BMBC The Production that Will Maximize Revenue Question

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funqnr

Mathematics

CUNY Bernard M Baruch College

Question Description

I'm working on a mathematics multi-part question and need an explanation and answer to help me learn.

Suppose that the price per unit in dollars of a cell phone production is modeled by p = $45 ? 0.0125x, where x is in thousands of phones produced, and the revenue represented by thousands of dollars is R = x ? p. Find the production level that will maximize revenue.

A soccer stadium holds 62,000 spectators. With a ticket price of $11, the average attendance has been 26,000. When the price dropped to $9, the average attendance rose to 31,000. Assuming that attendance is linearly related to ticket price, what ticket price would maximize revenue?

  1. The revenue function of a fictional cable company can be modeled by the polynomial function:

R(t) = ?0.037t4 + 1.414t3 ? 19.777t2 + 118.696t ? 205.332

where R represents the revenue in millions of dollars and t represents the year, with t = 1 corresponding to 2001. Using MS-Excel, create a graph of the Revenue Function and determine over which intervals is the revenue for the company increasing? Over which intervals is the revenue for the company decreasing?

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Explanation & Answer

View attached explanation and answer. Let me know if you have any questions.

Q1.
Given, 𝑝 = 45$ − 0.125𝑥
And, 𝑅 = 𝑥 × 𝑝 = 𝑥 × (45 − 0.125𝑥) = 45𝑥 − 0.125𝑥 2
For maximum revenue
𝑑𝑅
=0
𝑑𝑥
𝑑(45𝑥 − 0.125𝑥 2 )
𝑜𝑟,
=0
𝑑𝑥
𝑜𝑟, 45 − 2 × 0.125𝑥 = 0
𝑜𝑟, 45 − 0.25𝑥 = 0
45
𝑜𝑟, 𝑥 =
0.25
∴ 𝑥 = 180

Hence, the production that will maximize revenue is 180 thousand units of phone produced.

Q2)
Let, ticket price be x variable and no of spectator be y variable
Here, (𝑥1 , 𝑦1 ) =(11,26000) and (𝑥2 , 𝑦2 ) = (9,31000) is linearly related so, slope of line is:
𝑚=

𝑦2 − 𝑦1 31000 − 26000
=
= −2500
𝑥2 − 𝑥1
9 − 11

Hence, relation between price of ticket and no of spectators is:
𝑦 − 𝑦1 = 𝑚 ( 𝑥 − 𝑥1 )
𝑜𝑟, 𝑦 − 26000 = −2500(𝑥 − 11)
𝑜𝑟, 𝑦 − 26000 = −2500𝑥 + 275000
𝑜𝑟, 𝑦 = −2500𝑥 + 53500
Now, revenue (R) = 𝑡𝑖𝑐𝑘𝑒𝑡 𝑝𝑟𝑖𝑐𝑒 × 𝑛𝑜 𝑜𝑓 𝑠𝑝𝑒𝑐𝑡𝑎𝑡𝑜𝑟 = 𝑥 × 𝑦 = 𝑥 × (−2500𝑥 + 53500) =
−2500𝑥 2 + 53500𝑥
R will be maximum when,
𝑑𝑅
=0
𝑑𝑥
𝑑(−2500𝑥 2 + 53500𝑥)
𝑜𝑟,
=...

nawnaounggnenv2054 (455)
University of Virginia

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