Question

quotient rule: problem 2
(1 point)
suppose the number of cd players a retail chain is willing to sell per week at a price of p dollars is given by the function
s(p)=\frac{100p}{0.3p + 3}
find the supply and the instantaneous rate of change of the supply with respect to price when the price is 40 dollars.
s(40)=
s(40)=
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Explanation:

Step1: Calculate $S(40)$

Substitute $p = 40$ into $S(p)=\frac{100p}{0.3p + 3}$.
$S(40)=\frac{100\times40}{0.3\times40 + 3}=\frac{4000}{12 + 3}=\frac{4000}{15}=\frac{800}{3}\approx266.67$

Step2: Find the derivative of $S(p)$ using the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 100p$, $u'=100$, $v = 0.3p + 3$, $v'=0.3$.
$S'(p)=\frac{100(0.3p + 3)-100p\times0.3}{(0.3p + 3)^{2}}=\frac{30p+300 - 30p}{(0.3p + 3)^{2}}=\frac{300}{(0.3p + 3)^{2}}$

Step3: Calculate $S'(40)$

Substitute $p = 40$ into $S'(p)$.
$S'(40)=\frac{300}{(0.3\times40 + 3)^{2}}=\frac{300}{(12 + 3)^{2}}=\frac{300}{225}=\frac{4}{3}\approx1.33$

Answer:

$S(40)=\frac{800}{3}$
$S'(40)=\frac{4}{3}$