Calculus Examples

$p = 25 - 0.3 q$ , $q = 50$

Step 1

To find elasticity of demand, use the formula $E = | \frac{p}{q} \frac{d q}{d p} |$ .

Step 2

Substitute $50$ for $q$ in $p = 25 - 0.3 q$ and simplify to find $p$ .

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Step 2.1

Substitute $50$ for $q$ .

$p = 25 - 0.3 \cdot 50$

Step 2.2

Multiply $- 0.3$ by $50$ .

$p = 25 - 15$

Step 2.3

Subtract $15$ from $25$ .

$p = 10$

Step 3

Solve the demand function for $q$ .

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Step 3.1

Rewrite the equation as $25 - 0.3 q = p$ .

$25 - 0.3 q = p$

Step 3.2

Subtract $25$ from both sides of the equation.

$- 0.3 q = p - 25$

Step 3.3

Divide each term in $- 0.3 q = p - 25$ by $- 0.3$ and simplify.

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Step 3.3.1

Divide each term in $- 0.3 q = p - 25$ by $- 0.3$ .

$\frac{- 0.3 q}{- 0.3} = \frac{p}{- 0.3} + \frac{- 25}{- 0.3}$

Step 3.3.2

Simplify the left side.

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Step 3.3.2.1

Cancel the common factor of $- 0.3$ .

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Step 3.3.2.1.1

Cancel the common factor.

$\frac{- 0.3 q}{- 0.3} = \frac{p}{- 0.3} + \frac{- 25}{- 0.3}$

Step 3.3.2.1.2

Divide $q$ by $1$ .

$q = \frac{p}{- 0.3} + \frac{- 25}{- 0.3}$

Step 3.3.3

Simplify the right side.

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Step 3.3.3.1

Simplify each term.

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Step 3.3.3.1.1

Move the negative in front of the fraction.

$q = - \frac{p}{0.3} + \frac{- 25}{- 0.3}$

Step 3.3.3.1.2

Multiply by $1$ .

$q = - \frac{1 p}{0.3} + \frac{- 25}{- 0.3}$

Step 3.3.3.1.3

Factor $0.3$ out of $0.3$ .

$q = - \frac{1 p}{0.3 (1)} + \frac{- 25}{- 0.3}$

Step 3.3.3.1.4

Separate fractions.

$q = - (\frac{1}{0.3} \cdot \frac{p}{1}) + \frac{- 25}{- 0.3}$

Step 3.3.3.1.5

Divide $1$ by $0.3$ .

$q = - (3.\overline{3} \frac{p}{1}) + \frac{- 25}{- 0.3}$

Step 3.3.3.1.6

Divide $p$ by $1$ .

$q = - (3.\overline{3} p) + \frac{- 25}{- 0.3}$

Step 3.3.3.1.7

Multiply $3.\overline{3}$ by $- 1$ .

$q = - 3.\overline{3} p + \frac{- 25}{- 0.3}$

Step 3.3.3.1.8

Divide $- 25$ by $- 0.3$ .

$q = - 3.\overline{3} p + 83.\overline{3}$

Step 4

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

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Step 4.1

Differentiate the demand function.

$\frac{d q}{d p} = \frac{d}{d p} [- 3.\overline{3} p + 83.\overline{3}]$

Step 4.2

By the Sum Rule, the derivative of $- 3.\overline{3} p + 83.\overline{3}$ with respect to $p$ is $\frac{d}{d p} [- 3.\overline{3} p] + \frac{d}{d p} [83.\overline{3}]$ .

$\frac{d q}{d p} = \frac{d}{d p} [- 3.\overline{3} p] + \frac{d}{d p} [83.\overline{3}]$

Step 4.3

Evaluate $\frac{d}{d p} [- 3.\overline{3} p]$ .

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Step 4.3.1

Since $- 3.\overline{3}$ is constant with respect to $p$ , the derivative of $- 3.\overline{3} p$ with respect to $p$ is $- 3.\overline{3} \frac{d}{d p} [p]$ .

$\frac{d q}{d p} = - 3.\overline{3} \frac{d}{d p} [p] + \frac{d}{d p} [83.\overline{3}]$

Step 4.3.2

Differentiate using the Power Rule which states that $\frac{d}{d p} [p^{n}]$ is $n p^{n - 1}$ where $n = 1$ .

$\frac{d q}{d p} = - 3.\overline{3} \cdot 1 + \frac{d}{d p} [83.\overline{3}]$

Step 4.3.3

Multiply $- 3.\overline{3}$ by $1$ .

$\frac{d q}{d p} = - 3.\overline{3} + \frac{d}{d p} [83.\overline{3}]$

Step 4.4

Differentiate using the Constant Rule.

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Step 4.4.1

Since $83.\overline{3}$ is constant with respect to $p$ , the derivative of $83.\overline{3}$ with respect to $p$ is $0$ .

$\frac{d q}{d p} = - 3.\overline{3} + 0$

Step 4.4.2

Add $- 3.\overline{3}$ and $0$ .

$\frac{d q}{d p} = - 3.\overline{3}$

Step 5

Substitute into the formula for elasticity $E = | \frac{p}{q} \frac{d q}{d p} |$ and simplify.

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Step 5.1

Substitute $- 3.\overline{3}$ for $\frac{d q}{d p}$ .

$E = | \frac{p}{q} \cdot - 3.\overline{3} |$

Step 5.2

Substitute the values of $p$ and $q$ .

$E \approx | \frac{10}{50} \cdot - 3.\overline{3} |$

Step 5.3

Cancel the common factor of $10$ and $50$ .

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Step 5.3.1

Factor $10$ out of $10$ .

$E \approx | \frac{10 (1)}{50} \cdot - 3.\overline{3} |$

Step 5.3.2

Cancel the common factors.

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Step 5.3.2.1

Factor $10$ out of $50$ .

$E \approx | \frac{10 \cdot 1}{10 \cdot 5} \cdot - 3.\overline{3} |$

Step 5.3.2.2

Cancel the common factor.

$E \approx | \frac{10 \cdot 1}{10 \cdot 5} \cdot - 3.\overline{3} |$

Step 5.3.2.3

Rewrite the expression.

$E \approx | \frac{1}{5} \cdot - 3.\overline{3} |$

Step 5.4

Combine $\frac{1}{5}$ and $- 3.\overline{3}$ .

$E \approx | \frac{- 3.\overline{3}}{5} |$

Step 5.5

Divide $- 3.\overline{3}$ by $5$ .

$E \approx | - 0.\overline{6} |$

Step 5.6

The absolute value is the distance between a number and zero. The distance between $- 0.\overline{6}$ and $0$ is $0.\overline{6}$ .

$E \approx 0.66666666$

Step 6

Since $E < 1$ , the demand is inelastic.

$E \approx 0.66666666$

$Inelastic$

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