A clothing company has compiled data based on their records of annual jacket sales. They have found a relationship between the sales of one jacket and that jackets's price: q = -20s + 1200, where q = quantity sold and s = selling price of the item. The cost to produce each jacket is $10.

a. Using the Profit formula P = Total Revenue - Production Costs (in this case, P = sq - 10q), write an equation that could be used to find the selling price, s, that produces the maximum profit.

(Hint: Take the equation for q and substitute it into the equation for P, then simplify.)

Question

A clothing company has compiled data based on their records of annual jacket sales.  They have found a relationship between the sales of one jacket and that jackets's price: q = -20s + 1200, where q = quantity sold and s = selling price of the item.  The cost to produce each jacket is $10.

 

a.  Using the Profit formula P = Total Revenue - Production Costs (in this case, P = sq - 10q), write an equation that could be used to find the selling price, s, that produces the maximum profit.

(Hint: Take the equation for q and substitute it into the equation for P, then simplify.)

anonymous · Answer

b.  Using that equation, what selling price should they set to maximize profit?

(Hint: Think about the general shape of the graph from your answer in Part A.  What part of the graph represents the maximum?  How do you find it?)

anonymous · Answer

Are you taking calculus?

anonymous · Answer

This is an algebra course, why?

anonymous · Answer

Different way of solving it.

anonymous · Answer

You can teach me either one, as long as it works I'm fine with it.

anonymous · Answer

Always start with what you're given:

$$q = -20s + 1200$$$$P = sq - 10q = q(s - 10)$$
The instructions tell you to plug the equation for q into the equation for P, so go ahead and do that:

$$P = (-20s + 1200)(s -10)$$$$P = -20s^{2} + 1200s + 200s -12000$$$$P = -20s^{2} + 1400s -12000$$$$P = -20(s^{2} - 70s + 600)$$
That answers part a. You can solve b using derivatives (calculus) but you're asked to do it based on graphs, so I'll leave that up to you.

anonymous · Answer

Alright thanks I got it!

anonymous · Answer

In case you're curious, the derivative measures the change in the curve (slope) at any given point. At both the minimum and maximum (there isn't always both), the change, or slope, will be 0. So once you have the equation for the derivative, you just need to solve it for 0. 

In a problem like this, finding the equation for the derivative is easy because you only have one variable on the righthand side of your equation. You just take each power of s, multiply the constant for that power by the power, and subtract one from the power. So s^2 becomes 2s^1, or 2s. s becomes 1s^0, or 1, and a constant (no s) becomes 0:

$$-20(2s - 70 + 0) = -20(2s -70)$$
Now just solve this:

$$-20(2s -70) = 0$$
There's only one answer, so it's either a maximum or a minimum. Plug it into the original equation along with any other value (0 works nicely) to tell which.