Find the Maximum/Minimum Value -15x^2+1200x+30000

$- 15 x^{2} + 1200 x + 30000$

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 $b$ .

$x = - \frac{1200}{2 (- 15)}$

Remove parentheses.

$x = - \frac{1200}{2 (- 15)}$

Simplify $- \frac{1200}{2 (- 15)}$ .

Cancel the common factor of $1200$ and $2$ .

Factor $2$ out of $1200$ .

$x = - \frac{2 \cdot 600}{2 \cdot - 15}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 15$ .

$x = - \frac{2 \cdot 600}{2 (- 15)}$

Cancel the common factor.

$x = - \frac{2 \cdot 600}{2 \cdot - 15}$

Rewrite the expression.

$x = - \frac{600}{- 15}$

Cancel the common factor of $600$ and $- 15$ .

Factor $15$ out of $600$ .

$x = - \frac{15 (40)}{- 15}$

Move the negative one from the denominator of $\frac{40}{- 1}$ .

$x = - (- 1 \cdot 40)$

Multiply $- (- 1 \cdot 40)$ .

Multiply $- 1$ by $40$ .

$x = - - 40$

Multiply $- 1$ by $- 40$ .

$x = 40$

Step 3

Evaluate $f (40)$ .

Replace the variable $x$ with $40$ in the expression.

$f (40) = - 15 {(40)}^{2} + 1200 (40) + 30000$

Simplify the result.

Simplify each term.

Raise $40$ to the power of $2$ .

$f (40) = - 15 \cdot 1600 + 1200 (40) + 30000$

Multiply $- 15$ by $1600$ .

$f (40) = - 24000 + 1200 (40) + 30000$

Multiply $1200$ by $40$ .

$f (40) = - 24000 + 48000 + 30000$

Simplify by adding numbers.

Add $- 24000$ and $48000$ .

$f (40) = 24000 + 30000$

Add $24000$ and $30000$ .

$f (40) = 54000$

The final answer is $54000$ .

$54000$

Step 4

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

$(40, 54000)$

Step 5

Find the Maximum/Minimum Value -15x^2+1200x+30000

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