An oil producing country can sell 7 million barrels of oil a day at a price of $90$ per barrel. If each $1$ price increase will result in a sales decrease of $100,000$ barrels per day, what price will maximize the country's revenue? How many barrels will it sell at that price?
This is how I set it up:
$x$=each $1 price increase
$$P(x)=90+x,\ q(x)=7-100,000x$$ $$R(x)=(90+x)(7-100,000x)$$
Am I doing this right so far?
$\endgroup$ 21 Answer
$\begingroup$Yep! Your revenue function looks good: $$\text{revenue} = \text{price} \times \text{quantity}$$ The price is $90 + x$, where $x$ is the change in price, and the quantity is $7,000,000 - 100,000x$ (I'm going to assume that's what you meant when you wrote $(7-100,000x)$). Now just take the derivative of that function. $$r(x) = (90+x)(7,000,000 - 100,000x) = -100,000x^2 - 2,000,000x + 630,000,000 \Rightarrow r'(x) = -200,000x - 2,000,000$$. For $r(x)$ to be at either a maximum or a minimum at a given $x$, $r'(x)$ must be 0 at that point, so set $r'(x) = 0$ and see where this occurs (we'll call that point $x_1$). This doesn't guarantee that $r(x)$ is maximized at $x_1$, just that it can be, so you need to check to make sure it is. If $r'(x) > 0$ before $x_1$ and $<0$ after, you know $r(x)$ is increasing up to $x_1$ and decreasing after, which means it is in fact a local maximum. To find the global maximum, see if there is more than one $x$ for which $r(x)$ is locally maximized, and compare the values of $r(x)$ at all those points. The $x$ for which r(x) is highest of those points will be your global maximum, and that's where revenue is maximized. The revenue-maximizing price, therefore, is $90+x$, and the number of barrels sold is $7,000,000 - 100,000x$.
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