a company selling widgets has found that the number of items sold ( x ) depends upon the price ( p ) at…

Explanation:

Step1: Find the revenue function

Revenue (R = p\times x). Given (x=\frac{90000}{\sqrt{6p + 1}}), then (R(p)=p\times\frac{90000}{\sqrt{6p + 1}}=\frac{90000p}{(6p + 1)^{\frac{1}{2}}})

Step2: Use the quotient rule for differentiation

The quotient rule is (\left(\frac{u}{v}\right)'=\frac{u'v - uv'}{v^{2}}). Here (u = 90000p), (u'=90000) and (v=(6p + 1)^{\frac{1}{2}}), (v'=\frac{6}{2}(6p + 1)^{-\frac{1}{2}} = 3(6p + 1)^{-\frac{1}{2}})

[ \begin{align*} R'(p)&=\frac{90000(6p + 1)^{\frac{1}{2}}-90000p\times3(6p + 1)^{-\frac{1}{2}}}{6p + 1}\ &=\frac{90000(6p + 1)-270000p}{(6p + 1)^{\frac{3}{2}}}\ &=\frac{540000p+90000 - 270000p}{(6p + 1)^{\frac{3}{2}}}\ &=\frac{270000p + 90000}{(6p + 1)^{\frac{3}{2}}} \end{align*} ]

Step3: Use the chain - rule

We know that (\frac{dR}{dt}=\frac{dR}{dp}\times\frac{dp}{dt}). Given (\frac{dp}{dt}=3)

When (p = 190), first find (6p+1=6\times190 + 1=1141)

[ \begin{align*} \frac{dR}{dp}&=\frac{270000\times190+90000}{(1141)^{\frac{3}{2}}}\ &=\frac{51300000+90000}{(1141)^{\frac{3}{2}}}\ &=\frac{51390000}{1141\sqrt{1141}} \end{align*} ]

Then (\frac{dR}{dt}=\frac{dR}{dp}\times\frac{dp}{dt})

[ \begin{align*} \frac{dR}{dt}&=\frac{51390000}{1141\sqrt{1141}}\times3\ &=\frac{154170000}{1141\sqrt{1141}}\ &\approx\frac{154170000}{1141\times33.78}\ &\approx\frac{154170000}{38542.98}\ &\approx - 4000 \end{align*} ]

Answer:

The rate at which revenue is changing is approximately (-4000) dollars per month.