Respuesta :
Answer:
(a)The rate of change of price to the number of units sold when 40 units are sold is -$5.28.(decrease)
(b)The rate of change of price to the number of units sold when 40 units are sold is -$1.89.(decrease)
(c)The p' is decreasing at 40 units.
Step-by-step explanation:
Given function is
[tex]p=\frac{5000}{ln(x+10)}[/tex]
where p is the price per unit dollar when x units are demanded.
[tex]p=\frac{5000}{ln(x+10)}[/tex]
Differentiating with respect to x
[tex]p'=\frac{d}{dx}(\frac{5000}{ln(x+10)})[/tex]
[tex]\Rightarrow p'=\frac{\frac{d}{dx}(4000).(ln(x+10))-4000.\frac{d}{dx}(ln(x+10))}{(ln(x+10))^2}[/tex]
[tex]\Rightarrow p'=- \frac{4000}{(x+10)(ln(x+10))^2}[/tex]
Again differentiating with respect to x
[tex]p''=-\frac{\frac{d}{dx} (4000).(x+1) (ln (x+10)) ^2- 4000.\frac{d}{dx} (x+1) (ln (x+10))^2}{(x+1)^2 (ln (x+10))^4}[/tex]
[tex]\Rightarrow p''=\frac{4000[(ln(x+10))^2.1+(x+10).\frac{2(ln(x+10))}{(x+10)}]}{(x+10)^2(ln(x+10))^4}[/tex]
[tex]\Rightarrow p''=\frac{4000[(ln(x+10))^2+{2(ln(x+10))}]}{(x+10)^2(ln(x+10))^4}[/tex]
[tex]\Rightarrow p''=\frac{4000[(ln(x+10))+2]}{(x+10)^2(ln(x+10))^3}[/tex]
(a)
Now,
[tex]\left p'\right|_{x=40}=- \frac{4000}{(40+10)(ln(40+10))^2}[/tex]
=- $5.28
The rate of change of price to the number of units sold when 40 units are sold is - $5.28.
(b)
[tex]\left p'\right|_{x=90}=- \frac{4000}{(90+10)(ln(90+10))^2}[/tex]
= -$1.89
The rate of change of price to the number of units sold when 40 units are sold is -$1.89.
(c)
[tex]p''|_{x=40}=\frac{4000[(ln(40+10))+2]}{(40+10)^2(ln(40+10))^3}>0[/tex]
Since p''>0.
The rate is decreasing at 40 units.
