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Answer the following questions for the price-demand equation.
p + 0.01x = 50
(A) Express the demand x as a function of the price p. x =
The domain of this function is (Type an inequality or a compound inequality.)
(B) Express the revenue R as a function of price p. R(p) =
(C) Find the elasticity of demand. E(p). E(p) =
(D) For which values of p is the demand elastic? I
(E) For which values of p is revenue increasing? Decreasing?
(F) If P = 20 and the price is decreased, will revenue increase or decrease?
(G) If p = 30 and the price decreased, will revenue increase or decrease? Enter your answer in each of the answer boxes


Sagot :

Answer:

(A) x = 5,000 - 100·p

(B) R(p) = 5,000·p - 100·p²

(C) The elasticity of demand E(P) = -100·p/x

(D) The demand is elastic when P is larger than 0.01·x

(E) The revenue is decreasing for values of p < 25

(F) The revenue will decrease if P = 20 and the price decreases

(G) The revenue will increase until the price reaches 25

Step-by-step explanation:

The given price demand equation is, p + 0.01·x = 50

(A) The demand, 'x', as a function of price is therefore expressed as follows;

x = (50 - p)/0.01 = 5,000 - 100·p

∴ x = 5,000 - 100·p

(B) The revenue, 'R', as a function of price, R(p) = x·p

p = 50 - 0.01·x

∴ R = R(p) = x·p = x × (50 - 0.01·x) = 50·x - 0.01·x²

The revenue, R = 50·x - 0.01·x²

∴ R(P) = 50 × (5,000 - 100·p) - 0.01 × (5,000 - 100·p)² = 5,000·p - 100·p²

R(P) = 5,000·p - 100·p²

(C) The elasticity of demand E(P) = (dx/dp) × p/x

∴ E(P) = d(5,000 - 100·p)/dp × p/x = -100 × p/x = -100·p/x

E(P) = -100·p/x

(D) The demand is elastic when E(P) > -1

Therefore, we have;

-100·p/x > -1

-p > -0.01·x

p > 0.01·x

The demand is elastic when P is larger than 0.01·x

(E) The revenue is decreasing for the values of 'p', given as follows;

At maximum point, dR/dx = 50 - 0.02·x = 0

∴ x = 50/0.02 = 2,500

p = 50 - 0.01·x = 50 - 0.01 × 2500 = 25

Therefore, the revenue is decreasing for values of p < 25

(F) If the P = 20 and the price decreases, we have;

x = 5,000 - 100·p

∴ x = 5000 - 100 × 20 = 3,000

If the price decreases, the quantity, 'x' increases above 3,000 and the revenue decreases

Therefore the revenue will decrease if P = 20 and the price decreases

(G) If p = 30, we have;

x = 5000 - 100 × 30 = 2,000

Therefore, the revenue is increasing until p = 25

Therefore, if p = 30 and the price decreases, the revenue will increase until the price reaches 25