Question

20. If the price charged for a candy bar is p(x) cents, the x thousand candy bar will be sold in a certain city, where P(x) = 6- 20 (a) Find an expression for the total revenue for the sale of x thousands candy bar: 20 (b) Find the value of x that leads to maximum revenues: 20 (©) The maximum revenue:

Answer 1

One candy bar is sold for p(x) cents. Therefore, one candy bar is sold for 60 dollars. Therefore, one thousand candy bars are sold for 1000x760 10p(-x) dollars. Therefore, thousand candy bars are sold for z. · 10p(2) dollars. Let R(2) represent that amount. Then

R(2) = r.10p(2) = 10:0 (6 (6-5) = (60 60.C 5 5.62 4 dollars

$\text{For maximum value }R'(x)=0$

$\Rightarrow \left ( 60x-\frac{5x^2}{4} \right )' = 0$

$\Rightarrow 60\left (x \right )'-\frac{5}{4}\left (x^2 \right )' = 0$

$\Rightarrow 60\left (1 \right )-\frac{5}{4}\left (2x \right ) = 0$

$\Rightarrow 60 - \frac{5x}{2} = 0$

$\Rightarrow \frac{5x}{2} = 60$

$\Rightarrow x = \frac{60\times 2}{5}$

$\Rightarrow x = 24$

$\text{As from second derivative test, }R''(24) = -\frac{5}{2} <0\text{ then }x=24\text{ must be a maximum}$

$\text{We find}$

$R(24) = \left ( 60\left ( 24 \right )-\frac{5\left ( 24 \right )^2}{4} \right ) = 1440 - 720 = 720 \quad dollars$

For any doubts, do ask in the comments