**BACKGROUND**

First we need some basics (assuming everything is linear, we continue):

We logically know that:

Profit = What you make – What you spend

In math, that is:

**P = Revenue – Cost**

**(1) P = R – C**

And,

Revenue = price * quantity

**(2) R = px**

**(3) Cost** = **(variable cost)*x + (fixed cost)**

Now, there is a difference between big **P (profit) **and little **p (price or demand)**

We usually assume price is linear, so:

**(4) p = mx + b**

Everything in **BOLD** are things you must know!

OK…. Now let’s start deciphering the actual problem:

**Q: A manufacturer sells 150 tables a month at the price of $200 each. For each $1 decrease in price, he can sell 25 more tables. The tables cost $125 to make. Express monthly profit as a function of the price, draw a graph and estimate the optimal selling price.**

Let’s take the data out… Notice there is info on the price… I will write it as a data point: (quantity, price) = (x, p)

Point (a) (150, $200)

Point (b) (175, $199) [because for everyone $1 decrease, he sells 25 more!]

This is data to make our price equation from (4): **p = mx + b**

First, find m (the slope)…. m = (200 – 199) / (150 – 175) = – 1 / 25

So, **p = -1/25x + b**

Now, we need to solve for **b**. Plug in any point for x and p.. I will pick point (a):

200 = -1/25(150) + b

200 = -6 + b

206 = b

Therefore:

**p = -1/25x + 206**

**Above is the price function (or demand function, depending on what your teacher calls it)**

Now, Revenue = price * x

**R = p*x**

**R = (****-1/25x + 206)*x**

**R = ****-1/25x² + 206x**

**Above is your revenue equation.**

Cost = $125 per table

**C = 125x**

**Above is your cost equation.**

Phew! Now, Profit = Revenue – Cost

**P = ****(-1/25x² + 206x) – (125x)**

**P = ****-1/25x² + 81x**

**Above is your profit function. You are suposed to graph that, I will leave that to you.**

But, there is one more question…** Estimate the optimal selling price.** This essential means to find the price that maximizes profit. So, we must** maximize profit **(which means, take the derivative of profit and set it equal to zero, for starters):

**P = ****-1/25x² + 81x**

**P’ = -2/25x + 81**

**0 = -2/25x + 81**

**Now solve:**

**2/25x = 81**

**x = 1012.5**

So, what you have found is the number of tables to sell that maximizes profit (1012.5 tables). However, they want the price that maximizes profit! So, we know the price equation is:

**p = -1/25x + 206**

Now, plug in x = 1012.5 to find price (p):

**p = -1/25(1012.5) + 206**

**p = 165.5**

So, the price that maximizes your profit is $165.5

[Since this was a long problem, I need to check the answer for any minor algebra errors… I don’t have time to do that now, I will leave that to you and check it when I return to make any corrections]

Thank you that was short (efficient) but understandable (effective)

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Your instructions were very clear and easy to follow. I checked the accuracy of the algebra and found everything correct. Thank you.

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