Assuming that "elasticity" = P:
A. [tex]Q=90-0.1P^2\\50=90-0.1P^2\\-40=-0.1P^2\\400=P^2\\20=P\\\\\frac{dQ}{dP}=-0.2P\\E=\frac{P}{Q}*\frac{dQ}{dP}\\E=\frac{(20)}{(90)}(-0.2P)\\E=\frac{-2(20)}{45}\\E=\frac{-8}{9}[/tex] - Elasticity = -0.889
B. [tex]0>\frac{-8}{9}>-1>-\infty[/tex] - The demand is inelastic because the elasticity > -1.
(C). Set P and Q to 1 in two separate functions. If Q < P revenue will increase. If Q > P revenue will decrease.
[tex]Q=90-0.1P^2\\Q=90-0.1(1)^2\\Q=90-0.1\\Q=89.9\\\\(1)=90-0.1P^2\\-89=-0.1P^2\\890=P^2\\\sqrt{890}=P\\P=29.833\\Q>P\\(89.9)>(29.833)[/tex]
Q > P therefore revenue will decrease.
(D). [tex]Q=90-0.1P^2\\\frac{dQ}{dP}=-0.2P\\-0.2P=0\\P=0[/tex]
One obviously won't be able to maximize revenue if their price per unit, P, equals 0. Quantity of a product can only be sold in whole, so the closest integer to 90 is 89. The value of P that maximizes revenue is [tex]Q=90-0.1P^2\\(89)=90-0.1P^2\\-1=-0.1P^2\\10=P^2\\\sqrt{10}=P[/tex] - sqrt(10).
Therefore, the values of P and Q that maximize revenue are 3.162 and 89, respectively.
