Question

A store sells about 40 video game systems each month when it charges $200 per system. for each $10 increase in price, about 1 less system per month is sold. The store's revenue can be modeled by r(x)=(22+10x)(40-x), where x represents the number of price increases. How can the store maximize monthly revenue?

Answer 1

To maximize the store's monthly revenue from selling video game systems, we analyze the quadratic revenue function r(x)=(220+10x)(40-x) and look for the vertex, which indicates the most profitable number of $10 price increases.

Step-by-step explanation:

To maximize monthly revenue, the store must determine the number of $10 price increases that will result in the highest revenue for video game systems. We accomplish this by analyzing the revenue function r(x)=(220+10x)(40-x), where x is the number of $10 price increases. To find the maximum revenue, we need to find the vertex of the quadratic function, which is in the form of ax^2 + bx + c.

We know that the x-coordinate of the vertex is at -b/(2a), which gives us the most profitable number of price increases. The coefficients from the revenue function are a = -1 and b = 180, so the x-coordinate of the vertex would be -180/(2 * -1) = 90. Since the store cannot increase the price by $900 at once, we should use the next best option that is feasible; hence, we must check revenue values for integer numbers of price increases around this x-coordinate to find the maximum. As the function is a parabola opening downwards, the maximum integer x value before 90 will give us the maximum revenue.

To determine maximum revenue, we calculate values of r(x) at integers around 90, which are likely to be much lower due to the limitations of physical price increases and the reality of consumer demand. Therefore, a calculation or a graph of the function is needed to specifically identify the optimal price point.