Revenue and quadratic formula - for every x increase in price there are y fewer sales

Question

"When a shoe costs $\$80.00$, there are $300$ sales. Every $\$5.00$ increase in price will result in 10 fewer sales. Find the price that will maximize income."

I am able to solve the question just fine, but I am confused about the logic in setting up the quadratic formula.

The first step is to let $x$ be a $\$5$ increase in price, and you would plug in $x$ in the equation.

$$y = (80 + 5x)(300 - 10x)$$

The equation above is where I am confused. If $x$ is equal to one $\$5$ increase in price, then why wouldn't the equation be:

$$y = (80 + x)(300 - 2x)$$

Answer 1

$x$ is not the price increase, $\$5$ is the price increase.

$x$ is just the number of price increases (like, say, number of five-dollar bills customer would have to pay extra) and also the corresponding number of sales decreases (each decrease is $10$ sales).

One price increase brings one sales decrease, so:

$$ y = (80 + 5\cdot 1)(300 - 10\cdot 1). $$ Two price increases bring two sales decreases:

$$ y = (80 + 5\cdot 2)(300 - 10\cdot 2). $$

And so on.

Answer 2

Let say $x$ is the number of times you increase (or decrease) the price by $\$5$.

The revenue is the price $80+x\cdot 5$ by the number of sales.
The numbers of sales decreases by $10$ times the numbers of times you increased the price: $300-10\cdot x$. So the revenue is $$(80+5\cdot x)(300-10\cdot x)$$ $$= -50\cdot x^2+700\cdot x+24000$$ Now you want to maximize the revenue. Thus you do not need to care about the $24000$ (that won't change). So you want to maximize $$-50\cdot x^2+700\cdot x$$Dividing by $50$ $$=-x^2+14\cdot x$$ Since the sign of $-x^2$ is negative, the parabola curve has the shape of a $\bigcap$, thus finding the point where the derivative is zero would be an easy way to find the max, but I'm not sure you studied the derivatives yet^*...

So what we can do is find where the equation gives a zero result, and then simply take the middle point, that'll give the max value $$-x^2+14\cdot x$$ $$=(14-x)x$$ Giving zero for $x=0$ and $x=14$. Resulting as the max being the middle point, $$x=(14-0)/2=7$$

^{(from Wolfram alpha)}

^*(using derivatives: we find where the derivative gives zero. The derivative is $-100\cdot x+700$ which gives $0$ for $x=7$)

Answer 3

This is a little too long to be a comment, so I have posted an "answer". You can deal with the price increment in the way you suggest, but it would only alter the factor involving price. ir7 proposes thinking of $ \ x \ $ as the number of additional $ \ \$ 5 \ $ bills a purchaser would need to pay out; you could also treat it as the number $ \ u \ $ of such bills the price requests. In that case, you would need to give the "base price" for the shoes as $ \ u \ = \ 16 \ · ( \$ 5 ) \ \ . $ If the price is increased by $ \ 1 \ \ $ (which is five dollars), the number of pairs of shoes sold still declines by $ \ 10 \ \ . $ You would then write the factors as $ \ ( 16 + u ) · (300 - 10u) \ \ ; $ the revenue would still be maximized for $ \ u = 7 \ \ , $ just as the maximum is reached at $ \ x = 7 \ \ $ in the original formulation. At the end, to find the maximum revenue itself, you would need to convert $ \ (16 + 7)·(300 - 70) \ = \ 5290 \ $ five-dollar bills into $ \ \$ 26,450 \ \ . $

This comes to the same thing as taking the original factors $ \ (80 + x)·(300 - 10x) \ \ $ and extracting a factor of $ \ 5 \ $ to write the product as $ \ 5·\left(16 + \frac{x}{5} \right)·(300 - 10x) \ \ , $ which has no effect on the values associated with the maximum revenue.