Find the Maximum/Minimum Value 0.5x^2+23x-216+2600/x

$0.5 x^{2} + 23 x - 216 + \frac{2600}{x}$

Step 1

Find the first derivative of the function.

By the Sum Rule, the derivative of $0.5 x^{2} + 23 x - 216 + \frac{2600}{x}$ with respect to $x$ is $\frac{d}{d x} [0.5 x^{2}] + \frac{d}{d x} [23 x] + \frac{d}{d x} [- 216] + \frac{d}{d x} [\frac{2600}{x}]$ .

$\frac{d}{d x} [0.5 x^{2}] + \frac{d}{d x} [23 x] + \frac{d}{d x} [- 216] + \frac{d}{d x} [\frac{2600}{x}]$

Evaluate $\frac{d}{d x} [0.5 x^{2}]$ .

Since $0.5$ is constant with respect to $x$ , the derivative of $0.5 x^{2}$ with respect to $x$ is $0.5 \frac{d}{d x} [x^{2}]$ .

$0.5 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [23 x] + \frac{d}{d x} [- 216] + \frac{d}{d x} [\frac{2600}{x}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$0.5 (2 x) + \frac{d}{d x} [23 x] + \frac{d}{d x} [- 216] + \frac{d}{d x} [\frac{2600}{x}]$

Multiply $2$ by $0.5$ .

$1 x + \frac{d}{d x} [23 x] + \frac{d}{d x} [- 216] + \frac{d}{d x} [\frac{2600}{x}]$

Multiply $x$ by $1$ .

$x + \frac{d}{d x} [23 x] + \frac{d}{d x} [- 216] + \frac{d}{d x} [\frac{2600}{x}]$

Evaluate $\frac{d}{d x} [23 x]$ .

Since $23$ is constant with respect to $x$ , the derivative of $23 x$ with respect to $x$ is $23 \frac{d}{d x} [x]$ .

$x + 23 \frac{d}{d x} [x] + \frac{d}{d x} [- 216] + \frac{d}{d x} [\frac{2600}{x}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$x + 23 \cdot 1 + \frac{d}{d x} [- 216] + \frac{d}{d x} [\frac{2600}{x}]$

Multiply $23$ by $1$ .

$x + 23 + \frac{d}{d x} [- 216] + \frac{d}{d x} [\frac{2600}{x}]$

Since $- 216$ is constant with respect to $x$ , the derivative of $- 216$ with respect to $x$ is $0$ .

$x + 23 + 0 + \frac{d}{d x} [\frac{2600}{x}]$

Evaluate $\frac{d}{d x} [\frac{2600}{x}]$ .

Since $2600$ is constant with respect to $x$ , the derivative of $\frac{2600}{x}$ with respect to $x$ is $2600 \frac{d}{d x} [\frac{1}{x}]$ .

$x + 23 + 0 + 2600 \frac{d}{d x} [\frac{1}{x}]$

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$x + 23 + 0 + 2600 \frac{d}{d x} [x^{- 1}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 1$ .

$x + 23 + 0 + 2600 (- x^{- 2})$

Multiply $- 1$ by $2600$ .

$x + 23 + 0 - 2600 x^{- 2}$

Simplify.

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x + 23 + 0 - 2600 \frac{1}{x^{2}}$

Combine terms.

Add $x + 23$ and $0$ .

$x + 23 - 2600 \frac{1}{x^{2}}$

Combine $- 2600$ and $\frac{1}{x^{2}}$ .

$x + 23 + \frac{- 2600}{x^{2}}$

Move the negative in front of the fraction.

$x + 23 - \frac{2600}{x^{2}}$

Reorder terms.

$x - \frac{2600}{x^{2}} + 23$

Step 2

Find the second derivative of the function.

Differentiate.

By the Sum Rule, the derivative of $x - \frac{2600}{x^{2}} + 23$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- \frac{2600}{x^{2}}] + \frac{d}{d x} [23]$ .

$f'' (x) = \frac{d}{d x} (x) + \frac{d}{d x} (- \frac{2600}{x^{2}}) + \frac{d}{d x} (23)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = 1 + \frac{d}{d x} (- \frac{2600}{x^{2}}) + \frac{d}{d x} (23)$

Evaluate $\frac{d}{d x} [- \frac{2600}{x^{2}}]$ .

Since $- 2600$ is constant with respect to $x$ , the derivative of $- \frac{2600}{x^{2}}$ with respect to $x$ is $- 2600 \frac{d}{d x} [\frac{1}{x^{2}}]$ .

