Calculus Examples

$A (t) = 300 t e^{- 2.5 t}$

Step 1

Find the first derivative of the function.

Tap for more steps...

Step 1.1

Since $300$ is constant with respect to $t$ , the derivative of $300 t e^{- 2.5 t}$ with respect to $t$ is $300 \frac{d}{d t} [t e^{- 2.5 t}]$ .

$300 \frac{d}{d t} [t e^{- 2.5 t}]$

Step 1.2

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t$ and $g (t) = e^{- 2.5 t}$ .

$300 (t \frac{d}{d t} [e^{- 2.5 t}] + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 1.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - 2.5 t$ .

Tap for more steps...

Step 1.3.1

To apply the Chain Rule, set $u$ as $- 2.5 t$ .

$300 (t (\frac{d}{d u} [e^{u}] \frac{d}{d t} [- 2.5 t]) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 1.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$300 (t (e^{u} \frac{d}{d t} [- 2.5 t]) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 1.3.3

Replace all occurrences of $u$ with $- 2.5 t$ .

$300 (t (e^{- 2.5 t} \frac{d}{d t} [- 2.5 t]) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 1.4

Differentiate.

Tap for more steps...

Step 1.4.1

Since $- 2.5$ is constant with respect to $t$ , the derivative of $- 2.5 t$ with respect to $t$ is $- 2.5 \frac{d}{d t} [t]$ .

$300 (t (e^{- 2.5 t} (- 2.5 \frac{d}{d t} [t])) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 1.4.2

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

$300 (t (e^{- 2.5 t} (- 2.5 \cdot 1)) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 1.4.3

Simplify the expression.

Tap for more steps...

Step 1.4.3.1

Multiply $- 2.5$ by $1$ .

$300 (t (e^{- 2.5 t} \cdot - 2.5) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 1.4.3.2

Move $- 2.5$ to the left of $e^{- 2.5 t}$ .

$300 (t (- 2.5 \cdot e^{- 2.5 t}) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 1.4.4

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

$300 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t} \cdot 1)$

Step 1.4.5

Multiply $e^{- 2.5 t}$ by $1$ .

$300 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t})$

Step 1.5

Simplify.

Tap for more steps...

Step 1.5.1

Apply the distributive property.

$300 (t (- 2.5 e^{- 2.5 t})) + 300 e^{- 2.5 t}$

Step 1.5.2

Multiply $- 2.5$ by $300$ .

$- 750 t e^{- 2.5 t} + 300 e^{- 2.5 t}$

Step 1.5.3

Reorder terms.

$- 750 e^{- 2.5 t} t + 300 e^{- 2.5 t}$

Step 1.5.4

Reorder factors in $- 750 e^{- 2.5 t} t + 300 e^{- 2.5 t}$ .

$- 750 t e^{- 2.5 t} + 300 e^{- 2.5 t}$

Step 2

Find the second derivative of the function.

Tap for more steps...

Step 2.1

By the Sum Rule, the derivative of $- 750 t e^{- 2.5 t} + 300 e^{- 2.5 t}$ with respect to $t$ is $\frac{d}{d t} [- 750 t e^{- 2.5 t}] + \frac{d}{d t} [300 e^{- 2.5 t}]$ .

$A'' (t) = \frac{d}{d t} (- 750 t e^{- 2.5 t}) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2

Evaluate $\frac{d}{d t} [- 750 t e^{- 2.5 t}]$ .

Tap for more steps...

Step 2.2.1

Since $- 750$ is constant with respect to $t$ , the derivative of $- 750 t e^{- 2.5 t}$ with respect to $t$ is $- 750 \frac{d}{d t} [t e^{- 2.5 t}]$ .

$A'' (t) = - 750 \frac{d}{d t} (t e^{- 2.5 t}) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2.2

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t$ and $g (t) = e^{- 2.5 t}$ .

