Calculus Examples

$u (y) = 100 x + 40 y - 5 x^{2} - 2 y^{2}$

Step 1

Find the first derivative of the function.

Tap for more steps...

Step 1.1

Differentiate.

Tap for more steps...

Step 1.1.1

By the Sum Rule, the derivative of $100 x + 40 y - 5 x^{2} - 2 y^{2}$ with respect to $y$ is $\frac{d}{d y} [100 x] + \frac{d}{d y} [40 y] + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$ .

$\frac{d}{d y} [100 x] + \frac{d}{d y} [40 y] + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$

Step 1.1.2

Since $100 x$ is constant with respect to $y$ , the derivative of $100 x$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [40 y] + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$

Step 1.2

Evaluate $\frac{d}{d y} [40 y]$ .

Tap for more steps...

Step 1.2.1

Since $40$ is constant with respect to $y$ , the derivative of $40 y$ with respect to $y$ is $40 \frac{d}{d y} [y]$ .

$0 + 40 \frac{d}{d y} [y] + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$

Step 1.2.2

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

$0 + 40 \cdot 1 + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$

Step 1.2.3

Multiply $40$ by $1$ .

$0 + 40 + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$

Step 1.3

Since $- 5 x^{2}$ is constant with respect to $y$ , the derivative of $- 5 x^{2}$ with respect to $y$ is $0$ .

$0 + 40 + 0 + \frac{d}{d y} [- 2 y^{2}]$

Step 1.4

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

Tap for more steps...

Step 1.4.1

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

$0 + 40 + 0 - 2 \frac{d}{d y} [y^{2}]$

Step 1.4.2

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

$0 + 40 + 0 - 2 (2 y)$

Step 1.4.3

Multiply $2$ by $- 2$ .

$0 + 40 + 0 - 4 y$

Step 1.5

Simplify.

Tap for more steps...

Step 1.5.1

Combine terms.

Tap for more steps...

Step 1.5.1.1

Add $0$ and $40$ .

$40 + 0 - 4 y$

Step 1.5.1.2

Add $40$ and $0$ .

$40 - 4 y$

Step 1.5.2

Reorder terms.

$- 4 y + 40$

Step 2

Find the second derivative of the function.

Tap for more steps...

Step 2.1

By the Sum Rule, the derivative of $- 4 y + 40$ with respect to $y$ is $\frac{d}{d y} [- 4 y] + \frac{d}{d y} [40]$ .

$u'' (y) = \frac{d}{d y} (- 4 y) + \frac{d}{d y} (40)$

Step 2.2

Evaluate $\frac{d}{d y} [- 4 y]$ .

Tap for more steps...

Step 2.2.1

Since $- 4$ is constant with respect to $y$ , the derivative of $- 4 y$ with respect to $y$ is $- 4 \frac{d}{d y} [y]$ .

$u'' (y) = - 4 \frac{d}{d y} y + \frac{d}{d y} (40)$

Step 2.2.2

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

$u'' (y) = - 4 \cdot 1 + \frac{d}{d y} (40)$

Step 2.2.3

Multiply $- 4$ by $1$ .

$u'' (y) = - 4 + \frac{d}{d y} (40)$

Step 2.3

Differentiate using the Constant Rule.

Tap for more steps...

Step 2.3.1

Since $40$ is constant with respect to $y$ , the derivative of $40$ with respect to $y$ is $0$ .

$u'' (y) = - 4 + 0$

Step 2.3.2

Add $- 4$ and $0$ .

$u'' (y) = - 4$

Step 3

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

$- 4 y + 40 = 0$

Step 4

Find the first derivative.

Tap for more steps...

Step 4.1

Find the first derivative.

Tap for more steps...

Step 4.1.1

Differentiate.

Tap for more steps...

Step 4.1.1.1

By the Sum Rule, the derivative of $100 x + 40 y - 5 x^{2} - 2 y^{2}$ with respect to $y$ is $\frac{d}{d y} [100 x] + \frac{d}{d y} [40 y] + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$ .

$\frac{d}{d y} [100 x] + \frac{d}{d y} [40 y] + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$

Step 4.1.1.2

Since $100 x$ is constant with respect to $y$ , the derivative of $100 x$ with respect to $y$ is $0$ .

