I was having a bit of trouble with a math question regarding a "linear demand equation".
The problem asks: "A company can sell $30$ products at a price of \$$20$ per product. For every dollar in price increase they sell 2 fewer products per day. Find a linear equation for the amount expected to sell, $A$, as a function of price, $p$."
The equation turned out (I hope) to be: $$A = (-2/1)p + 30$$
However, it then asks with the equation $$R = pA$$ at what price will revenue ($R$) be highest, what is the highest price, and how many widgets are sold at highest. Basically, I need to find the vertex, but this is where I'm stuck. I'm not really sure how to proceed from here, as I always have trouble finding the vertex without a graph. What I tried was inputting a random price like so: $$R = 2(-2/1(2)+30)$$ which gave me $56$, but that doesn't fit. Could anyone point me in the right direction? I struggle really badly with these.
Thanks!
Best Answer
The amount $A$ sold at price $p\ge 20$ should be $(-2)(p-20)+30$. For the problem says that for every dollar in price increase the amount sold decreases by $2$.
So $A=70-2p$, and therefore $R=70p-2p^2$.
The curve with equation $y=70x-2x^2$ is a downward-facing parabola. We can find the vertex by completing the square, or by noting it is halfway between the roots of $70x-2x^2=0$.
Remark: We end up with the fact that $70p-2p^2$ is maximized at $p=\frac{35}{2}$. This is below the price of $20$.
There are two possible conclusions. If the price/demand relationship holds below $p=20$, then we maximize revenue at $p=\frac{35}{2}$.
However, if the relationship does not hold below $20$, or for some reason going below $20$ is forbidden, then maximum revenue is at [price $20$.