Calculus Examples

D (p) = 200 - p^{2}

$D (p) = 200 - p^{2}$ ,

p = 10

$p = 10$

Step 1

Write

D (p) = 200 - p^{2}

$D (p) = 200 - p^{2}$ as an equation.

q = 200 - p^{2}

$q = 200 - p^{2}$

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

10

$10$ for

p

$p$ in

q = 200 - p^{2}

$q = 200 - p^{2}$ and simplify to find

q

$q$ .

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Step 3.1

Substitute

10

$10$ for

p

$p$ .

q = 200 - 10^{2}

$q = 200 - 10^{2}$

Step 3.2

Simplify each term.

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Step 3.2.1

Raise

10

$10$ to the power of

2

$2$ .

q = 200 - 1 \cdot 100

$q = 200 - 1 \cdot 100$

Step 3.2.2

Multiply

- 1

$- 1$ by

100

$100$ .

q = 200 - 100

$q = 200 - 100$

q = 200 - 100

$q = 200 - 100$

Step 3.3

Subtract

100

$100$ from

200

$200$ .

q = 100

$q = 100$

q = 100

$q = 100$

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} [200 - p^{2}]

$\frac{d q}{d p} = \frac{d}{d p} [200 - p^{2}]$

Step 4.2

Differentiate.

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Step 4.2.1

By the Sum Rule, the derivative of

200 - p^{2}

$200 - p^{2}$ with respect to

p

$p$ is

\frac{d}{d p} [200] + \frac{d}{d p} [- p^{2}]

$\frac{d}{d p} [200] + \frac{d}{d p} [- p^{2}]$ .

\frac{d q}{d p} = \frac{d}{d p} [200] + \frac{d}{d p} [- p^{2}]

$\frac{d q}{d p} = \frac{d}{d p} [200] + \frac{d}{d p} [- p^{2}]$

Step 4.2.2

Since

200

$200$ is constant with respect to

p

$p$ , the derivative of

200

$200$ with respect to

p

$p$ is

0

$0$ .

\frac{d q}{d p} = 0 + \frac{d}{d p} [- p^{2}]

$\frac{d q}{d p} = 0 + \frac{d}{d p} [- p^{2}]$

\frac{d q}{d p} = 0 + \frac{d}{d p} [- p^{2}]

$\frac{d q}{d p} = 0 + \frac{d}{d p} [- p^{2}]$

Step 4.3

Evaluate

\frac{d}{d p} [- p^{2}]

$\frac{d}{d p} [- p^{2}]$ .

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Step 4.3.1

Since

- 1

$- 1$ is constant with respect to

p

$p$ , the derivative of

- p^{2}

$- p^{2}$ with respect to

p

$p$ is

- \frac{d}{d p} [p^{2}]

$- \frac{d}{d p} [p^{2}]$ .

\frac{d q}{d p} = 0 - \frac{d}{d p} [p^{2}]

$\frac{d q}{d p} = 0 - \frac{d}{d p} [p^{2}]$

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 = 2

$n = 2$ .

\frac{d q}{d p} = 0 - (2 p)

$\frac{d q}{d p} = 0 - (2 p)$

Step 4.3.3

Multiply

2

$2$ by

- 1

$- 1$ .

\frac{d q}{d p} = 0 - 2 p

$\frac{d q}{d p} = 0 - 2 p$

\frac{d q}{d p} = 0 - 2 p

$\frac{d q}{d p} = 0 - 2 p$

Step 4.4

Subtract

2 p

$2 p$ from

0

$0$ .

\frac{d q}{d p} = - 2 p

$\frac{d q}{d p} = - 2 p$

\frac{d q}{d p} = - 2 p

$\frac{d q}{d p} = - 2 p$

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

- 2 p

$- 2 p$ for

\frac{d q}{d p}

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

E = | \frac{p}{q} (- 2 p) |

$E = | \frac{p}{q} (- 2 p) |$

Step 5.2

Substitute the values of

p

$p$ and

q

$q$ .

E = | \frac{10}{100} (- 2 \cdot 10) |

$E = | \frac{10}{100} (- 2 \cdot 10) |$

Step 5.3

Cancel the common factor of

10

$10$ and

100

$100$ .

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Step 5.3.1

Factor

10

$10$ out of

10

$10$ .

E = | \frac{10 (1)}{100} (- 2 \cdot 10) |

$E = | \frac{10 (1)}{100} (- 2 \cdot 10) |$

Step 5.3.2

Cancel the common factors.

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Step 5.3.2.1

Factor

10

$10$ out of

100

$100$ .

E = | \frac{10 \cdot 1}{10 \cdot 10} (- 2 \cdot 10) |

$E = | \frac{10 \cdot 1}{10 \cdot 10} (- 2 \cdot 10) |$

Step 5.3.2.2

Cancel the common factor.

E = | \frac{10 \cdot 1}{10 \cdot 10} (- 2 \cdot 10) |

$E = | \frac{10 \cdot 1}{10 \cdot 10} (- 2 \cdot 10) |$

Step 5.3.2.3

Rewrite the expression.

E = | \frac{1}{10} (- 2 \cdot 10) |

$E = | \frac{1}{10} (- 2 \cdot 10) |$

E = | \frac{1}{10} (- 2 \cdot 10) |

$E = | \frac{1}{10} (- 2 \cdot 10) |$

E = | \frac{1}{10} (- 2 \cdot 10) |

$E = | \frac{1}{10} (- 2 \cdot 10) |$

Step 5.4

Multiply

- 2

$- 2$ by

10

$10$ .

E = | \frac{1}{10} \cdot - 20 |

$E = | \frac{1}{10} \cdot - 20 |$

Step 5.5

Cancel the common factor of

10

$10$ .

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Step 5.5.1

Factor

10

$10$ out of

- 20

$- 20$ .

E = | \frac{1}{10} \cdot (10 (- 2)) |

$E = | \frac{1}{10} \cdot (10 (- 2)) |$

Step 5.5.2

Cancel the common factor.

E = | \frac{1}{10} \cdot (10 \cdot - 2) |

$E = | \frac{1}{10} \cdot (10 \cdot - 2) |$

Step 5.5.3

Rewrite the expression.

E = | - 2 |

$E = | - 2 |$

E = | - 2 |

$E = | - 2 |$

Step 5.6

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

- 2

$- 2$ and

0

$0$ is

2

$2$ .

E = 2

$E = 2$

E = 2

$E = 2$

Step 6

Since

E > 1

$E > 1$ , the demand is elastic.

E = 2

$E = 2$

Elastic

$Elastic$

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