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

         

Answer 2

Following are the responses to the given question:

For point a:

$\to p(x)=300-(\frac{x}{20})\\\\\to revenue \ R(x)=p\times x\\\\\to revenue\ R(x)=300x -(\frac{x^2}{20})$

for maximum revenue $\frac{dR}{dx} =0 ,$

$\to 300-(\frac{2x}{20})=0\\\\\to \frac{x}{10}=300\\\\\to x=3000$

maximum revenue:

$\to R(3000)=300\times 3000 -(\frac{3000^2}{20})$

$\to R(3000)=\ 450000$

For point b:

$\to \text{profit =revenue -cost}$

$\to P(x)=300x -(\frac{x^2}{20})-72000-60x\\\\\to P(x)=240x -(\frac{x^2}{20})-72000$

for maximum cost$\frac{dP}{dx} =0$

$\to 240 -(\frac{2x}{20})=0\\\\\to x=240\times 10\\\\\to x=2400\\\\\to p(2400)=300−(\frac{2400}{20})=180\\\\\to P(2400)=240\times 2400 -(\frac{2400^2}{20})-72000 \\\\\to P(2400)=216000$

Maximum profit = $\ 216000$ when $\bold{2400 \ sets}$ were manufacture and solded for $\bold{\ 180}$ each.

For point c:

$\to \text{profit =revenue -cost -tax}\\\\\to P(x)=300x -(\frac{x^2}{20})-72000-60x-55x\\\\\to P(x)=185x -(\frac{x^2}{20})-72000$

for maximum cost $\frac{dP}{dx} =0$

$\to 185-(\frac{x^2}{20})=0\\\\\to x=185\times 10\\\\\to x=1850\\\\\to p(1850)=300−(\frac{1850}{20})=207.5\\\\\to P(1850)=185\times 1850 -(\frac{1850}{20})-72000\\\\\to P(1850)=\ 99125$

In each taxed set at $\bold{\ 55}$ so, the maximum profit = $\bold{\ 99125}$. If 1850 sets are manufacture and sold as each $\bold{\ 207.5}$.

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