Calculus Examples

D (p) = 1500 - 12 p

$D (p) = 1500 - 12 p$ ,

p = 100

$p = 100$

Step 1

Write

D (p) = 1500 - 12 p

$D (p) = 1500 - 12 p$ as an equation.

q = 1500 - 12 p

$q = 1500 - 12 p$

Step 2

To find elasticity of demand, use the formula

E = ∣ ∣ ∣ \frac{p}{q} \frac{d q}{d p} ∣ ∣ ∣

$E = | \frac{p}{q} \frac{d q}{d p} |$ .

Step 3

Substitute

100

$100$ for

p

$p$ in

q = 1500 - 12 p

$q = 1500 - 12 p$ and simplify to find

q

$q$ .

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Step 3.1

Substitute

100

$100$ for

p

$p$ .

q = 1500 - 12 \cdot 100

$q = 1500 - 12 \cdot 100$

Step 3.2

Multiply

- 12

$- 12$ by

100

$100$ .

q = 1500 - 1200

$q = 1500 - 1200$

Step 3.3

Subtract

1200

$1200$ from

1500

$1500$ .

q = 300

$q = 300$

q = 300

$q = 300$

Step 4

Find

\frac{d q}{d p}

$\frac{d q}{d p}$ by differentiating the demand function.

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Step 4.1

Differentiate the demand function.

\frac{d q}{d p} = \frac{d}{d p} [1500 - 12 p]

$\frac{d q}{d p} = \frac{d}{d p} [1500 - 12 p]$

Step 4.2

Differentiate.

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Step 4.2.1

By the Sum Rule, the derivative of

1500 - 12 p

$1500 - 12 p$ with respect to

p

$p$ is

\frac{d}{d p} [1500] + \frac{d}{d p} [- 12 p]

$\frac{d}{d p} [1500] + \frac{d}{d p} [- 12 p]$ .

\frac{d q}{d p} = \frac{d}{d p} [1500] + \frac{d}{d p} [- 12 p]

$\frac{d q}{d p} = \frac{d}{d p} [1500] + \frac{d}{d p} [- 12 p]$

Step 4.2.2

Since

1500

$1500$ is constant with respect to

p

$p$ , the derivative of

1500

$1500$ with respect to

p

$p$ is

0

$0$ .

\frac{d q}{d p} = 0 + \frac{d}{d p} [- 12 p]

$\frac{d q}{d p} = 0 + \frac{d}{d p} [- 12 p]$

\frac{d q}{d p} = 0 + \frac{d}{d p} [- 12 p]

$\frac{d q}{d p} = 0 + \frac{d}{d p} [- 12 p]$

Step 4.3

Evaluate

\frac{d}{d p} [- 12 p]

$\frac{d}{d p} [- 12 p]$ .

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Step 4.3.1

Since

- 12

$- 12$ is constant with respect to

p

$p$ , the derivative of

- 12 p

$- 12 p$ with respect to

p

$p$ is

- 12 \frac{d}{d p} [p]

$- 12 \frac{d}{d p} [p]$ .

\frac{d q}{d p} = 0 - 12 \frac{d}{d p} [p]

$\frac{d q}{d p} = 0 - 12 \frac{d}{d p} [p]$

Step 4.3.2

Differentiate using the Power Rule which states that

\frac{d}{d p} [p^{n}]

$\frac{d}{d p} [p^{n}]$ is

n p^{n - 1}

$n p^{n - 1}$ where

n = 1

$n = 1$ .

\frac{d q}{d p} = 0 - 12 \cdot 1

$\frac{d q}{d p} = 0 - 12 \cdot 1$

Step 4.3.3

Multiply

- 12

$- 12$ by

1

$1$ .

\frac{d q}{d p} = 0 - 12

$\frac{d q}{d p} = 0 - 12$

\frac{d q}{d p} = 0 - 12

$\frac{d q}{d p} = 0 - 12$

Step 4.4

Subtract

12

$12$ from

0

$0$ .

\frac{d q}{d p} = - 12

$\frac{d q}{d p} = - 12$

\frac{d q}{d p} = - 12

$\frac{d q}{d p} = - 12$

Step 5

Substitute into the formula for elasticity

E = ∣ ∣ ∣ \frac{p}{q} \frac{d q}{d p} ∣ ∣ ∣

$E = | \frac{p}{q} \frac{d q}{d p} |$ and simplify.

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Step 5.1

Substitute

$- 12$ for

$\frac{d q}{d p}$ .

$E = | \frac{p}{q} \cdot - 12 |$

Step 5.2

Substitute the values of

$p$ and

$q$ .

$E = | \frac{100}{300} \cdot - 12 |$

Step 5.3

Cancel the common factor of

$12$ .

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Step 5.3.1

Factor

$12$ out of

$300$ .

$E = | \frac{100}{12 (25)} \cdot - 12 |$

Step 5.3.2

Factor

$12$ out of

$- 12$ .

$E = | \frac{100}{12 \cdot 25} \cdot (12 \cdot - 1) |$

Step 5.3.3

Cancel the common factor.

$E = | \frac{100}{12 \cdot 25} \cdot (12 \cdot - 1) |$

Step 5.3.4

Rewrite the expression.

$E = | \frac{100}{25} \cdot - 1 |$

Step 5.4

Combine

$\frac{100}{25}$ and

$- 1$ .

$E = | \frac{100 \cdot - 1}{25} |$

Step 5.5

Multiply

$100$ by

$- 1$ .

$E = | \frac{- 100}{25} |$

Step 5.6

Divide

$- 100$ by

$25$ .

$E = | - 4 |$

Step 5.7

The absolute value is the distance between a number and zero. The distance between

$- 4$ and

$0$ is

$4$ .

$E = 4$

Step 6

Since

$E > 1$ , the demand is elastic.

$E = 4$

$Elastic$

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