Find the Maximum/Minimum Value -5x^2+100x+4000

$- 5 x^{2} + 100 x + 4000$

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 $b$ .

$x = - \frac{100}{2 (- 5)}$

Remove parentheses.

$x = - \frac{100}{2 (- 5)}$

Simplify $- \frac{100}{2 (- 5)}$ .

Cancel the common factor of $100$ and $2$ .

Factor $2$ out of $100$ .

$x = - \frac{2 \cdot 50}{2 \cdot - 5}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 5$ .

$x = - \frac{2 \cdot 50}{2 (- 5)}$

Cancel the common factor.

$x = - \frac{2 \cdot 50}{2 \cdot - 5}$

Rewrite the expression.

$x = - \frac{50}{- 5}$

Cancel the common factor of $50$ and $- 5$ .

Factor $5$ out of $50$ .

$x = - \frac{5 (10)}{- 5}$

Move the negative one from the denominator of $\frac{10}{- 1}$ .

$x = - (- 1 \cdot 10)$

Multiply $- (- 1 \cdot 10)$ .

Multiply $- 1$ by $10$ .

$x = - - 10$

Multiply $- 1$ by $- 10$ .

$x = 10$

Step 3

Evaluate $f (10)$ .

Replace the variable $x$ with $10$ in the expression.

$f (10) = - 5 {(10)}^{2} + 100 (10) + 4000$

Simplify the result.

Simplify each term.

Raise $10$ to the power of $2$ .

$f (10) = - 5 \cdot 100 + 100 (10) + 4000$

Multiply $- 5$ by $100$ .

$f (10) = - 500 + 100 (10) + 4000$

Multiply $100$ by $10$ .

$f (10) = - 500 + 1000 + 4000$

Simplify by adding numbers.

Add $- 500$ and $1000$ .

$f (10) = 500 + 4000$

Add $500$ and $4000$ .

$f (10) = 4500$

The final answer is $4500$ .

$4500$

Step 4

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

$(10, 4500)$

Step 5

Find the Maximum/Minimum Value -5x^2+100x+4000

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