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2 years ago

14

Which number line can be used to find the distance between (-1, 2) and (-5, 2)?

1 answer:

Step2247 [10]2 years ago

5 0

The second line since it have marks on -1 and on 2

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2 years ago

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A company manufactures and sells x television sets per month. The monthly cost and price-demand equations are C(x)=72,000+60x

myrzilka [38]

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

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3 years ago