The problem in the image asks about maximizing revenue for a retailer selling backpacks. The revenue function is given by:

$R(z) = z(126 - z)$ where $z$ is the price of a backpack in dollars, and the goal is to determine the price $z$ that maximizes the revenue.

Steps to solve:

1. Revenue Function: The revenue $R(z)$ is defined as the product of the price $z$ and the number of backpacks sold, $126 - z$, meaning $R(z) = z(126 - z)$.

2. Expand the Function: $R(z) = 126z - z^2$ This is a quadratic equation in the form $R(z) = -z^2 + 126z$, which represents a downward-opening parabola.

3. Find the Vertex: The maximum value of a quadratic function $ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$. In this case:

   • $a = -1$ and $b = 126$, so the maximum occurs at: $z = -\frac{126}{2(-1)} = \frac{126}{2} = 63$ Thus, the selling price $z = 63$ dollars will give the maximum revenue.

4. Find the Maximum Revenue: Substitute $z = 63$ into the revenue function: $R(63) = 63(126 - 63) = 63 \times 63 = 3969$ Therefore, the maximum revenue is $3969$ dollars.

Final Answers:

• The selling price that gives maximum revenue is 63 dollars.
• The maximum revenue is 3969 dollars.

Let me know if you'd like more details, or if you have any further questions!

Here are 5 related questions:

1. What if the price of backpacks increased by $5? How would that affect the maximum revenue?
2. Can this same quadratic method be applied to any price-quantity relationship?
3. How can you find the break-even point for revenue?
4. How does the coefficient of the quadratic term affect the shape of the parabola?
5. Could other external factors (like cost of materials) affect this revenue model?

Tip: The vertex formula $z = -\frac{b}{2a}$ is a quick way to find the maximum or minimum of any quadratic function.