Find the Local Maxima and Minima h(t)=60t-16t^2

$h (t) = 60 t - 16 t^{2}$

Find the first derivative of the function.

By the Sum Rule, the derivative of $60 t - 16 t^{2}$ with respect to $t$ is $\frac{d}{d t} [60 t] + \frac{d}{d t} [- 16 t^{2}]$ .

$\frac{d}{d t} [60 t] + \frac{d}{d t} [- 16 t^{2}]$

Evaluate $\frac{d}{d t} [60 t]$ .

Since $60$ is constant with respect to $t$ , the derivative of $60 t$ with respect to $t$ is $60 \frac{d}{d t} [t]$ .

$60 \frac{d}{d t} [t] + \frac{d}{d t} [- 16 t^{2}]$

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

$60 \cdot 1 + \frac{d}{d t} [- 16 t^{2}]$

Multiply $60$ by $1$ .

$60 + \frac{d}{d t} [- 16 t^{2}]$

Evaluate $\frac{d}{d t} [- 16 t^{2}]$ .

Since $- 16$ is constant with respect to $t$ , the derivative of $- 16 t^{2}$ with respect to $t$ is $- 16 \frac{d}{d t} [t^{2}]$ .

$60 - 16 \frac{d}{d t} [t^{2}]$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$60 - 16 (2 t)$

Multiply $2$ by $- 16$ .

$60 - 32 t$

Reorder terms.

$- 32 t + 60$

Find the second derivative of the function.

By the Sum Rule, the derivative of $- 32 t + 60$ with respect to $t$ is $\frac{d}{d t} [- 32 t] + \frac{d}{d t} [60]$ .

$h'' (t) = \frac{d}{d t} (- 32 t) + \frac{d}{d t} (60)$

Evaluate $\frac{d}{d t} [- 32 t]$ .

Since $- 32$ is constant with respect to $t$ , the derivative of $- 32 t$ with respect to $t$ is $- 32 \frac{d}{d t} [t]$ .

$h'' (t) = - 32 \frac{d}{d t} t + \frac{d}{d t} (60)$

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

$h'' (t) = - 32 \cdot 1 + \frac{d}{d t} (60)$

Multiply $- 32$ by $1$ .

$h'' (t) = - 32 + \frac{d}{d t} (60)$

Differentiate using the Constant Rule.

Since $60$ is constant with respect to $t$ , the derivative of $60$ with respect to $t$ is $0$ .

$h'' (t) = - 32 + 0$

Add $- 32$ and $0$ .

$h'' (t) = - 32$

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

$- 32 t + 60 = 0$

Find the first derivative.

By the Sum Rule, the derivative of $60 t - 16 t^{2}$ with respect to $t$ is $\frac{d}{d t} [60 t] + \frac{d}{d t} [- 16 t^{2}]$ .

$f' (t) = \frac{d}{d t} (60 t) + \frac{d}{d t} (- 16 t^{2})$

Evaluate $\frac{d}{d t} [60 t]$ .

Since $60$ is constant with respect to $t$ , the derivative of $60 t$ with respect to $t$ is $60 \frac{d}{d t} [t]$ .

$f' (t) = 60 \frac{d}{d t} (t) + \frac{d}{d t} (- 16 t^{2})$

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

$f' (t) = 60 \cdot 1 + \frac{d}{d t} (- 16 t^{2})$

Multiply $60$ by $1$ .

$f' (t) = 60 + \frac{d}{d t} (- 16 t^{2})$

Evaluate $\frac{d}{d t} [- 16 t^{2}]$ .

Since $- 16$ is constant with respect to $t$ , the derivative of $- 16 t^{2}$ with respect to $t$ is $- 16 \frac{d}{d t} [t^{2}]$ .

$f' (t) = 60 - 16 \frac{d}{d t} t^{2}$

Differentiate using the Power Rule which states that $\frac{d}{d t} [t^{n}]$ is $n t^{n - 1}$ where $n = 2$ .

$f' (t) = 60 - 16 (2 t)$

Multiply $2$ by $- 16$ .

$f' (t) = 60 - 32 t$

Reorder terms.

$f' (t) = - 32 t + 60$

The first derivative of $h (t)$ with respect to $t$ is $- 32 t + 60$ .

$- 32 t + 60$

Set the first derivative equal to $0$ then solve the equation $- 32 t + 60 = 0$ .

Set the first derivative equal to $0$ .

$- 32 t + 60 = 0$

Subtract $60$ from both sides of the equation.

$- 32 t = - 60$

Divide each term in $- 32 t = - 60$ by $- 32$ and simplify.

Divide each term in $- 32 t = - 60$ by $- 32$ .

$\frac{- 32 t}{- 32} = \frac{- 60}{- 32}$

Simplify the left side.

Cancel the common factor of $- 32$ .

