MyUSF Thursday 11/05 Part 1 of 1 X Course Home X mathxl.com/Student/PlayerHomework.aspx?homeworkld-576815529&questionid=1&flushed MAC2233 Business Calcu

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Question

MyUSF Thursday 11/05 Part 1 of 1 X Course Home X mathxl.com/Student/PlayerHomework.aspx?homeworkld-576815529&questionid=1&flushed MAC2233 Business Calculus, Sections 021-026 Fall 2020 Homework: HW 15: Price Elasticity of Demand Score: 0 of 2 2 pts 2 of 3 (0 comp Bus Econ 6.3.21 For the following demand function, find a E, and b. the values of a cif any) at which total revenue is maximized q=39,000 -3p2 a. Determine the elasticity of demand, E. E-(Type an expression using pas the variable) Enter your answer in the answer box and then click Check Answer 1 par Clear All - remaining Book

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Expert Answer

Step 1)

Elasticity of demand is given by,

$E = -\\frac{p}{q}\\cdot \\frac{dq}{dp}$

we have q = 39,000 - 3p² hence,

$E = -\\frac{p}{39,000-3p^2}\\cdot \\frac{d}{dp}(39,000-3p^2)$

$E = -\\frac{p}{39,000-3p^2}\\cdot (0-6p)$

$E = -\\frac{p}{39,000-3p^2}\\cdot (-6p)$

$E = \\frac{6p^2}{39,000-3p^2}$

Step 2)

we know that total revenue ia maximized when E = 1

we have,

$E = \\frac{6p^2}{39,000-3p^2}$

Hence we can write,

$\\frac{6p^2}{39,000-3p^2}=1$

6p^2=39,000-3p^2

9p^2=39,000

$p^2=\\frac{13,000}{3}$

$p=\\pm \\sqrt{\\frac{13,000}{3}}$

price cannot be negative hence we have,

$p= \\sqrt{\\frac{13,000}{3}}$

we have,

q = 39,000 -3p^2

Hence we can write values of q at which revenue is maximum is given by,

$q = 39,000 -3\\left ( \\sqrt{\\frac{13,000}{3}} \\right )^2$

$q = 39000 -3\\left ( \\frac{13,000}{3} \\right )$

q = 39,000 -13,000

q = 26,000

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