$A'' (t) = - 750 (t \frac{d}{d t} (e^{- 2.5 t}) + e^{- 2.5 t} \frac{d}{d t} (t)) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - 2.5 t$ .

Tap for more steps...

Step 2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $- 2.5 t$ .

$A'' (t) = - 750 (t (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d t} (- 2.5 t)) + e^{- 2.5 t} \frac{d}{d t} (t)) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$A'' (t) = - 750 (t (e^{u_{1}} \frac{d}{d t} (- 2.5 t)) + e^{- 2.5 t} \frac{d}{d t} (t)) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2.3.3

Replace all occurrences of $u_{1}$ with $- 2.5 t$ .

$A'' (t) = - 750 (t (e^{- 2.5 t} \frac{d}{d t} (- 2.5 t)) + e^{- 2.5 t} \frac{d}{d t} (t)) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2.4

Since $- 2.5$ is constant with respect to $t$ , the derivative of $- 2.5 t$ with respect to $t$ is $- 2.5 \frac{d}{d t} [t]$ .

$A'' (t) = - 750 (t (e^{- 2.5 t} (- 2.5 \frac{d}{d t} t)) + e^{- 2.5 t} \frac{d}{d t} (t)) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2.5

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

$A'' (t) = - 750 (t (e^{- 2.5 t} (- 2.5 \cdot 1)) + e^{- 2.5 t} \frac{d}{d t} (t)) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2.6

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

$A'' (t) = - 750 (t (e^{- 2.5 t} (- 2.5 \cdot 1)) + e^{- 2.5 t} \cdot 1) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2.7

Multiply $- 2.5$ by $1$ .

$A'' (t) = - 750 (t (e^{- 2.5 t} \cdot - 2.5) + e^{- 2.5 t} \cdot 1) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2.8

Move $- 2.5$ to the left of $e^{- 2.5 t}$ .

$A'' (t) = - 750 (t (- 2.5 \cdot e^{- 2.5 t}) + e^{- 2.5 t} \cdot 1) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.2.9

Multiply $e^{- 2.5 t}$ by $1$ .

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t}) + \frac{d}{d t} (300 e^{- 2.5 t})$

Step 2.3

Evaluate $\frac{d}{d t} [300 e^{- 2.5 t}]$ .

Tap for more steps...

Step 2.3.1

Since $300$ is constant with respect to $t$ , the derivative of $300 e^{- 2.5 t}$ with respect to $t$ is $300 \frac{d}{d t} [e^{- 2.5 t}]$ .

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t}) + 300 \frac{d}{d t} (e^{- 2.5 t})$

Step 2.3.2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - 2.5 t$ .

Tap for more steps...

Step 2.3.2.1

To apply the Chain Rule, set $u_{2}$ as $- 2.5 t$ .

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t}) + 300 (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d t} (- 2.5 t))$

Step 2.3.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t}) + 300 (e^{u_{2}} \frac{d}{d t} (- 2.5 t))$

Step 2.3.2.3

Replace all occurrences of $u_{2}$ with $- 2.5 t$ .

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t}) + 300 (e^{- 2.5 t} \frac{d}{d t} (- 2.5 t))$

Step 2.3.3

Since $- 2.5$ is constant with respect to $t$ , the derivative of $- 2.5 t$ with respect to $t$ is $- 2.5 \frac{d}{d t} [t]$ .

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t}) + 300 (e^{- 2.5 t} (- 2.5 \frac{d}{d t} t))$

Step 2.3.4

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

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t}) + 300 (e^{- 2.5 t} (- 2.5 \cdot 1))$

Step 2.3.5

Multiply $- 2.5$ by $1$ .

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t}) + 300 (e^{- 2.5 t} \cdot - 2.5)$

Step 2.3.6

Move $- 2.5$ to the left of $e^{- 2.5 t}$ .

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t}) + 300 (- 2.5 \cdot e^{- 2.5 t})$

Step 2.3.7

Multiply $- 2.5$ by $300$ .

