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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
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