As the selling price of a product increases, demand decreases. Fewer people will buy a product if it is more expensive. This is a very important concept. The demand equation for new iSmarts is given by p = 124 0.08x , where p is the unit price (in dollars) of the iSmart and x is the number of units produced. We can determine the total revenue (money earned for selling x iSmarts) by multiplying the number of items we sell, and the price p. The total revenue obtained by producing and selling x iSmarts is given by R = xp = 124z 0.08z2. It makes sense that this function is quadratic. The company can price the iSmarts so high that no one will buy them and the revenue will be zero. The can also set the price to $0, but the revenue will still be $0, no matter how many iSmarts they "sell". In between, the graph is a parabola opening downward. Determine prices p that would keep revenue above 7450 dollars. Give your answers to 2 decimal places. The revenue will be above 7450 if price is between $' |{Iower price) and $' |(higher price) How many iSmarts will be sold at these prices? Give the nearest whole number of iISmarts. After all, who would buy a fractiorr of an iSmart? At the lower price, iISmarts will be sold and at the higher price, iSmarts will be sold. What price will maximize revenue, how many iSmarts is this, and what is the maximum revenue? The company will maximize revenue if they sell the iSmarts for $ |. They will sell ' iSmarts and will have a revenue of $|