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t}) - 750 e^{- 2.5 t}$

Step 2.4

Simplify.

Tap for more steps...

Step 2.4.1

Apply the distributive property.

$A'' (t) = - 750 (t (- 2.5 e^{- 2.5 t})) - 750 e^{- 2.5 t} - 750 e^{- 2.5 t}$

Step 2.4.2

Combine terms.

Tap for more steps...

Step 2.4.2.1

Multiply $- 2.5$ by $- 750$ .

$A'' (t) = 1875 (t (e^{- 2.5 t})) - 750 e^{- 2.5 t} - 750 e^{- 2.5 t}$

Step 2.4.2.2

Subtract $750 e^{- 2.5 t}$ from $- 750 e^{- 2.5 t}$ .

$A'' (t) = 1875 t e^{- 2.5 t} - 1500 e^{- 2.5 t}$

Step 2.4.3

Reorder terms.

$A'' (t) = 1875 e^{- 2.5 t} t - 1500 e^{- 2.5 t}$

Step 2.4.4

Reorder factors in $1875 e^{- 2.5 t} t - 1500 e^{- 2.5 t}$ .

$A'' (t) = 1875 t e^{- 2.5 t} - 1500 e^{- 2.5 t}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- 750 t e^{- 2.5 t} + 300 e^{- 2.5 t} = 0$

Step 4

Find the first derivative.

Tap for more steps...

Step 4.1

Find the first derivative.

Tap for more steps...

Step 4.1.1

Since $300$ is constant with respect to $t$ , the derivative of $300 t e^{- 2.5 t}$ with respect to $t$ is $300 \frac{d}{d t} [t e^{- 2.5 t}]$ .

$300 \frac{d}{d t} [t e^{- 2.5 t}]$

Step 4.1.2

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t$ and $g (t) = e^{- 2.5 t}$ .

$300 (t \frac{d}{d t} [e^{- 2.5 t}] + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 4.1.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - 2.5 t$ .

Tap for more steps...

Step 4.1.3.1

To apply the Chain Rule, set $u$ as $- 2.5 t$ .

$300 (t (\frac{d}{d u} [e^{u}] \frac{d}{d t} [- 2.5 t]) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 4.1.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$300 (t (e^{u} \frac{d}{d t} [- 2.5 t]) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 4.1.3.3

Replace all occurrences of $u$ with $- 2.5 t$ .

$300 (t (e^{- 2.5 t} \frac{d}{d t} [- 2.5 t]) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 4.1.4

Differentiate.

Tap for more steps...

Step 4.1.4.1

Since $- 2.5$ is constant with respect to $t$ , the derivative of $- 2.5 t$ with respect to $t$ is $- 2.5 \frac{d}{d t} [t]$ .

$300 (t (e^{- 2.5 t} (- 2.5 \frac{d}{d t} [t])) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 4.1.4.2

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

$300 (t (e^{- 2.5 t} (- 2.5 \cdot 1)) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 4.1.4.3

Simplify the expression.

Tap for more steps...

Step 4.1.4.3.1

Multiply $- 2.5$ by $1$ .

$300 (t (e^{- 2.5 t} \cdot - 2.5) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 4.1.4.3.2

Move $- 2.5$ to the left of $e^{- 2.5 t}$ .

$300 (t (- 2.5 \cdot e^{- 2.5 t}) + e^{- 2.5 t} \frac{d}{d t} [t])$

Step 4.1.4.4

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

$300 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t} \cdot 1)$

Step 4.1.4.5

Multiply $e^{- 2.5 t}$ by $1$ .

$300 (t (- 2.5 e^{- 2.5 t}) + e^{- 2.5 t})$

Step 4.1.5

Simplify.

Tap for more steps...

Step 4.1.5.1

Apply the distributive property.

$300 (t (- 2.5 e^{- 2.5 t})) + 300 e^{- 2.5 t}$

Step 4.1.5.2

Multiply $- 2.5$ by $300$ .

