Q.(a) A firm has the marginal revenue function given by $MR=\dfrac{a}{(x+b)^{2}}-c$. Where $x$ is the output and $a,b,c$ are constants. Show that the demand function is given by $x=\dfrac{a}{b(p+c)}-b$

(a) $R=\int MR\,dx$ with $R(0)=0$ gives $p=a/[b(x+b)]-c$, hence $x=a/[b(p+c)]-b$. (b) $Z=-5$ exceeds the $1\%$ critical value $2.58$, so the claim is rejected.

Part (a) — Demand from marginal revenue $MR=\dfrac{a}{(x+b)^{2}}-c.$

$R=\int MR\,dx=\int\!\left[\frac{a}{(x+b)^{2}}-c\right]dx=-\frac{a}{x+b}-cx+k.$

Revenue is zero when output is zero, so $R=0$ at $x=0$: $0=-\dfrac{a}{b}+k\Rightarrow k=\dfrac{a}{b}.$

$R=\frac{a}{b}-\frac{a}{x+b}-cx=\frac{a[(x+b)-b]}{b(x+b)}-cx=\frac{ax}{b(x+b)}-cx.$

Price = average revenue $p=\dfrac{R}{x}$:

$p=\frac{a}{b(x+b)}-c\Rightarrow p+c=\frac{a}{b(x+b)}\Rightarrow b(x+b)=\frac{a}{p+c}.$

$x+b=\frac{a}{b(p+c)}\Rightarrow x=\frac{a}{b(p+c)}-b.$

Hence the demand function is $x=\dfrac{a}{b(p+c)}-b.$

Part (b) — Test of a single mean. $\mu=400,\ \sigma=20,\ n=100,\ \bar x=390.$

$H_0:\mu=400$ (claim true) vs. $H_1:\mu eq400.$ …