Question

a company selling widgets has found that the number of items sold ( x ) depends upon the price ( p ) at which, theyre sold, according the equation ( x=\frac{90000}{sqrt{6 p + 1}} ).
due to inflation and increasing health benefit costs, the company has been increasing the price by ( $ 3 ) per month. find the rate at which revenue is changing when the company is selling widgets at ( $ 190 ) each.
( square ) dollars per month

Explanation:

Step1: Find the revenue function

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

Step2: Differentiate the revenue function using the product rule

The product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u = 90000p\) and \(v=(6p + 1)^{-\frac{1}{2}}\)
\(u^\prime=90000\)
\(v^\prime=-\frac{1}{2}(6p + 1)^{-\frac{3}{2}}\times6=- 3(6p + 1)^{-\frac{3}{2}}\)
\(R^\prime(p)=90000(6p + 1)^{-\frac{1}{2}}+90000p\times(-3)(6p + 1)^{-\frac{3}{2}}\)
\(R^\prime(p)=\frac{90000}{\sqrt{6p + 1}}-\frac{270000p}{(6p + 1)^{\frac{3}{2}}}\)
Factor out \(\frac{90000}{(6p + 1)^{\frac{3}{2}}}\):
\(R^\prime(p)=\frac{90000(6p + 1)-270000p}{(6p + 1)^{\frac{3}{2}}}=\frac{540000p+90000 - 270000p}{(6p + 1)^{\frac{3}{2}}}=\frac{270000p + 90000}{(6p + 1)^{\frac{3}{2}}}\)

Step3: Find the value of \(p\) and substitute into \(R^\prime(p)\)

Given \(p = 190\), then \(6p+1=6\times190 + 1=1141\)
\(R^\prime(190)=\frac{270000\times190+90000}{(1141)^{\frac{3}{2}}}\)
First, calculate the numerator: \(270000\times190+90000=51300000+90000 = 51390000\)
\((1141)^{\frac{3}{2}}=\sqrt{1141^3}\approx\sqrt{(1141)^2\times1141}=1141\sqrt{1141}\approx1141\times33.78=38542.98\)
\(R^\prime(190)=\frac{51390000}{38542.98}\approx - 1333.33\)

Answer:

\(-1333.33\)