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The question you're working on involves finding the instantaneous rate of change of demand with respect to price for a given demand function:

\[
q = \frac{256}{\sqrt{p}} - 1
\]

where $ q $ represents demand and $ p $ represents price in dollars.

### Part (a):
You need to find $ q'(p) $ and evaluate it at $ p = 16 $.

1. **Derivative of $ q $ with respect to $ p $:**

The given function is $ q(p) = \frac{256}{\sqrt{p}} - 1 $. We need to apply the chain rule to differentiate this expression.

\[
q(p) = 256 \cdot p^{-\frac{1}{2}} - 1
\]

Taking the derivative of $ q(p) $ with respect to $ p $:

\[
q'(p) = 256 \cdot \left( -\frac{1}{2} \right) p^{-\frac{3}{2}} = -\frac{128}{p^{\frac{3}{2}}}
\]

2. **Evaluate $ q'(p) $ at $ p = 16 $:**

\[
q'(16) = -\frac{128}{16^{\frac{3}{2}}}
\]

Since $ 16^{\frac{3}{2}} = 64 $, we have:

\[
q'(16) = -\frac{128}{64} = -2
\]

So, the instantaneous rate of change of demand when the price is $ 16 $ dollars is $ -2 $.

### Part (b):
You need to do the same for $ p = 64 $.

Using the derivative formula we found earlier:

\[
q'(p) = -\frac{128}{p^{\frac{3}{2}}}
\]

Evaluate $ q'(64) $:

\[
q'(64) = -\frac{128}{64^{\frac{3}{2}}}
\]

Since $ 64^{\frac{3}{2}} = 512 $, we have:

\[
q'(64) = -\frac{128}{512} = -\frac{1}{4}
\]

So, the instantaneous rate of change of demand when the price is $ 64 $ dollars is $ -\frac{1}{4} $.

---

### Interpretation:
- For **part (a)**: The rate of change at $ p = 16 $ is $ -2 $, meaning for every $1 increase in price, the demand drops by 2 units.
- For **part (b)**: The rate of change at $ p = 64 $ is $ -\frac{1}{4} $, meaning for every $1 increase in price, the demand drops by $ \frac{1}{4} $ of a unit.

Let me know if you'd like further details or explanations!

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Here are some related questions:
1. How do we apply the chain rule to find the derivative of functions with radicals?
2. Why does the demand drop more significantly when the price is lower?
3. What is the meaning of the negative sign in the rate of change?
4. How would the interpretation change if the rate of change were positive?
5. How can the instantaneous rate of change be used in business or economics to optimize pricing strategies?

**Tip:** Understanding the derivative helps analyze sensitivity—how much one variable (like demand) responds to changes in another (like price).

The demand for q units of a product depends on the price p (in dollars) according to q = 256 / sqrt(p) - 1, for p > 0. Find and explain the meaning of the instantaneous rate of change of demand with respect to price when the price is as follows: (a) $16, (b) $64.

Learn how to find the instantaneous rate of change of demand with respect to price for the function q = 256 / sqrt(p) - 1. This problem covers key calculus concepts like derivatives and the chain rule, suitable for undergraduate students in Calculus I.

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