A cosmetics company is planning the introduction and promotion of a new lipstick line. The marketing research department has found that the demand in a particular city is given approximately by p=8ex, 0≤x≤2, where x thousand lipsticks were sold per week at a price of p dollars each. At what price will the weekly revenue be maximized?

Question

Business

Morgan Washington

19 November, 20:15

A cosmetics company is planning the introduction and promotion of a new lipstick line. The marketing research department has found that the demand in a particular city is given approximately by p=8ex, 0≤x≤2, where x thousand lipsticks were sold per week at a price of p dollars each. At what price will the weekly revenue be maximized?

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Answers (1)

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Ivan Beltran · Answer 1

p = 59.11 dollars

Explanation:

Given

Price: p (x) = 8eˣ (0 ≤ x ≤ 2)

Revenue; R = x*p = 8xeˣ

p = ? when R be at maximum

We can apply

dR/dx = d (x*p) / dx = 0

⇒ d (8xeˣ) / dx = 8 * (1*eˣ + x*eˣ) = 0

⇒ eˣ * (1 + x) = 0 ⇒ x = - 1

as x = - 1 ∉ [0, 2]

then, we have

p (0) = 8e⁰ = 8

R = 0*8 = 0

If x = 1

p (1) = 8e¹ ≈ 21.74

R = 1*21.74 = 21.74

If x = 2

p (2) = 8e² ≈ 59.11

R = 2*59.11 = 118.22

Implies that, R (x) is maximum at x = 2.

Thus, the price that maximize the revenue of the company is 59.11 dollars.