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

Explanation:

Step1: Recall the revenue formula

Revenue (R = p\times x). Given (x=\frac{70000}{\sqrt{4p + 1}}), then (R=\frac{70000p}{\sqrt{4p + 1}}).

Step2: Use the quotient - rule for differentiation

The quotient rule is ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}). Let (u = 70000p) and (v=(4p + 1)^{\frac{1}{2}}). Then (u^\prime=70000) and (v^\prime=\frac{4}{2}(4p + 1)^{-\frac{1}{2}} = 2(4p + 1)^{-\frac{1}{2}}).

[ \begin{align*} R^\prime&=\frac{70000(4p + 1)^{\frac{1}{2}}-70000p\times2(4p + 1)^{-\frac{1}{2}}}{4p + 1}\ &=\frac{70000(4p + 1)-140000p}{(4p + 1)^{\frac{3}{2}}}\ &=\frac{280000p+70000 - 140000p}{(4p + 1)^{\frac{3}{2}}}\ &=\frac{140000p + 70000}{(4p + 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) (price is increasing by $3 per month).

When (p = 120), first find ((4p + 1)=(4\times120+1)=481)

[ \begin{align*} \frac{dR}{dp}&=\frac{140000\times120+70000}{481^{\frac{3}{2}}}\ &=\frac{16800000+70000}{481^{\frac{3}{2}}}\ &=\frac{16870000}{481^{\frac{3}{2}}} \end{align*} ]

Then (\frac{dR}{dt}=\frac{16870000}{481^{\frac{3}{2}}}\times3)

[ \begin{align*} 481^{\frac{3}{2}}&=(481^{\frac{1}{2}})^3=\sqrt{481}^3\approx21.93^3\approx10639.7 \end{align*} ]

[ \begin{align*} \frac{dR}{dt}&=\frac{16870000\times3}{10639.7}\ &=\frac{50610000}{10639.7}\approx - 4757 \end{align*} ]

Answer:

(-4757)