Question:
A company finds that the cost of producing $x$ calculators is $C(x)=100+10x+0.01x^2$ dollars. The price for which each calculator can be sold at production level $x$ is $p(x) = 26 -0.1x$ dollars. (The equation for $p(x)$ is called the demand equation.) Assuming that all the calculators produced are sold, find the revenue function, $R(x)$. How many calculators should be manufactured in order to maximise the profits the company makes?
Thinking:
$R(x) = p(x) - C(x)$, therefore $R(x) = -0.01x^2-10.1x-74$. In order to maximise profits, $R'(x)$ has to be found and set to $0$ to extract $x$, which is the number of calculators.
So, if $R(x)$ is correct, then $R'(x)=-0.02x-10.1$. But, when $R'(x)=0$, $x=-505$.
Note:
But I am not sure, if the equation for $R(x)$ is correct and if the method for finding the number of calculators is correct. Also, not sure what a negative answer for $x$ means in this context. Does it mean that $505$ less calculators are to be manufactured in order to maximise the profit, or is $R(x)$ wrong?
EDIT: (after solution given)
$R(x)=x\times p(x)$, therefore $R(x)=26x-0.1x^2$. Now, $P(x)=R(x)-C(x)=-0.11x^2+16x-100$. Find $P'(x)$ and get $P'(x)=-0.22x+16$, so $x=\dfrac{16}{0.22}=72.7$.
Your equation for "revenue" is wrong in two ways. Revenue is the amount of money the company gets from sales. If the company makes $10$ calculators, for example, the price for each calculator is $p(10) = 25,$ and the revenue from selling $10$ calculators at that price is $$ 10 \times 25 = 250.$$
In general, revenue is simply (number of units sold) $\times$ (price per unit). In terms of your problem statement, the revenue is $$ R(x) = x \times p(x).$$
Cost should never be included in the revenue function. After you have finished computing revenue, then you can subtract cost to find the profit.