answer:Let's denote the demand as D and the profit as P. We have two cases to consider: Case 1: D <= 100 units In this case, the store purchases the product for 30 per unit. The profit per unit is 50 - 30 = 20. So, the total profit is P = 20D. Case 2: D > 100 units In this case, the store purchases the first 100 units for 30 per unit and the remaining (D - 100) units for 40 per unit. The profit per unit for the first 100 units is 20, and the profit per unit for the remaining units is 50 - 40 = 10. So, the total profit is P = 20 * 100 + 10(D - 100) = 2000 + 10D - 1000 = 10D + 1000. Now, we need to find the optimal demand (D) that maximizes the profit (P). For Case 1 (D <= 100 units), the profit function is P = 20D. The maximum profit occurs when D = 100, so P = 20 * 100 = 2000. For Case 2 (D > 100 units), the profit function is P = 10D + 1000. The maximum profit occurs when D = 150 (the maximum demand), so P = 10 * 150 + 1000 = 2500. Since the profit in Case 2 (P = 2500) is greater than the profit in Case 1 (P = 2000), the store should aim for a demand greater than 100 units to maximize its profit. Therefore, the optimal pricing of the product is 50 per unit, which will result in a maximum profit of 2500 when the demand reaches 150 units per week.

answer:To find the optimal price to maximize profit, we first need to find the revenue function. Revenue is the product of price (P) and quantity (Q). So, the revenue function R(P) can be found by multiplying the demand function by the price: R(P) = P * Q = P * (400 - 4P) Now, we need to find the profit function. Profit is the difference between revenue and cost. Since the cost to manufacture each product is 40, the cost function is C(Q) = 40Q. We can substitute the demand function for Q: C(P) = 40(400 - 4P) The profit function is the difference between the revenue function and the cost function: Π(P) = R(P) - C(P) = P(400 - 4P) - 40(400 - 4P) Now, we need to find the critical points of the profit function by taking the first derivative with respect to P and setting it to zero: Π'(P) = d(Π(P))/dP = 0 Π'(P) = (400 - 8P) - 160 = 0 Solving for P: 240 - 8P = 0 8P = 240 P = 30 The optimal price for the company to sell the product is 30 to maximize its profit.

answer:Let's denote the price per unit as P and the number of units sold as Q. We are given that when P = 80, Q = 500. We are also given that for every 5 increase in price, the demand decreases by 50 units. We can express this relationship as follows: Q = 500 - 50 * (P - 80) / 5 Simplifying the equation: Q = 500 - 10 * (P - 80) Now, we want to find the optimal price P to maximize revenue. Revenue (R) is the product of price (P) and quantity (Q): R = P * Q Substitute the expression for Q from above: R = P * (500 - 10 * (P - 80)) Expanding the equation: R = 500P - 10P^2 + 8000 This is a quadratic equation, and to find the maximum revenue, we need to find the vertex of the parabola. The x-coordinate of the vertex (in this case, the optimal price P) can be found using the formula: P = -b / 2a In our equation, a = -10 and b = 500: P = -500 / (2 * -10) P = 500 / 20 P = 25 However, this result indicates the change in price from the initial price of 80. So, the optimal price is: Optimal Price = Initial Price + Change in Price Optimal Price = 80 + 25 Optimal Price = 105 Therefore, the company should charge 105 per unit to maximize their revenue.

answer:Let's denote the price of the product as P and the number of units sold as Q. We are given that the initial price is 50 and the initial quantity is 100 units. We are also given that for every dollar increase in price, the demand decreases by 2 units. So, if the price increases by x dollars, the new price will be P = 50 + x, and the new quantity will be Q = 100 - 2x. Revenue (R) is the product of price and quantity, so R = P * Q. Substituting the expressions for P and Q, we get: R = (50 + x) * (100 - 2x) Expanding this expression, we get: R = 5000 - 100x + 100x - 2x^2 R = -2x^2 + 5000 To find the optimal price that maximizes revenue, we need to find the vertex of this quadratic function. The x-coordinate of the vertex is given by the formula: x = -b / 2a In our case, a = -2 and b = 0, so: x = -0 / (2 * -2) = 0 Since x represents the increase in price, the optimal price is: P = 50 + x = 50 + 0 = 50 So, the optimal price to maximize revenue is 50. However, this result indicates that there is no increase in price that would lead to a higher revenue, given the constraints of the problem. The store owner should maintain the current price of 50 to maximize revenue.