$f'' (x) = 1 - 2600 \frac{d}{d x} \frac{1}{x^{2}} + \frac{d}{d x} (23)$

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

$f'' (x) = 1 - 2600 \frac{d}{d x} {(x^{2})}^{- 1} + \frac{d}{d x} (23)$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2}$ .

To apply the Chain Rule, set $u$ as $x^{2}$ .

$f'' (x) = 1 - 2600 (\frac{d}{d u} (u^{- 1}) \frac{d}{d x} (x^{2})) + \frac{d}{d x} (23)$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = - 1$ .

$f'' (x) = 1 - 2600 (- u^{- 2} \frac{d}{d x} x^{2}) + \frac{d}{d x} (23)$

Replace all occurrences of $u$ with $x^{2}$ .

$f'' (x) = 1 - 2600 (- {(x^{2})}^{- 2} \frac{d}{d x} x^{2}) + \frac{d}{d x} (23)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = 1 - 2600 (- {(x^{2})}^{- 2} (2 x)) + \frac{d}{d x} (23)$

Multiply the exponents in ${(x^{2})}^{- 2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f'' (x) = 1 - 2600 (- x^{2 \cdot - 2} (2 x)) + \frac{d}{d x} (23)$

Multiply $2$ by $- 2$ .

$f'' (x) = 1 - 2600 (- x^{- 4} (2 x)) + \frac{d}{d x} (23)$

Multiply $2$ by $- 1$ .

$f'' (x) = 1 - 2600 (- 2 x^{- 4} x) + \frac{d}{d x} (23)$

Raise $x$ to the power of $1$ .

$f'' (x) = 1 - 2600 (- 2 (x \cdot x^{- 4})) + \frac{d}{d x} (23)$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 1 - 2600 (- 2 x^{1 - 4}) + \frac{d}{d x} (23)$

Subtract $4$ from $1$ .

$f'' (x) = 1 - 2600 (- 2 x^{- 3}) + \frac{d}{d x} (23)$

Multiply $- 2$ by $- 2600$ .

$f'' (x) = 1 + 5200 x^{- 3} + \frac{d}{d x} (23)$

Since $23$ is constant with respect to $x$ , the derivative of $23$ with respect to $x$ is $0$ .

$f'' (x) = 1 + 5200 x^{- 3} + 0$

Simplify.

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f'' (x) = 1 + 5200 (\frac{1}{x^{3}}) + 0$

Combine terms.

Combine $5200$ and $\frac{1}{x^{3}}$ .

$f'' (x) = 1 + \frac{5200}{x^{3}} + 0$

Add $1 + \frac{5200}{x^{3}}$ and $0$ .

$f'' (x) = 1 + \frac{5200}{x^{3}}$

Reorder terms.

$f'' (x) = \frac{5200}{x^{3}} + 1$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$x - \frac{2600}{x^{2}} + 23 = 0$

Step 4

Find the first derivative.

$f' (x) = \frac{d}{d x} (0.5 x^{2}) + \frac{d}{d x} (23 x) + \frac{d}{d x} (- 216) + \frac{d}{d x} (\frac{2600}{x})$

Evaluate $\frac{d}{d x} [0.5 x^{2}]$ .

Since $0.5$ is constant with respect to $x$ , the derivative of $0.5 x^{2}$ with respect to $x$ is $0.5 \frac{d}{d x} [x^{2}]$ .

$f' (x) = 0.5 \frac{d}{d x} (x^{2}) + \frac{d}{d x} (23 x) + \frac{d}{d x} (- 216) + \frac{d}{d x} (\frac{2600}{x})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f' (x) = 0.5 (2 x) + \frac{d}{d x} (23 x) + \frac{d}{d x} (- 216) + \frac{d}{d x} (\frac{2600}{x})$

Multiply $2$ by $0.5$ .

$f' (x) = 1 x + \frac{d}{d x} (23 x) + \frac{d}{d x} (- 216) + \frac{d}{d x} (\frac{2600}{x})$

Multiply $x$ by $1$ .

$f' (x) = x + \frac{d}{d x} (23 x) + \frac{d}{d x} (- 216) + \frac{d}{d x} (\frac{2600}{x})$

Evaluate $\frac{d}{d x} [23 x]$ .

Since $23$ is constant with respect to $x$ , the derivative of $23 x$ with respect to $x$ is $23 \frac{d}{d x} [x]$ .

$f' (x) = x + 23 \frac{d}{d x} (x) + \frac{d}{d x} (- 216) + \frac{d}{d x} (\frac{2600}{x})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f' (x) = x + 23 \cdot 1 + \frac{d}{d x} (- 216) + \frac{d}{d x} (\frac{2600}{x})$

Multiply $23$ by $1$ .

