Find the Maximum/Minimum Value h(T)=112t-16t^2

$h (T) = 112 t - 16 t^{2}$

Step 1

Find the first derivative of the function.

By the Sum Rule, the derivative of $112 t - 16 t^{2}$ with respect to $T$ is $\frac{d}{d T} [112 t] + \frac{d}{d T} [- 16 t^{2}]$ .

$\frac{d}{d T} [112 t] + \frac{d}{d T} [- 16 t^{2}]$

Since $112 t$ is constant with respect to $T$ , the derivative of $112 t$ with respect to $T$ is $0$ .

$0 + \frac{d}{d T} [- 16 t^{2}]$

Since $- 16 t^{2}$ is constant with respect to $T$ , the derivative of $- 16 t^{2}$ with respect to $T$ is $0$ .

$0 + 0$

Add $0$ and $0$ .

$0$

Step 2

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

$h'' (T) = 0$

Step 3

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

$0 = 0$

Step 4

Find the first derivative.

By the Sum Rule, the derivative of $112 t - 16 t^{2}$ with respect to $t$ is $\frac{d}{d t} [112 t] + \frac{d}{d t} [- 16 t^{2}]$ .

$f' (t) = \frac{d}{d t} (112 t) + \frac{d}{d t} (- 16 t^{2})$

Evaluate $\frac{d}{d t} [112 t]$ .

Since $112$ is constant with respect to $t$ , the derivative of $112 t$ with respect to $t$ is $112 \frac{d}{d t} [t]$ .

$f' (t) = 112 \frac{d}{d t} (t) + \frac{d}{d t} (- 16 t^{2})$

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

$f' (t) = 112 \cdot 1 + \frac{d}{d t} (- 16 t^{2})$

Multiply $112$ by $1$ .

$f' (t) = 112 + \frac{d}{d t} (- 16 t^{2})$

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

$f' (t) = 112 - 16 \frac{d}{d t} t^{2}$

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

$f' (t) = 112 - 16 (2 t)$

Multiply $2$ by $- 16$ .

$f' (t) = 112 - 32 t$

Reorder terms.

$f' (t) = - 32 t + 112$

The first derivative of $h (T)$ with respect to $T$ is $- 32 t + 112$ .

$- 32 t + 112$

Step 5

Set the first derivative equal to $0$ then solve the equation $- 32 t + 112 = 0$ .

Set the first derivative equal to $0$ .

$- 32 t + 112 = 0$

Subtract $112$ from both sides of the equation.

$- 32 t = - 112$

Divide each term in $- 32 t = - 112$ by $- 32$ and simplify.

Divide each term in $- 32 t = - 112$ by $- 32$ .

$\frac{- 32 t}{- 32} = \frac{- 112}{- 32}$

Simplify the left side.

Cancel the common factor of $- 32$ .

Cancel the common factor.

$\frac{- 32 t}{- 32} = \frac{- 112}{- 32}$

Divide $t$ by $1$ .

$t = \frac{- 112}{- 32}$

Simplify the right side.

Cancel the common factor of $- 112$ and $- 32$ .

Factor $- 16$ out of $- 112$ .

$t = \frac{- 16 (7)}{- 32}$

Cancel the common factors.

Factor $- 16$ out of $- 32$ .

$t = \frac{- 16 \cdot 7}{- 16 \cdot 2}$

Cancel the common factor.

$t = \frac{- 16 \cdot 7}{- 16 \cdot 2}$

Rewrite the expression.

$t = \frac{7}{2}$

Step 6

Critical points to evaluate.

$T = \frac{7}{2}$

Step 7

Evaluate the second derivative at $T = \frac{7}{2}$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$0$

Step 8

Since there is at least one point with $0$ or undefined second derivative, apply the first derivative test.

Split $(- \infty, \infty)$ into separate intervals around the $T$ values that make the first derivative $0$ or undefined.

$(- \infty, \frac{7}{2}) \cup (\frac{7}{2}, \infty)$

Substitute any number, such as $0$ , from the interval $(- \infty, \frac{7}{2})$ in the first derivative $0$ to check if the result is negative or positive.

Replace the variable $T$ with $0$ in the expression.

$h' (0) = 0$

The final answer is $0$ .

$0$

Substitute any number, such as $6$ , from the interval $(\frac{7}{2}, \infty)$ in the first derivative $0$ to check if the result is negative or positive.

Replace the variable $T$ with $6$ in the expression.

$h' (6) = 0$

The final answer is $0$ .

$0$

Since the first derivative did not change signs around $T = \frac{7}{2}$ , this is not a local maximum or minimum.

Not a local maximum or minimum

No local maxima or minima found for $h (T) = 112 t - 16 t^{2}$ .

No local maxima or minima

Step 9

Find the Maximum/Minimum Value h(T)=112t-16t^2

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