SOLUTION: The price - demand equation and the costfunction for the production of honey is given, respectively, by x = 5,000 - 100p and C(x) = 2,500 + 4x+ 0.01x2 where x is the number of

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Question 1197579: The price - demand equation and the costfunction for the production of
honey is given, respectively, by
x = 5,000 - 100p and C(x) = 2,500 + 4x+ 0.01x2
where x is the number of bottles that can be sold at a price of $p per
bottle and C(x) is the total cost (in dollars) of producing x bottles.
a) Express the price p as a function of the demand x, and find the domain of
this function.
b) Find the marginal cost.
c) Find the revenue function and state its domain.
d) Find the marginal revenue.
e) Find R′(2,000) and R′(3,000) and interpret these quantities.
f) Find the profit function in terms of x.
g) Find the marginal profit.
h) Find P′(1,000) and P′(1,500) and interpret these quantities.
Answer by onyulee(41)   (Show Source): You can put this solution on YOUR website!

**a) Express the price p as a function of the demand x, and find the domain of this function.**
* Given: x = 5,000 - 100p
* Solve for p:
* 100p = 5,000 - x
* p = (5,000 - x) / 100
* p = 50 - 0.01x
* **Domain:**
* The demand (x) must be non-negative (you can't sell a negative number of bottles).
* The price (p) must also be non-negative.
*
* 50 - 0.01x ≥ 0
* 0.01x ≤ 50
* x ≤ 5,000
* **Therefore, the domain of the price function is 0 ≤ x ≤ 5,000**
**b) Find the marginal cost.**
* **Marginal Cost:** The derivative of the cost function with respect to x.
* C(x) = 2,500 + 4x + 0.01x^2
* C'(x) = 4 + 0.02x
**c) Find the revenue function and state its domain.**
* **Revenue (R) = Price (p) * Quantity (x)**
* R(x) = p * x
* R(x) = (50 - 0.01x) * x
* R(x) = 50x - 0.01x^2
* **Domain:** The domain of the revenue function is the same as the domain of the price function, which is 0 ≤ x ≤ 5,000.
**d) Find the marginal revenue.**
* **Marginal Revenue:** The derivative of the revenue function with respect to x.
* R(x) = 50x - 0.01x^2
* R'(x) = 50 - 0.02x
**e) Find R′(2,000) and R′(3,000) and interpret these quantities.**
* R'(2,000) = 50 - 0.02 * 2,000 = 50 - 40 = 10
* When producing 2,000 bottles, the revenue is increasing at a rate of $10 per additional bottle.
* R'(3,000) = 50 - 0.02 * 3,000 = 50 - 60 = -10
* When producing 3,000 bottles, the revenue is decreasing at a rate of $10 per additional bottle.
**f) Find the profit function in terms of x.**
* **Profit (P) = Revenue (R) - Cost (C)**
* P(x) = R(x) - C(x)
* P(x) = (50x - 0.01x^2) - (2,500 + 4x + 0.01x^2)
* P(x) = 50x - 0.01x^2 - 2,500 - 4x - 0.01x^2
* P(x) = 46x - 0.02x^2 - 2,500
**g) Find the marginal profit.**
* **Marginal Profit:** The derivative of the profit function with respect to x.
* P(x) = 46x - 0.02x^2 - 2,500
* P'(x) = 46 - 0.04x
**h) Find P′(1,000) and P′(1,500) and interpret these quantities.**
* P'(1,000) = 46 - 0.04 * 1,000 = 46 - 40 = 6
* When producing 1,000 bottles, the profit is increasing at a rate of $6 per additional bottle.
* P'(1,500) = 46 - 0.04 * 1,500 = 46 - 60 = -14
* When producing 1,500 bottles, the profit is decreasing at a rate of $14 per additional bottle.
**a) Express the price p as a function of the demand x, and find the domain of this function.**
* **Solve the price-demand equation for p:**
* x = 5,000 - 100p
* 100p = 5,000 - x
* p = (5,000 - x) / 100
* p = 50 - 0.01x
* **Domain of the price function:**
* The demand (x) must be non-negative (you cannot sell a negative number of bottles).
* 0 ≤ x ≤ 5,000 (The maximum demand occurs when the price is 0)
**b) Find the marginal cost.**
* **Marginal Cost (MC):** The derivative of the cost function with respect to x.
* MC = C'(x) = d/dx (2,500 + 4x + 0.01x²)
* MC = 4 + 0.02x
**c) Find the revenue function and state its domain.**
* **Revenue (R):** Price per unit * Number of units sold
* R(x) = p * x
* R(x) = (50 - 0.01x) * x
* R(x) = 50x - 0.01x²
* **Domain of the revenue function:**
* Same as the domain of the price function: 0 ≤ x ≤ 5,000
**d) Find the marginal revenue.**
* **Marginal Revenue (MR):** The derivative of the revenue function with respect to x.
* MR = R'(x) = d/dx (50x - 0.01x²)
* MR = 50 - 0.02x
**e) Find R′(2,000) and R′(3,000) and interpret these quantities.**
* **R'(2,000):**
* R'(2,000) = 50 - 0.02 * 2,000 = 50 - 40 = 10
* Interpretation: When 2,000 bottles are sold, the revenue is increasing at a rate of $10 per additional bottle sold.
* **R'(3,000):**
* R'(3,000) = 50 - 0.02 * 3,000 = 50 - 60 = -10
* Interpretation: When 3,000 bottles are sold, the revenue is decreasing at a rate of $10 per additional bottle sold.
**f) Find the profit function in terms of x.**
* **Profit (P):** Revenue - Cost
* P(x) = R(x) - C(x)
* P(x) = (50x - 0.01x²) - (2,500 + 4x + 0.01x²)
* P(x) = 46x - 0.02x² - 2,500
**g) Find the marginal profit.**
* **Marginal Profit (MP):** The derivative of the profit function with respect to x.
* MP = P'(x) = d/dx (46x - 0.02x² - 2,500)
* MP = 46 - 0.04x
**h) Find P′(1,000) and P′(1,500) and interpret these quantities.**
* **P'(1,000):**
* P'(1,000) = 46 - 0.04 * 1,000 = 46 - 40 = 6
* Interpretation: When 1,000 bottles are sold, the profit is increasing at a rate of $6 per additional bottle sold.
* **P'(1,500):**
* P'(1,500) = 46 - 0.04 * 1,500 = 46 - 60 = -14
* Interpretation: When 1,500 bottles are sold, the profit is decreasing at a rate of $14 per additional bottle sold.
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