Math Problem Statement

The demand for a product is given by

D(p) =

1000

√p

− 1

(a) What is the average rate of change of demand when p increases from 1 to 25?

(b) What is the instantaneous rate of change of demand at p = 25? The marginal

demand function at p = 25? The slope of the tangent line to demand function at

p = 25?

(c) Find the equation of tangent line to demand function at p = 25?

Solution

To solve the given problem, let's break it down step by step.

Given:

The demand function is: D(p)=1000p1D(p) = \frac{1000}{\sqrt{p}} - 1

Part (a): Average Rate of Change of Demand from p=1p = 1 to p=25p = 25

The average rate of change of the demand function D(p)D(p) from p=1p = 1 to p=25p = 25 is given by:

Average Rate of Change=D(25)D(1)251\text{Average Rate of Change} = \frac{D(25) - D(1)}{25 - 1}

Let's compute D(25)D(25) and D(1)D(1):

D(25)=1000251=100051=2001=199D(25) = \frac{1000}{\sqrt{25}} - 1 = \frac{1000}{5} - 1 = 200 - 1 = 199

D(1)=100011=100011=10001=999D(1) = \frac{1000}{\sqrt{1}} - 1 = \frac{1000}{1} - 1 = 1000 - 1 = 999

Now, compute the average rate of change:

Average Rate of Change=199999251=80024=33.33\text{Average Rate of Change} = \frac{199 - 999}{25 - 1} = \frac{-800}{24} = -33.33

So, the average rate of change of demand when pp increases from 1 to 25 is 33.33-33.33.


Part (b): Instantaneous Rate of Change of Demand at p=25p = 25

The instantaneous rate of change of demand at p=25p = 25 is the derivative of D(p)D(p) evaluated at p=25p = 25. We will first find the derivative of the demand function D(p)D(p).

D(p)=1000p1=1000p121D(p) = \frac{1000}{\sqrt{p}} - 1 = 1000p^{-\frac{1}{2}} - 1

Differentiate D(p)D(p) with respect to pp:

= -\frac{500}{p^{\frac{3}{2}}}$$ Now, evaluate $$D'(p)$$ at $$p = 25$$: $$D'(25) = -\frac{500}{25^{\frac{3}{2}}} = -\frac{500}{125} = -4$$ So, the instantaneous rate of change of demand at $$p = 25$$ (or the marginal demand function at $$p = 25$$) is $$-4$$. This is also the slope of the tangent line to the demand function at $$p = 25$$. --- ### Part (c): Equation of the Tangent Line to the Demand Function at $$p = 25$$ The equation of the tangent line at $$p = 25$$ can be written in point-slope form: $$y - D(25) = D'(25)(p - 25)$$ We already know $$D(25) = 199$$ and $$D'(25) = -4$$, so the equation of the tangent line is: $$y - 199 = -4(p - 25)$$ Simplify the equation: $$y - 199 = -4p + 100$$ $$y = -4p + 299$$ Thus, the equation of the tangent line to the demand function at $$p = 25$$ is: $$y = -4p + 299$$ --- ### Summary: 1. The average rate of change of demand from $$p = 1$$ to $$p = 25$$ is $$-33.33$$. 2. The instantaneous rate of change of demand at $$p = 25$$ is $$-4$$. 3. The equation of the tangent line to the demand function at $$p = 25$$ is $$y = -4p + 299$$. --- Would you like further explanation or clarification on any part of the solution? ### 5 Related Questions: 1. What does the negative average rate of change indicate about the demand as price increases? 2. How can the marginal demand function help in understanding the pricing strategy? 3. How would the equation of the tangent line change for different values of $$p$$? 4. How can we interpret the slope of the tangent line in the context of demand and price? 5. What other methods can be used to approximate instantaneous rates of change in economics? ### Tip: The marginal demand function is a useful tool for businesses because it shows how small changes in price can impact the demand for a product.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Rates of Change
Tangent Lines

Formulas

Average Rate of Change: (D(b) - D(a)) / (b - a)
Derivative of a function: D'(p) = -500/p^(3/2)
Tangent line equation: y - D(p) = D'(p)(p - p0)

Theorems

Derivative as the instantaneous rate of change
Point-slope form of a linear equation

Suitable Grade Level

College level (Calculus I)