Question

A company manufactures and sells x television sets per month. The monthly cost and​ price-demand equations are ​C(x)=72,000+60x and p(x)=300−(x/20​),
0l≤x≤6000.


​(A) Find the maximum revenue.

(B) Find the maximum​ profit, the production level that will realize the maximum​ profit, and the price the company should charge for each television set.

​(C) If the government decides to tax the company ​$55 for each set it​ produces, how many sets should the company manufacture each month to maximize its​ profit? What is the maximum​ profit? What should the company charge for each​ set?

Answer 1

Answer:

Step-by-step explanation:

Given

Cost Price $c(x)=72000+60x$

Price $p(x)=300-\frac{x}{20}$

Revenue generated $R(x)=P(x)\times x$

where x=no of units

$R(x)=300x-\frac{x^2}{20}$

To get maxima and minima differentiate R(x)

$\frac{\mathrm{d} R(x)}{\mathrm{d} x}=0$

$\frac{\mathrm{d} R(x)}{\mathrm{d} x}=300-2\times \frac{x}{20}=0$

$300=2\times \frac{x}{20}$

$x=3000$

maximum Revenue $R(x)=(300-\frac{300}{20})\times 300=4,50,000$

(b)Profit=Revenue - cost

$Profit=xp(x)-c(x)$

$Profit=300x-\frac{x^2}{20}-72000-60x$

$Profit(z)=240x-\frac{x^2}{20}-72000$

differentiate Profit to get maximum value

$\frac{\mathrm{d} z}{\mathrm{d} x}=240-2\times \frac{x}{20}$

$x=2400$

maximum Profit $z=2,16,000$

(c)Now company decided to tax the company $ 55 for each set

Profit $(z_1)=xp(x)-c(x)-55x$

$z_1=300x-\frac{x^2}{20}-72000x-60x^2-55x$

$z_1=185x-\frac{x^2}{20}-72,000$

differentiate Profit to get maximum value

$\frac{\mathrm{d} z_1}{\mathrm{d} x}=0$

$\frac{\mathrm{d} z_1}{\mathrm{d} x}=185-\frac{2x}{20}=0$

$x=1850$

$P(z_1\ at\ x=1850)=99125$

company should charge 207.5 $ for each set