increase he would lose 2 sales per day. What should the selling price be to maximize profits?

And what should the equation be?

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In order to figure out what the new selling price needs to be in order to maximize profits we need to start by figuring out how increasing the cost per necklace decreases the number sold.

I would start out by letting `x ` equal the amount we are going to increase our price and let `p` equal our total profit.

The amount of necklaces sold would be equal to `30-2x`.

This describes the condition of losing 2 sales per dollar increase (`x `).

The cost of each necklace would be ` 10+x `.

This represents an increase of ` x` dollars per necklace.

Therefore the total profit (` p ` ) would be equal to our new price times how many we sell each day.

This would be represented by the equation:

` (10+x)(30-2x)=p `

If we multiply the terms on the left hand side of the equation and combine like terms we get:

`-2x^2+10x+300=p `

In order to find ` x` (the amount to increase our cost) we need to see where the maximum is for our equation.

This is a graph of our equation for profit. From the graph you can see that profit is maximized at somewhere between 2 and 3 dollars. I will zoom in to get a closer look.

Now, in order to see where are maximum is exactly we need to find where the slope is equal to 0. You can find this by finding the first derivative of our equation for profit, and finding its zero.

The first derivative would be ` -4y+10=p' ` .

Solving for where `p'=0` would give us:

`-4y+10=0 `

`-4y=-10 `

`4y=10 `

`y=5/2 `

Therefore in order to maximize our profits, we would need to increase our selling price by $2.50 for a total cost of $12.50.

This would increase our profits from $300.00 per day to $312.50 per day, selling them at $12.50.

When Terry was selling necklaces on the beach last summer the number of necklaces sold averaged 30 per day with each sold for $10. Terry's average daily profit was 30*10 = $300.

Terry plans to increase prices this year and a survey has revealed that an increase in price of $1 would reduce sales by 2. If the price is increased by x, the number of items sold decreases by 2x. The total profit made as a function of x is P(x) = (10 + x)*(30 - 2x)

= 300 + 30x - 20x - 2x^2

= -2x^2 + 10x + 300

P(x) has to be maximized. To determine the value of x at which P'(x) is maximum solve P'(x) = 0 for x.

P'(x) = -4x + 10

-4x + 10 = 0

=> x = 2.5

If the price is kept at $12.5, the profits made are maximized.

To come up with an equation we will use 2 variables. x and y. X will represent cost and Y will represent your money earned, or profit. The problem says that each necklace was sold for $10. If you were going to increase the price you would be adding a X amount of dollars which is represented as 10+x.

The problem also says that you sold an average of 30 necklaces a day. However if you increase the price you will lose about 2 sales a day. This would be written as 30-2x. You will need to multiply these two parts together to get your profits.

y=(10+x)(30-2x)

y=300-20x+30x-2x^2

y=-2x^2+10x+300

To find the maximum you need to find where the slope reaches 0 by setting y equal to 0 and solving for x.

0=-4x+10

-10=-4x

-10/-4=x

2.5=x

This means that the price can increase by $2.5, having your selling price be **$12.5** and maximizing your profits.