Linear Demand Equation - Max Revenue 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!
 A: 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 $20$.  
A: Your equation for $A$ is not quite correct. You have the slope correct, but since you are given $A = 30$ when $p = \$20$ you need to solve
$$30 = -2(20) + b$$
for $b$. This gives $b = 70$ so $$A = -2p + 70$$
You now have $$R = pA = p(-2p + 70) = -2p^2 + 70p$$ You can find the derivative of $R$ with respect to $p$, set it to $0$, and solve for $p$. This gives you the price that maximizes the revenue, which you can use in your equations for $R$ and $A$ to find the maximum revenue and number of widgets are sold at that maximum.
