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laiz [17]
2 years ago
6

An average of 70,000 people visit Riverside Park each day in the summer. The park charges $12.00 for admission. Consultants pred

ict that for each $1.00 increase in the entrance price, the park would lose an average of 2,500 customers per day. (a) Express the daily revenue from ticket sales, R as a function of the number of $1.00 price increases, x. R=f(x)= (70000-2500x)(12+x) equation editorEquation Editor (b) What ticket price maximizes the revenue from ticket sales?
Mathematics
1 answer:
s2008m [1.1K]2 years ago
6 0

Answer:

The price that maximizes the revenue is $20

Step-by-step explanation:

(a) Express the daily revenue from ticket sales, R as a function of the number of $1.00 price increases,

R(x) = (70000-2500x)(12+x)

(b) What ticket price maximizes the revenue from ticket sales?

The ticket price that maximizes the revenue is associated with the x of the vertex of the parabola.

R(x) = (70000-2500x)(12+x) = 840000 + 70000x - 30000x - 2500x² =

-2500x² + 40000x + 840000

x vertex = -b/2a = -40000/2.(-2500) = 8

12+8 = 20

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Answer : 96

x – y = 16 --------> equation 1

\frac{1}{8} x +\frac{1}{2} y = 52

x is the higher grade and y is the lower grade

We solve the first equation for y

x - y = 16

-y = 16 -x ( divide each term by -1)

y = -16 + x

Now substitute y in second equation

\frac{1}{8} x +\frac{1}{2} ( -16 + x ) = 52

\frac{1}{8} x - 8 +   \frac{1}{2} x = 52

\frac{1}{8} x  +   \frac{1}{2} x - 8 = 52

Take common denominator to combine fractions

\frac{1}{8} x  +   \frac{4}{8} x  -8 = 52

\frac{5}{8} x - 8  = 52

Add 8 on both sides

\frac{5}{8} x   = 60

Multiply both sides by \frac{8}{5}

x = 96

We know x is the higher grade

96 is the higher grade of Jose’s two tests.

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