Cancel the common factor.

$\frac{- 32 t}{- 32} = \frac{- 60}{- 32}$

Divide $t$ by $1$ .

$t = \frac{- 60}{- 32}$

Simplify the right side.

Cancel the common factor of $- 60$ and $- 32$ .

Factor $- 4$ out of $- 60$ .

$t = \frac{- 4 (15)}{- 32}$

Cancel the common factors.

Factor $- 4$ out of $- 32$ .

$t = \frac{- 4 \cdot 15}{- 4 \cdot 8}$

Cancel the common factor.

$t = \frac{- 4 \cdot 15}{- 4 \cdot 8}$

Rewrite the expression.

$t = \frac{15}{8}$

Find the values where the derivative is undefined.

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.

Critical points to evaluate.

$t = \frac{15}{8}$

Evaluate the second derivative at $t = \frac{15}{8}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 32$

$t = \frac{15}{8}$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$t = \frac{15}{8}$ is a local maximum

Find the y-value when $t = \frac{15}{8}$ .

Replace the variable $t$ with $\frac{15}{8}$ in the expression.

$h (\frac{15}{8}) = 60 (\frac{15}{8}) - 16 {(\frac{15}{8})}^{2}$

Simplify the result.

Simplify each term.

Cancel the common factor of $4$ .

Factor $4$ out of $60$ .

$h (\frac{15}{8}) = 4 (15) (\frac{15}{8}) - 16 {(\frac{15}{8})}^{2}$

Factor $4$ out of $8$ .

$h (\frac{15}{8}) = 4 \cdot (15 (\frac{15}{4 \cdot 2})) - 16 {(\frac{15}{8})}^{2}$

Cancel the common factor.

$h (\frac{15}{8}) = 4 \cdot (15 (\frac{15}{4 \cdot 2})) - 16 {(\frac{15}{8})}^{2}$

Rewrite the expression.

$h (\frac{15}{8}) = 15 (\frac{15}{2}) - 16 {(\frac{15}{8})}^{2}$

Combine $15$ and $\frac{15}{2}$ .

$h (\frac{15}{8}) = \frac{15 \cdot 15}{2} - 16 {(\frac{15}{8})}^{2}$

Multiply $15$ by $15$ .

$h (\frac{15}{8}) = \frac{225}{2} - 16 {(\frac{15}{8})}^{2}$

Apply the product rule to $\frac{15}{8}$ .

$h (\frac{15}{8}) = \frac{225}{2} - 16 \frac{15^{2}}{8^{2}}$

Raise $15$ to the power of $2$ .

$h (\frac{15}{8}) = \frac{225}{2} - 16 \frac{225}{8^{2}}$

Raise $8$ to the power of $2$ .

$h (\frac{15}{8}) = \frac{225}{2} - 16 (\frac{225}{64})$

Cancel the common factor of $16$ .

Factor $16$ out of $- 16$ .

$h (\frac{15}{8}) = \frac{225}{2} + 16 (- 1) (\frac{225}{64})$

Factor $16$ out of $64$ .

$h (\frac{15}{8}) = \frac{225}{2} + 16 \cdot (- 1 \frac{225}{16 \cdot 4})$

Cancel the common factor.

$h (\frac{15}{8}) = \frac{225}{2} + 16 \cdot (- 1 \frac{225}{16 \cdot 4})$

Rewrite the expression.

$h (\frac{15}{8}) = \frac{225}{2} - 1 (\frac{225}{4})$

Rewrite $- 1 (\frac{225}{4})$ as $- (\frac{225}{4})$ .

$h (\frac{15}{8}) = \frac{225}{2} - \frac{225}{4}$

To write $\frac{225}{2}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$h (\frac{15}{8}) = \frac{225}{2} \cdot \frac{2}{2} - \frac{225}{4}$

Write each expression with a common denominator of $4$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{225}{2}$ by $\frac{2}{2}$ .

$h (\frac{15}{8}) = \frac{225 \cdot 2}{2 \cdot 2} - \frac{225}{4}$

Multiply $2$ by $2$ .

$h (\frac{15}{8}) = \frac{225 \cdot 2}{4} - \frac{225}{4}$

Combine the numerators over the common denominator.

$h (\frac{15}{8}) = \frac{225 \cdot 2 - 225}{4}$

Simplify the numerator.

Multiply $225$ by $2$ .

$h (\frac{15}{8}) = \frac{450 - 225}{4}$

Subtract $225$ from $450$ .

$h (\frac{15}{8}) = \frac{225}{4}$

The final answer is $\frac{225}{4}$ .

$y = \frac{225}{4}$

These are the local extrema for $h (t) = 60 t - 16 t^{2}$ .

$(\frac{15}{8}, \frac{225}{4})$ is a local maxima

Find the Local Maxima and Minima h(t)=60t-16t^2

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