$- 750 t e^{- 2.5 t} + 300 e^{- 2.5 t}$

Step 4.1.5.3

Reorder terms.

$- 750 e^{- 2.5 t} t + 300 e^{- 2.5 t}$

Step 4.1.5.4

Reorder factors in $- 750 e^{- 2.5 t} t + 300 e^{- 2.5 t}$ .

$f' (t) = - 750 t e^{- 2.5 t} + 300 e^{- 2.5 t}$

Step 4.2

The first derivative of $A (t)$ with respect to $t$ is $- 750 t e^{- 2.5 t} + 300 e^{- 2.5 t}$ .

$- 750 t e^{- 2.5 t} + 300 e^{- 2.5 t}$

Step 5

Set the first derivative equal to $0$ then solve the equation $- 750 t e^{- 2.5 t} + 300 e^{- 2.5 t} = 0$ .

Tap for more steps...

Step 5.1

Set the first derivative equal to $0$ .

$- 750 t e^{- 2.5 t} + 300 e^{- 2.5 t} = 0$

Step 5.2

Factor $150 e^{- 2.5 t}$ out of $- 750 t e^{- 2.5 t} + 300 e^{- 2.5 t}$ .

Tap for more steps...

Step 5.2.1

Factor $150 e^{- 2.5 t}$ out of $- 750 t e^{- 2.5 t}$ .

$150 e^{- 2.5 t} (- 5 t) + 300 e^{- 2.5 t} = 0$

Step 5.2.2

Factor $150 e^{- 2.5 t}$ out of $300 e^{- 2.5 t}$ .

$150 e^{- 2.5 t} (- 5 t) + 150 e^{- 2.5 t} (2) = 0$

Step 5.2.3

Factor $150 e^{- 2.5 t}$ out of $150 e^{- 2.5 t} (- 5 t) + 150 e^{- 2.5 t} (2)$ .

$150 e^{- 2.5 t} (- 5 t + 2) = 0$

Step 5.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$e^{- 2.5 t} = 0$

$- 5 t + 2 = 0$

Step 5.4

Set $e^{- 2.5 t}$ equal to $0$ and solve for $t$ .

Tap for more steps...

Step 5.4.1

Set $e^{- 2.5 t}$ equal to $0$ .

$e^{- 2.5 t} = 0$

Step 5.4.2

Solve $e^{- 2.5 t} = 0$ for $t$ .

Tap for more steps...

Step 5.4.2.1

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- 2.5 t}) = \ln (0)$

Step 5.4.2.2

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

Step 5.4.2.3

There is no solution for $e^{- 2.5 t} = 0$

No solution

Step 5.5

Set $- 5 t + 2$ equal to $0$ and solve for $t$ .

Tap for more steps...

Step 5.5.1

Set $- 5 t + 2$ equal to $0$ .

$- 5 t + 2 = 0$

Step 5.5.2

Solve $- 5 t + 2 = 0$ for $t$ .

Tap for more steps...

Step 5.5.2.1

Subtract $2$ from both sides of the equation.

$- 5 t = - 2$

Step 5.5.2.2

Divide each term in $- 5 t = - 2$ by $- 5$ and simplify.

Tap for more steps...

Step 5.5.2.2.1

Divide each term in $- 5 t = - 2$ by $- 5$ .

$\frac{- 5 t}{- 5} = \frac{- 2}{- 5}$

Step 5.5.2.2.2

Simplify the left side.

Tap for more steps...

Step 5.5.2.2.2.1

Cancel the common factor of $- 5$ .

Tap for more steps...

Step 5.5.2.2.2.1.1

Cancel the common factor.

$\frac{- 5 t}{- 5} = \frac{- 2}{- 5}$

Step 5.5.2.2.2.1.2

Divide $t$ by $1$ .

