Math Problem Statement

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.

Solution

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=256p1q = \frac{256}{\sqrt{p}} - 1

where qq represents demand and pp represents price in dollars.

Part (a):

You need to find q(p)q'(p) and evaluate it at p=16p = 16.

  1. Derivative of qq with respect to pp:

The given function is q(p)=256p1q(p) = \frac{256}{\sqrt{p}} - 1. We need to apply the chain rule to differentiate this expression.

q(p)=256p121q(p) = 256 \cdot p^{-\frac{1}{2}} - 1

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

q(p)=256(12)p32=128p32q'(p) = 256 \cdot \left( -\frac{1}{2} \right) p^{-\frac{3}{2}} = -\frac{128}{p^{\frac{3}{2}}}

  1. Evaluate q(p)q'(p) at p=16p = 16:

q(16)=1281632q'(16) = -\frac{128}{16^{\frac{3}{2}}}

Since 1632=6416^{\frac{3}{2}} = 64, we have:

q(16)=12864=2q'(16) = -\frac{128}{64} = -2

So, the instantaneous rate of change of demand when the price is 1616 dollars is 2-2.

Part (b):

You need to do the same for p=64p = 64.

Using the derivative formula we found earlier:

q(p)=128p32q'(p) = -\frac{128}{p^{\frac{3}{2}}}

Evaluate q(64)q'(64):

q(64)=1286432q'(64) = -\frac{128}{64^{\frac{3}{2}}}

Since 6432=51264^{\frac{3}{2}} = 512, we have:

q(64)=128512=14q'(64) = -\frac{128}{512} = -\frac{1}{4}

So, the instantaneous rate of change of demand when the price is 6464 dollars is 14-\frac{1}{4}.


Interpretation:

  • For part (a): The rate of change at p=16p = 16 is 2-2, meaning for every $1 increase in price, the demand drops by 2 units.
  • For part (b): The rate of change at p=64p = 64 is 14-\frac{1}{4}, meaning for every $1 increase in price, the demand drops by 14\frac{1}{4} of a unit.

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


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).

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Instantaneous Rate of Change
Demand Function

Formulas

q(p) = 256 / sqrt(p) - 1
q'(p) = -128 / p^(3/2)

Theorems

Chain Rule of Differentiation

Suitable Grade Level

Undergraduate (Calculus I)