Find the Maximum/Minimum Value p=40000+300x-10x^2

$p = 40000 + 300 x - 10 x^{2}$

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 $b$ .

$x = - \frac{300}{2 (- 10)}$

Remove parentheses.

$x = - \frac{300}{2 (- 10)}$

Simplify $- \frac{300}{2 (- 10)}$ .

Cancel the common factor of $300$ and $2$ .

Factor $2$ out of $300$ .

$x = - \frac{2 \cdot 150}{2 \cdot - 10}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 10$ .

$x = - \frac{2 \cdot 150}{2 (- 10)}$

Cancel the common factor.

$x = - \frac{2 \cdot 150}{2 \cdot - 10}$

Rewrite the expression.

$x = - \frac{150}{- 10}$

Cancel the common factor of $150$ and $- 10$ .

Factor $10$ out of $150$ .

$x = - \frac{10 (15)}{- 10}$

Move the negative one from the denominator of $\frac{15}{- 1}$ .

$x = - (- 1 \cdot 15)$

Multiply $- (- 1 \cdot 15)$ .

Multiply $- 1$ by $15$ .

$x = - - 15$

Multiply $- 1$ by $- 15$ .

$x = 15$

Step 3

Evaluate $f (15)$ .

Replace the variable $x$ with $15$ in the expression.

$f (15) = 40000 + 300 (15) - 10 {(15)}^{2}$

Simplify the result.

Simplify each term.

Multiply $300$ by $15$ .

$f (15) = 40000 + 4500 - 10 {(15)}^{2}$

Raise $15$ to the power of $2$ .

$f (15) = 40000 + 4500 - 10 \cdot 225$

Multiply $- 10$ by $225$ .

$f (15) = 40000 + 4500 - 2250$

Simplify by adding and subtracting.

Add $40000$ and $4500$ .

$f (15) = 44500 - 2250$

Subtract $2250$ from $44500$ .

$f (15) = 42250$

The final answer is $42250$ .

$42250$

Step 4

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

$(15, 42250)$

Step 5

Find the Maximum/Minimum Value p=40000+300x-10x^2

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