$f' (x) = x + 23 + \frac{d}{d x} (- 216) + \frac{d}{d x} (\frac{2600}{x})$

Since $- 216$ is constant with respect to $x$ , the derivative of $- 216$ with respect to $x$ is $0$ .

$f' (x) = x + 23 + 0 + \frac{d}{d x} (\frac{2600}{x})$

Evaluate $\frac{d}{d x} [\frac{2600}{x}]$ .

Since $2600$ is constant with respect to $x$ , the derivative of $\frac{2600}{x}$ with respect to $x$ is $2600 \frac{d}{d x} [\frac{1}{x}]$ .

$f' (x) = x + 23 + 0 + 2600 \frac{d}{d x} (\frac{1}{x})$

Rewrite $\frac{1}{x}$ as $x^{- 1}$ .

$f' (x) = x + 23 + 0 + 2600 \frac{d}{d x} (x^{- 1})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = - 1$ .

$f' (x) = x + 23 + 0 + 2600 (- x^{- 2})$

Multiply $- 1$ by $2600$ .

$f' (x) = x + 23 + 0 - 2600 x^{- 2}$

Simplify.

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f' (x) = x + 23 + 0 - 2600 \frac{1}{x^{2}}$

Combine terms.

Add $x + 23$ and $0$ .

$f' (x) = x + 23 - 2600 \frac{1}{x^{2}}$

Combine $- 2600$ and $\frac{1}{x^{2}}$ .

$f' (x) = x + 23 + \frac{- 2600}{x^{2}}$

Move the negative in front of the fraction.

$f' (x) = x + 23 - \frac{2600}{x^{2}}$

Reorder terms.

$f' (x) = x - \frac{2600}{x^{2}} + 23$

The first derivative of $f (x)$ with respect to $x$ is $x - \frac{2600}{x^{2}} + 23$ .

$x - \frac{2600}{x^{2}} + 23$

Step 5

Set the first derivative equal to $0$ then solve the equation $x - \frac{2600}{x^{2}} + 23 = 0$ .

Set the first derivative equal to $0$ .

$x - \frac{2600}{x^{2}} + 23 = 0$

Graph each side of the equation. The solution is the x-value of the point of intersection.

$x \approx 9.01216529$

Step 6

Find the values where the derivative is undefined.

Set the denominator in $\frac{2600}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Solve for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Step 7

Critical points to evaluate.

$x = 9.01216529$

Step 8

Evaluate the second derivative at $x = 9.01216529$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$\frac{5200}{{(9.01216529)}^{3}} + 1$

Step 9

Evaluate the second derivative.

Simplify each term.

Raise $9.01216529$ to the power of $3$ .

$\frac{5200}{731.96016423} + 1$

Divide $5200$ by $731.96016423$ .

$7.10421175 + 1$

Add $7.10421175$ and $1$ .

$8.10421175$

Step 10

$x = 9.01216529$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 9.01216529$ is a local minimum

Step 11

Find the y-value when $x = 9.01216529$ .

Replace the variable $x$ with $9.01216529$ in the expression.

$f (9.01216529) = 0.5 {(9.01216529)}^{2} + 23 (9.01216529) - 216 + \frac{2600}{9.01216529}$

Simplify the result.

Simplify each term.

Raise $9.01216529$ to the power of $2$ .

$f (9.01216529) = 0.5 \cdot 81.21912329 + 23 (9.01216529) - 216 + \frac{2600}{9.01216529}$

Multiply $0.5$ by $81.21912329$ .

$f (9.01216529) = 40.60956164 + 23 (9.01216529) - 216 + \frac{2600}{9.01216529}$

Multiply $23$ by $9.01216529$ .

$f (9.01216529) = 40.60956164 + 207.27980177 - 216 + \frac{2600}{9.01216529}$

Divide $2600$ by $9.01216529$ .

$f (9.01216529) = 40.60956164 + 207.27980177 - 216 + 288.49892506$

Simplify by adding and subtracting.

Add $40.60956164$ and $207.27980177$ .

$f (9.01216529) = 247.88936342 - 216 + 288.49892506$

Subtract $216$ from $247.88936342$ .

$f (9.01216529) = 31.88936342 + 288.49892506$

Add $31.88936342$ and $288.49892506$ .

$f (9.01216529) = 320.38828848$

The final answer is $320.38828848$ .

$y = 320.38828848$

Step 12

These are the local extrema for $f (x) = 0.5 x^{2} + 23 x - 216 + \frac{2600}{x}$ .

$(9.01216529, 320.38828848)$ is a local minima

Step 13

Find the Maximum/Minimum Value 0.5x^2+23x-216+2600/x

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