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Lelu [443]
3 years ago
5

PLEASEE HELPP!!! DUE TODAYY!!!

Mathematics
1 answer:
Alika [10]3 years ago
4 0

Answer:

b 5/12

Step-by-step explanation:

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A sailboat travels a distance of 2 1/2 miles in 1/6 of an hour.

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You operate a gaming Web site, www.mudbeast.net, where users must pay a small fee to log on. When you charged $3 the demand was
Doss [256]

Answer:

A) The linear relation between price and demand is:

d=-550x+2750

The revenue R is:

R=-550x^2+2750x

B) The profit functionP is:

P=-550x^2+2750x-30

C) The largest monthly profit is obtained with a log-on fee of $2.5 per month. This corresponds to a profit of $3407.5.

Step-by-step explanation:

We have a site where the number of log-ons depends on our monthly fee. A linear relation is established between the price (log-on fee) and the number of log-ons.

We have two points for this linear relationship:

  • At price x=3, the demand is d=1100.
  • At price x=2.5, the demand is d=1375.

We will model the relation:

d=mx+b

We can calculate the slope m as:

m=\dfrac{\Delta d}{\Delta x}=\dfrac{d_2-d_1}{x_2-x_1}=\dfrac{1375-1100}{2.5-3}\\\\\\m=\dfrac{275}{-0.5}=-550

Then, replacing one point in the linear equation, we can calculate the intercept b:

d_1=mx_1+b\\\\1100=(-550)\cdot 3+b\\\\1100=-1650+b\\\\b=1100+1650=2750

Then, the linear relation between demand and price is:

d=-550x+2750

The revenue R can be expressed as the multiplication of the price and the demand:

R=x\cdot d=x(-550x+2750)=-550x^2+2750x

If we have a fixed cost of $30 per month, the profit P is:

P=R-FC=-550x^2+2750x-30

We can maximize the profit by deriving the profit function and making it equal to zero.

\dfrac{dP}{dx}=0\\\\\\\dfrac{dP}{dx}=-550(2x)+2750(1)=0\\\\\\-1100x+2750=0\\\\x=\dfrac{2750}{1100}=2.5

This corresponds to a profit of:

P(2.5)=-550(2.5)^2+2750(2.5)-30\\\\P(2.5)=-550\cdot 6.25+6875-30\\\\P(2.5)=-3437.5+6875-30\\\\P(2.5)=3407.5

5 0
3 years ago
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