$t = \frac{- 2}{- 5}$

Step 5.5.2.2.3

Simplify the right side.

Tap for more steps...

Step 5.5.2.2.3.1

Dividing two negative values results in a positive value.

$t = \frac{2}{5}$

Step 5.6

The final solution is all the values that make $150 e^{- 2.5 t} (- 5 t + 2) = 0$ true.

$t = \frac{2}{5}$

Step 6

Find the values where the derivative is undefined.

Tap for more steps...

Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$t = \frac{2}{5}$

Step 8

Evaluate the second derivative at $t = \frac{2}{5}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$1875 (\frac{2}{5}) e^{- 2.5 (\frac{2}{5})} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9

Evaluate the second derivative.

Tap for more steps...

Step 9.1

Simplify each term.

Tap for more steps...

Step 9.1.1

Cancel the common factor of $5$ .

Tap for more steps...

Step 9.1.1.1

Factor $5$ out of $1875$ .

$5 (375) \frac{2}{5} e^{- 2.5 (\frac{2}{5})} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.1.2

Cancel the common factor.

$5 \cdot 375 \frac{2}{5} e^{- 2.5 (\frac{2}{5})} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.1.3

Rewrite the expression.

$375 \cdot 2 e^{- 2.5 (\frac{2}{5})} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.2

Multiply $375$ by $2$ .

$750 e^{- 2.5 (\frac{2}{5})} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.3

Cancel the common factor of $2.5$ .

Tap for more steps...

Step 9.1.3.1

Factor $2.5$ out of $- 2.5$ .

$750 e^{2.5 (- 1) \frac{2}{5}} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.3.2

Factor $2.5$ out of $5$ .

$750 e^{2.5 \cdot - 1 \frac{2}{2.5 \cdot 2}} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.3.3

Cancel the common factor.

$750 e^{2.5 \cdot - 1 \frac{2}{2.5 \cdot 2}} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.3.4

Rewrite the expression.

$750 e^{- 1 (\frac{2}{2})} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 9.1.4.1

Cancel the common factor.

$750 e^{- 1 (\frac{2}{2})} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.4.2

Rewrite the expression.

$750 e^{- 1 \cdot 1} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.5

Multiply $- 1$ by $1$ .

$750 e^{- 1} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.6

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$750 \frac{1}{e} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.7

Combine $750$ and $\frac{1}{e}$ .

$\frac{750}{e} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.8

Replace $e$ with an approximation.

$\frac{750}{2.71828182} - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.9

Divide $750$ by $2.71828182$ .

$275.90958087 - 1500 e^{- 2.5 (\frac{2}{5})}$

Step 9.1.10

Cancel the common factor of $2.5$ .

Tap for more steps...

Step 9.1.10.1

Factor $2.5$ out of $- 2.5$ .

$275.90958087 - 1500 e^{2.5 (- 1) \frac{2}{5}}$

Step 9.1.10.2

Factor $2.5$ out of $5$ .

$275.90958087 - 1500 e^{2.5 \cdot - 1 \frac{2}{2.5 \cdot 2}}$

Step 9.1.10.3

Cancel the common factor.

$275.90958087 - 1500 e^{2.5 \cdot - 1 \frac{2}{2.5 \cdot 2}}$

Step 9.1.10.4

Rewrite the expression.

$275.90958087 - 1500 e^{- 1 (\frac{2}{2})}$

Step 9.1.11

Cancel the common factor of $2$ .

Tap for more steps...

Step 9.1.11.1

Cancel the common factor.

$275.90958087 - 1500 e^{- 1 (\frac{2}{2})}$

Step 9.1.11.2

Rewrite the expression.

$275.90958087 - 1500 e^{- 1 \cdot 1}$

Step 9.1.12

Multiply $- 1$ by $1$ .

$275.90958087 - 1500 e^{- 1}$

Step 9.1.13

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$275.90958087 - 1500 \frac{1}{e}$

Step 9.1.14

Combine $- 1500$ and $\frac{1}{e}$ .

