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3 years ago

Which is the composite function for the price a loyalty

Mathematics

1 answer:

FromTheMoon [43]3 years ago

7 0

Answer:

$(d)\ L(S(x)) = 0.7225x$

Step-by-step explanation:

Given

$S(x) = 0.85x$ ----- Sales price

$L(S) = 0.85S$ ---- Loyalty member discount

Required

This is represented as: L(S(x))

So, we have:

$L(S) = 0.85S$

Substitute S for S(x)

$L(S(x)) = 0.85S(x)$

Substitute $S(x) = 0.85x$

$L(S(x)) = 0.85*0.85x$

$L(S(x)) = 0.7225x$

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Find the percent of increase from 40 gallons to 89 gallons. Round to the nearest tenth of a percent if necessary.

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1 year ago

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4 years ago

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6 0

3 years ago

Before the pandemic cancelled sports, a baseball team played home games in a stadium that holds up to 50,000 spectators. When ti

spayn [35]

Answer:

(a) $D(x)=-2,500x+60,000$

(b) $R(x)=60,000x-2500x^2$

(d)Optimal ticket price: $12

Maximum Revenue:$360,000

Step-by-step explanation:

The stadium holds up to 50,000 spectators.

When ticket prices were set at $12, the average attendance was 30,000.

When the ticket prices were on sale for $10, the average attendance was 35,000.

(a)The number of people that will buy tickets when they are priced at x dollars per ticket = D(x)

Since D(x) is a linear function of the form y=mx+b, we first find the slope using the points (12,30000) and (10,35000).

$\text{Slope, m}=\dfrac{30000-35000}{12-10}=-2500$

Therefore, we have:

$y=-2500x+b$

At point (12,30000)

$30000=-2500(12)+b\\b=30000+30000\\b=60000$

Therefore:

$D(x)=-2,500x+60,000$

(b)Revenue

$R(x)=x \cdot D(x) \implies R(x)=x(-2,500x+60,000)\\\\R(x)=60,000x-2500x^2$

(c)To find the critical values for R(x), we take the derivative and solve by setting it equal to zero.

$R(x)=60,000x-2500x^2\\R'(x)=60,000-5,000x\\60,000-5,000x=0\\60,000=5,000x\\x=12$

The critical value of R(x) is x=12.

(d)If the possible range of ticket prices (in dollars) is given by the interval [1,24]

Using the closed interval method, we evaluate R(x) at x=1, 12 and 24.

$R(x)=60,000x-2500x^2\\R(1)=60,000(1)-2500(1)^2=\$57,500\\R(12)=60,000(12)-2500(12)^2=\$360,000\\R(24)=60,000(24)-2500(24)^2=\$0$

Therefore:

Optimal ticket price:$12
Maximum Revenue:$360,000

3 0

3 years ago