Find the Maximum/Minimum Value a=38x-x^2

$a = 38 x - x^{2}$

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 $f (- \frac{b}{2 a})$ .

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

Step 2

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

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

$x = - \frac{38}{2 (- 1)}$

Remove parentheses.

$x = - \frac{38}{2 (- 1)}$

Simplify $- \frac{38}{2 (- 1)}$ .

Cancel the common factor of $38$ and $2$ .

Factor $2$ out of $38$ .

$x = - \frac{2 \cdot 19}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{19}{- 1}$ .

$x = - (- 1 \cdot 19)$

Multiply $- (- 1 \cdot 19)$ .

Multiply $- 1$ by $19$ .

$x = - - 19$

Multiply $- 1$ by $- 19$ .

$x = 19$

Step 3

Evaluate $f (19)$ .

Replace the variable $x$ with $19$ in the expression.

$f (19) = 38 (19) - {(19)}^{2}$

Simplify the result.

Simplify each term.

Multiply $38$ by $19$ .

$f (19) = 722 - {(19)}^{2}$

Raise $19$ to the power of $2$ .

$f (19) = 722 - 1 \cdot 361$

Multiply $- 1$ by $361$ .

$f (19) = 722 - 361$

Subtract $361$ from $722$ .

$f (19) = 361$

The final answer is $361$ .

$361$

Step 4

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

$(19, 361)$

Step 5

Find the Maximum/Minimum Value a=38x-x^2

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