Find the Maximum/Minimum Value f(x)=232x-5500-x^2

$f (x) = 232 x - 5500 - x^{2}$

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 $b$ .

$x = - \frac{232}{2 (- 1)}$

Remove parentheses.

$x = - \frac{232}{2 (- 1)}$

Simplify $- \frac{232}{2 (- 1)}$ .

Cancel the common factor of $232$ and $2$ .

Factor $2$ out of $232$ .

$x = - \frac{2 \cdot 116}{2 \cdot - 1}$

Move the negative one from the denominator of $\frac{116}{- 1}$ .

$x = - (- 1 \cdot 116)$

Multiply $- (- 1 \cdot 116)$ .

Multiply $- 1$ by $116$ .

$x = - - 116$

Multiply $- 1$ by $- 116$ .

$x = 116$

Step 3

Evaluate $f (116)$ .

Replace the variable $x$ with $116$ in the expression.

$f (116) = 232 (116) - 5500 - {(116)}^{2}$

Simplify the result.

Simplify each term.

Multiply $232$ by $116$ .

$f (116) = 26912 - 5500 - {(116)}^{2}$

Raise $116$ to the power of $2$ .

$f (116) = 26912 - 5500 - 1 \cdot 13456$

Multiply $- 1$ by $13456$ .

$f (116) = 26912 - 5500 - 13456$

Simplify by subtracting numbers.

Subtract $5500$ from $26912$ .

$f (116) = 21412 - 13456$

Subtract $13456$ from $21412$ .

$f (116) = 7956$

The final answer is $7956$ .

$7956$

Step 4

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

$(116, 7956)$

Step 5

Find the Maximum/Minimum Value f(x)=232x-5500-x^2

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