The problem in your image involves finding the selling price that will maximize revenue using a quadratic function.

Problem Details:

The revenue $R(z)$ generated by selling backpacks is given by the quadratic function: $R(z) = -z^2 + 126z$ where $z$ is the selling price of each backpack.

The task is to:

1. Find the selling price $z$ that maximizes revenue.
2. Identify the units for the maximum revenue.
3. Calculate the maximum revenue value.

Solution:

Since the function $R(z) = -z^2 + 126z$ is quadratic with a negative leading coefficient, it opens downward, and thus has a maximum point at its vertex.

Step 1: Find the Selling Price for Maximum Revenue

The vertex of a quadratic function $ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$.

Here:

• $a = -1$
• $b = 126$

So, $z = -\frac{126}{2 \times -1} = \frac{126}{2} = 63$

Thus, the selling price that maximizes revenue is $z = 63$ dollars.

Step 2: Calculate the Maximum Revenue

To find the maximum revenue, substitute $z = 63$ into the revenue function $R(z)$: $R(63) = -(63)^2 + 126 \times 63$ $= -3969 + 7938 = 3969$

Therefore, the maximum revenue is $3969$.

Step 3: Identify the Units

• For the selling price: dollars.
• For the maximum revenue: dollars.

Summary:

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

Let me know if you want further details or have any questions.

Here are some related questions:

1. How would the maximum revenue change if the function's coefficient of $z^2$ were different?
2. What would happen to the selling price for maximum revenue if $R(z)$ had a positive leading coefficient?
3. How can we verify if a revenue function is maximized or minimized at its vertex?
4. How do changes in price impact the quantity sold in this quadratic model?
5. What would the revenue be if the selling price was set below the optimal price?

Tip: For any quadratic function $ax^2 + bx + c$ representing real-world scenarios like revenue or profit, the vertex formula helps find maximum or minimum values efficiently.