$275.90958087 + \frac{- 1500}{e}$

Step 9.1.15

Move the negative in front of the fraction.

$275.90958087 - \frac{1500}{e}$

Step 9.1.16

Replace $e$ with an approximation.

$275.90958087 - \frac{1500}{2.71828182}$

Step 9.1.17

Divide $1500$ by $2.71828182$ .

$275.90958087 - 1 \cdot 551.81916175$

Step 9.1.18

Multiply $- 1$ by $551.81916175$ .

$275.90958087 - 551.81916175$

Step 9.2

Subtract $551.81916175$ from $275.90958087$ .

$- 275.90958087$

Step 10

$t = \frac{2}{5}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$t = \frac{2}{5}$ is a local maximum

Step 11

Find the y-value when $t = \frac{2}{5}$ .

Tap for more steps...

Step 11.1

Replace the variable $t$ with $\frac{2}{5}$ in the expression.

$A (\frac{2}{5}) = 300 (\frac{2}{5}) \cdot e^{- 2.5 (\frac{2}{5})}$

Step 11.2

Simplify the result.

Tap for more steps...

Step 11.2.1

Cancel the common factor of $5$ .

Tap for more steps...

Step 11.2.1.1

Factor $5$ out of $300$ .

$A (\frac{2}{5}) = 5 (60) (\frac{2}{5}) \cdot e^{- 2.5 (\frac{2}{5})}$

Step 11.2.1.2

Cancel the common factor.

$A (\frac{2}{5}) = 5 \cdot (60 (\frac{2}{5})) \cdot e^{- 2.5 (\frac{2}{5})}$

Step 11.2.1.3

Rewrite the expression.

$A (\frac{2}{5}) = 60 \cdot 2 \cdot e^{- 2.5 (\frac{2}{5})}$

Step 11.2.2

Multiply $60$ by $2$ .

$A (\frac{2}{5}) = 120 \cdot e^{- 2.5 (\frac{2}{5})}$

Step 11.2.3

Cancel the common factor of $2.5$ .

Tap for more steps...

Step 11.2.3.1

Factor $2.5$ out of $- 2.5$ .

$A (\frac{2}{5}) = 120 \cdot e^{2.5 (- 1) (\frac{2}{5})}$

Step 11.2.3.2

Factor $2.5$ out of $5$ .

$A (\frac{2}{5}) = 120 \cdot e^{2.5 \cdot (- 1 \frac{2}{2.5 \cdot 2})}$

Step 11.2.3.3

Cancel the common factor.

$A (\frac{2}{5}) = 120 \cdot e^{2.5 \cdot (- 1 \frac{2}{2.5 \cdot 2})}$

Step 11.2.3.4

Rewrite the expression.

$A (\frac{2}{5}) = 120 \cdot e^{- 1 (\frac{2}{2})}$

Step 11.2.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 11.2.4.1

Cancel the common factor.

$A (\frac{2}{5}) = 120 \cdot e^{- 1 (\frac{2}{2})}$

Step 11.2.4.2

Rewrite the expression.

$A (\frac{2}{5}) = 120 \cdot e^{- 1 \cdot 1}$

Step 11.2.5

Multiply $- 1$ by $1$ .

$A (\frac{2}{5}) = 120 \cdot e^{- 1}$

Step 11.2.6

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$A (\frac{2}{5}) = 120 \cdot \frac{1}{e}$

Step 11.2.7

Combine $120$ and $\frac{1}{e}$ .

$A (\frac{2}{5}) = \frac{120}{e}$

Step 11.2.8

The final answer is $\frac{120}{e}$ .

$y = \frac{120}{e}$

Step 12

These are the local extrema for $A (t) = 300 t e^{- 2.5 t}$ .

$(\frac{2}{5}, \frac{120}{e})$ is a local maxima

Step 13