10. -/1 points details my notes tanapcalc10 4.4.057. maximizing revenue the quantity demanded each month of…

Explanation:

Step1: Define the revenue function

Revenue $R(x)=p\times x$, substituting $p = \frac{57}{0.01x^{2}+1}$ into it, we get $R(x)=\frac{57x}{0.01x^{2}+1}$, where $0\leq x\leq20$.

Step2: Differentiate $R(x)$ using the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 57x$, $u^\prime=57$, $v = 0.01x^{2}+1$, $v^\prime = 0.02x$. So $R^\prime(x)=\frac{57(0.01x^{2}+1)-57x(0.02x)}{(0.01x^{2}+1)^{2}}=\frac{0.57x^{2}+57 - 1.14x^{2}}{(0.01x^{2}+1)^{2}}=\frac{57 - 0.57x^{2}}{(0.01x^{2}+1)^{2}}$.

Step3: Set $R^\prime(x)=0$ to find critical points

Set $\frac{57 - 0.57x^{2}}{(0.01x^{2}+1)^{2}}=0$. Since the denominator $(0.01x^{2}+1)^{2}>0$ for all real $x$, we solve $57 - 0.57x^{2}=0$. Rearranging gives $0.57x^{2}=57$, then $x^{2}=100$, so $x = 10$ or $x=- 10$. But since $0\leq x\leq20$, we discard $x=-10$.

Step4: Check if $x = 10$ is a maximum

We can use the second - derivative test or simply note the behavior of the function. Since $R(x)$ is a continuous function on $[0,20]$ and we have a single critical point $x = 10$ in the interval $[0,20]$.

Answer:

$10000$