Respuesta :
Answer: a) [tex]E(x)=\dfrac{-5945}{(30x+11)(5x+200)}[/tex], b) 0.7975, demand is inelastic, c) increase.
Step-by-step explanation:
Since we have given that
[tex]D(x)=\dfrac{5x+200}{30+11}[/tex]
So, derivative w.r.t x would be
[tex]D'(x)=\dfrac{5(30x+11)-30(5x+200)}{(30x+11)^2}\\\\D'(x)=\dfrac{150x+55-150x-6000}{(30x+11)^2}\\\\D'(x)=\dfrac{5945}{(30x+11)^2}[/tex]
As we know that
[tex]E(x)=\dfrac{-xD'(x)}{D(x)}\\\\\\E(x)=\dfrac{\dfrac{-(-)5945x}{(30x+11)^2}}{\dfrac{5x+200}{30x+11}}\\\\\\E(x)=\dfrac{5945x}{(30x+11)(5x+200)}[/tex]
(b) Find the elasticity when x = 2.
So, we put x = 2, we get that
[tex]E(2)=\dfrac{5945\times 2}{(30(2)+11)((5(2)+200))}\\\\E(2)=\dfrac{11890}{(60+11)(10+200)}\\\\E(2)=\dfrac{11890}{71\times 210}\\\\E(2)=\dfrac{11890}{14910}\\\\E(2)=0.7975[/tex]
Since, 0.7975 < 1, so the demand is inelastic.
(c) At $2 per plant, will a small increase in price cause the total revenue to increase or decrease?
The total revenue will also increase with increase in price.
As total revenue = [tex]price\times quantity[/tex]
Hence, a) [tex]E(x)=\dfrac{-5945}{(30x+11)(5x+200)}[/tex], b) 0.7975, demand is inelastic, c) increase.
