i. find P first
"1600 - 2840 = -20P, P = 62"
Point elasticity of "\\frac{P}{Q} \\times F'" (demand function)
"F'(demand) = -20"
"\\frac{62\n}{1600 }\\times (-20) = -0.775," thus inelastic but not completely inelastic
"Ed = 0" Perfectly inelastic demand
"- 1 < Ed < 0" Inelastic or relatively inelastic demand
"Ed = - 1" Unit elastic, unit elasticity, unitary elasticity, or unitarily elastic demand
-"8 < Ed < - 1" Elastic or relatively elastic demand
"Ed = - 8" Perfectly elastic demand
ii. Total revenue:"P \\times (2840 - 20P)" (the last term stems from "q=2840-20p," revenue "= P \\times Q)"
Total revenue: "-20P^2 + 2840P"
Average revenue: "\\frac{-20P^2 + 2840P}{ Q}" or it can be expressed as "\\frac{(-20P^2 + 2840P)}{(2840-20P)}"
Marginal revenue = derivative of total revenue with respect to P
"F'(revenue) = -40P + 2840"
iii. maximize revenue
occurs when first derivative = 0
F'(total revenue) "= -40P + 2840," is equal to 0 when"P = \\frac{2840}{40} = 71" . Plus 71 into the first function from A)
Total revenue "= 71 \\times (2840 - 20\\times 71) = \\$100,820"
b. The total revenue will be the highest when the marginal revenue=0
From "0=2840-40P" we can get "P=71"
So the maximum total revenue "=2840\\times 71-20\\times 71^2=100820"
c. "E=-20\\times \\frac{62}{1600}=-0.775<1," so the point th point of demand is inelastic at this quantity.
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