Find the Maximum/Minimum Value h(t)=72t-16t^2

$h (t) = 72 t - 16 t^{2}$

Step 1

The maximum of a quadratic function occurs at $x = - \frac{b}{2 a}$ . If $a$ is negative, the maximum value of the function is $f (- \frac{b}{2 a})$ .

$f_{max}$ $t = a t^{2} + b t + c$ occurs at $t = - \frac{b}{2 a}$

Step 2

Find the value of $t = - \frac{b}{2 a}$ .

Substitute in the values of $a$ and $b$ .

$t = - \frac{72}{2 (- 16)}$

Remove parentheses.

$t = - \frac{72}{2 (- 16)}$

Simplify $- \frac{72}{2 (- 16)}$ .

Cancel the common factor of $72$ and $2$ .

Factor $2$ out of $72$ .

$t = - \frac{2 \cdot 36}{2 \cdot - 16}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 16$ .

$t = - \frac{2 \cdot 36}{2 (- 16)}$

Cancel the common factor.

$t = - \frac{2 \cdot 36}{2 \cdot - 16}$

Rewrite the expression.

$t = - \frac{36}{- 16}$

Cancel the common factor of $36$ and $- 16$ .

Factor $4$ out of $36$ .

$t = - \frac{4 (9)}{- 16}$

Cancel the common factors.

Factor $4$ out of $- 16$ .

$t = - \frac{4 \cdot 9}{4 \cdot - 4}$

Cancel the common factor.

$t = - \frac{4 \cdot 9}{4 \cdot - 4}$

Rewrite the expression.

$t = - \frac{9}{- 4}$

Move the negative in front of the fraction.

$t = - - \frac{9}{4}$

Multiply $- - \frac{9}{4}$ .

Multiply $- 1$ by $- 1$ .

$t = 1 (\frac{9}{4})$

Multiply $\frac{9}{4}$ by $1$ .

$t = \frac{9}{4}$

Step 3

Evaluate $f (\frac{9}{4})$ .

Replace the variable $t$ with $\frac{9}{4}$ in the expression.

$h (\frac{9}{4}) = 72 (\frac{9}{4}) - 16 {(\frac{9}{4})}^{2}$

Simplify the result.

Simplify each term.

Cancel the common factor of $4$ .

Factor $4$ out of $72$ .

$h (\frac{9}{4}) = 4 (18) (\frac{9}{4}) - 16 {(\frac{9}{4})}^{2}$

Cancel the common factor.

$h (\frac{9}{4}) = 4 \cdot (18 (\frac{9}{4})) - 16 {(\frac{9}{4})}^{2}$

Rewrite the expression.

$h (\frac{9}{4}) = 18 \cdot 9 - 16 {(\frac{9}{4})}^{2}$

Multiply $18$ by $9$ .

$h (\frac{9}{4}) = 162 - 16 {(\frac{9}{4})}^{2}$

Apply the product rule to $\frac{9}{4}$ .

$h (\frac{9}{4}) = 162 - 16 \frac{9^{2}}{4^{2}}$

Raise $9$ to the power of $2$ .

$h (\frac{9}{4}) = 162 - 16 \frac{81}{4^{2}}$

Raise $4$ to the power of $2$ .

$h (\frac{9}{4}) = 162 - 16 (\frac{81}{16})$

Cancel the common factor of $16$ .

Factor $16$ out of $- 16$ .

$h (\frac{9}{4}) = 162 + 16 (- 1) (\frac{81}{16})$

Cancel the common factor.

$h (\frac{9}{4}) = 162 + 16 \cdot (- 1 (\frac{81}{16}))$

Rewrite the expression.

$h (\frac{9}{4}) = 162 - 1 \cdot 81$

Multiply $- 1$ by $81$ .

$h (\frac{9}{4}) = 162 - 81$

Subtract $81$ from $162$ .

$h (\frac{9}{4}) = 81$

The final answer is $81$ .

$81$

Step 4

Use the $x$ and $y$ values to find where the maximum occurs.

$(\frac{9}{4}, 81)$

Step 5

Find the Maximum/Minimum Value h(t)=72t-16t^2

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