a company selling widgets has found that the number of items sold ( x ) depends upon the price ( p ) at…
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}} ).\ndue 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.\n( square ) dollars per month
Answer
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)