Find the Maximum/Minimum Value F(x)=-5x^2+280x-3600

$F (x) = - 5 x^{2} + 280 x - 3600$

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 $b$ .

$x = - \frac{280}{2 (- 5)}$

Remove parentheses.

$x = - \frac{280}{2 (- 5)}$

Simplify $- \frac{280}{2 (- 5)}$ .

Cancel the common factor of $280$ and $2$ .

Factor $2$ out of $280$ .

$x = - \frac{2 \cdot 140}{2 \cdot - 5}$

Cancel the common factors.

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

$x = - \frac{2 \cdot 140}{2 (- 5)}$

Cancel the common factor.

$x = - \frac{2 \cdot 140}{2 \cdot - 5}$

Rewrite the expression.

$x = - \frac{140}{- 5}$

Cancel the common factor of $140$ and $- 5$ .

Factor $5$ out of $140$ .

$x = - \frac{5 (28)}{- 5}$

Move the negative one from the denominator of $\frac{28}{- 1}$ .

$x = - (- 1 \cdot 28)$

Multiply $- (- 1 \cdot 28)$ .

Multiply $- 1$ by $28$ .

$x = - - 28$

Multiply $- 1$ by $- 28$ .

$x = 28$

Step 3

Evaluate $f (28)$ .

Replace the variable $x$ with $28$ in the expression.

$F (28) = - 5 {(28)}^{2} + 280 (28) - 3600$

Simplify the result.

Simplify each term.

Raise $28$ to the power of $2$ .

$F (28) = - 5 \cdot 784 + 280 (28) - 3600$

Multiply $- 5$ by $784$ .

$F (28) = - 3920 + 280 (28) - 3600$

Multiply $280$ by $28$ .

$F (28) = - 3920 + 7840 - 3600$

Simplify by adding and subtracting.

Add $- 3920$ and $7840$ .

$F (28) = 3920 - 3600$

Subtract $3600$ from $3920$ .

$F (28) = 320$

The final answer is $320$ .

$320$

Step 4

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

$(28, 320)$

Step 5

Find the Maximum/Minimum Value F(x)=-5x^2+280x-3600

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