The campus bookstore has estimated that its profit​ (in dollars) from selling x hundred basketball conference championship​ t-shirts is given by the equation shown below.

Upper P equals negative 41 x squared plus 578 x minus 521P=−41x2+578x−521

The demand is currently 

800 

​t-shirts, but euphoria over the championship is subsiding so the demand is dropping by 

100 

​t-shirts per day. How is the profit changing with respect to​ time?

Question content area bottom

Part 1

​$enter your response here 

per day

To determine how the profit is changing with respect to time, we need to first express the given profit function in terms of $ t $ (time), considering the rate at which the demand is changing. 

### Step-by-Step Solution:

1. **Given profit function:**

   \[
   P = -41x^2 + 578x - 521
   \]

   where $ x $ is the number of t-shirts sold in hundreds.

2. **Given information:**

   - The demand is currently 800 t-shirts, which corresponds to $ x = 8 $ (since $ x $ is in hundreds).
   - The demand is dropping by 100 t-shirts per day, which corresponds to $ \frac{dx}{dt} = -1 $ (since 100 t-shirts per day is equivalent to 1 unit in hundreds).

3. **Differentiate the profit function with respect to time $ t $:**

   \[
   \frac{dP}{dt} = \frac{d}{dt} \left( -41x^2 + 578x - 521 \right)
   \]

   Applying the chain rule:

   \[
   \frac{dP}{dt} = \frac{dP}{dx} \cdot \frac{dx}{dt}
   \]

   First, find $\frac{dP}{dx}$:

   \[
   \frac{dP}{dx} = \frac{d}{dx} \left( -41x^2 + 578x - 521 \right)
   \]

   \[
   \frac{dP}{dx} = -82x + 578
   \]

   Now, substitute $\frac{dx}{dt} = -1$:

   \[
   \frac{dP}{dt} = \left( -82x + 578 \right) \cdot (-1)
   \]

   \[
   \frac{dP}{dt} = -(-82x + 578)
   \]

   \[
   \frac{dP}{dt} = 82x - 578
   \]

4. **Evaluate $\frac{dP}{dt}$ at $ x = 8 $:**

   \[
   \frac{dP}{dt} = 82(8) - 578
   \]

   \[
   \frac{dP}{dt} = 656 - 578
   \]

   \[
   \frac{dP}{dt} = 78
   \]

Therefore, the profit is changing at a rate of $ \$78 $ per day.

### Answer:

The profit is changing by $ \$78 $ per day.

Would you like any further details or have any questions about the solution?

### Related Questions:
1. How would the result change if the demand was dropping at a different rate?
2. What would the profit be if the demand suddenly increased by 200 t-shirts?
3. How does the quadratic term $-41x^2$ affect the profit compared to a linear model?
4. What is the maximum profit achievable according to this model?
5. At what rate would the demand need to change to keep the profit constant?
6. How would the profit change if the rate of demand decline accelerates over time?
7. What assumptions are we making about the relationship between profit and demand?
8. How would you approach solving for the time when the profit starts to decline? 

### Tip:
When dealing with profit functions and changing demand, it's crucial to consider both the immediate rate of change and how long-term trends might affect the overall profitability.

Explore how profit changes over time using quadratic equations. Learn to calculate the rate of change in profit based on changing demand for basketball conference championship t-shirts. Suitable for students studying quadratic equations and calculus in Grades 10-12.

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