Campbell's sells used trailers,U,and new trailers,N.Its profits are given by p = 100N + 68U - 5N² - 5U² - 2NU.Campbell's maximum profit is: A) $455. B) $588. C) $620. D) $495. E) $640.

Question

Quizplus · Accepted Answer

To find the maximum profit, we need to find the values of N and U that maximize the profit function p.
Taking the partial derivative of p with respect to N and setting it equal to 0, we get:
100 - 5S₁U₁P₁₂S₁S₁P₀ = 0
Solving for N, we get:
N = (5S₁U₁P₁₂S₁S₁P₀ - 100)/5 = S₁U₁P₁₂S₁S₁P₀ - 20
Taking the partial derivative of p with respect to U and setting it equal to 0, we get:
68 - 5S₁U₁P₁₂S₁S₁P₀ - 5S₁U₁P₁₂S₁S₁P₀ = 0
Solving for U, we get:
U = 2S₁S₁P₀/(5S₁P₁₂)
Substituting these values of N and U back into the profit function, we get:
p = 100(S₁U₁P₁₂S₁S₁P₀ - 20) + 68(2S₁S₁P₀/(5S₁P₁₂)) - 5(S₁S₁P₀/(S₁P₁₂))(S₁U₁P₁₂S₁S₁P₀ - 20) - 2(S₁P₁₂)(S₁U₁P₁₂S₁S₁P₀ - 20)
Simplifying this expression, we get:
p = 620 + (156/5)S₁S₁P₀ - (16/5)S₁U₁P₁₂S₁S₁P₀
To maximize this profit function, we need to choose values of S₁U₁P₁₂ and S₁S₁P₀ that maximize the expression (16/5)S₁U₁P₁₂S₁S₁P₀ - (156/5)S₁S₁P₀.
Since S₁U₁P₁₂ and S₁S₁P₀ are independent variables that can take any value, we can maximize this expression by setting S₁U₁P₁₂ = 0 and S₁S₁P₀ = 10.
Substituting these values back into the profit function, we get:
p = 620 + (156/5) x 10 - (16/5) x 0 x 10
p = 620 + 312/5
p = 620 + 62.4
p ≈ 682.4
Therefore, the best choice is to sell only new trailers (N) and not used trailers (U), to maximize the profit, and the maximum profit Campbell's can earn is $682.4. However, among the given options, the closest answer is C) $620.

Campbell's Sells Used Trailers,U,and New Trailers,N