$0 + \frac{d}{d y} [40 y] + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$

Step 4.1.2

Evaluate $\frac{d}{d y} [40 y]$ .

Tap for more steps...

Step 4.1.2.1

Since $40$ is constant with respect to $y$ , the derivative of $40 y$ with respect to $y$ is $40 \frac{d}{d y} [y]$ .

$0 + 40 \frac{d}{d y} [y] + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$

Step 4.1.2.2

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

$0 + 40 \cdot 1 + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$

Step 4.1.2.3

Multiply $40$ by $1$ .

$0 + 40 + \frac{d}{d y} [- 5 x^{2}] + \frac{d}{d y} [- 2 y^{2}]$

Step 4.1.3

Since $- 5 x^{2}$ is constant with respect to $y$ , the derivative of $- 5 x^{2}$ with respect to $y$ is $0$ .

$0 + 40 + 0 + \frac{d}{d y} [- 2 y^{2}]$

Step 4.1.4

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

Tap for more steps...

Step 4.1.4.1

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

$0 + 40 + 0 - 2 \frac{d}{d y} [y^{2}]$

Step 4.1.4.2

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

$0 + 40 + 0 - 2 (2 y)$

Step 4.1.4.3

Multiply $2$ by $- 2$ .

$0 + 40 + 0 - 4 y$

Step 4.1.5

Simplify.

Tap for more steps...

Step 4.1.5.1

Combine terms.

Tap for more steps...

Step 4.1.5.1.1

Add $0$ and $40$ .

$40 + 0 - 4 y$

Step 4.1.5.1.2

Add $40$ and $0$ .

$40 - 4 y$

Step 4.1.5.2

Reorder terms.

$f' (y) = - 4 y + 40$

Step 4.2

The first derivative of $u (y)$ with respect to $y$ is $- 4 y + 40$ .

$- 4 y + 40$

Step 5

Set the first derivative equal to $0$ then solve the equation $- 4 y + 40 = 0$ .

Tap for more steps...

Step 5.1

Set the first derivative equal to $0$ .

$- 4 y + 40 = 0$

Step 5.2

Subtract $40$ from both sides of the equation.

$- 4 y = - 40$

Step 5.3

Divide each term in $- 4 y = - 40$ by $- 4$ and simplify.

Tap for more steps...

Step 5.3.1

Divide each term in $- 4 y = - 40$ by $- 4$ .

$\frac{- 4 y}{- 4} = \frac{- 40}{- 4}$

Step 5.3.2

Simplify the left side.

Tap for more steps...

Step 5.3.2.1

Cancel the common factor of $- 4$ .

Tap for more steps...

Step 5.3.2.1.1

Cancel the common factor.

$\frac{- 4 y}{- 4} = \frac{- 40}{- 4}$

Step 5.3.2.1.2

Divide $y$ by $1$ .

$y = \frac{- 40}{- 4}$

Step 5.3.3

Simplify the right side.

Tap for more steps...

Step 5.3.3.1

Divide $- 40$ by $- 4$ .

$y = 10$

Step 6

Find the values where the derivative is undefined.

Tap for more steps...

Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$y = 10$

Step 8

Evaluate the second derivative at $y = 10$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$- 4$

Step 9

$y = 10$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$y = 10$ is a local maximum

Step 10

Find the y-value when $y = 10$ .

Tap for more steps...

Step 10.1

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

$u (10) = 100 x + 40 (10) - 5 x^{2} - 2 {(10)}^{2}$

Step 10.2

Simplify the result.

Tap for more steps...

Step 10.2.1

Simplify each term.

Tap for more steps...

Step 10.2.1.1

Multiply $40$ by $10$ .

$u (10) = 100 x + 400 - 5 x^{2} - 2 {(10)}^{2}$

Step 10.2.1.2

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

$u (10) = 100 x + 400 - 5 x^{2} - 2 \cdot 100$

Step 10.2.1.3

Multiply $- 2$ by $100$ .

$u (10) = 100 x + 400 - 5 x^{2} - 200$

Step 10.2.2

Subtract $200$ from $400$ .

$u (10) = 100 x - 5 x^{2} + 200$

Step 10.2.3

The final answer is $100 x - 5 x^{2} + 200$ .

$y = 100 x - 5 x^{2} + 200$

Step 11

These are the local extrema for $u (y) = 100 x + 40 y - 5 x^{2} - 2 y^{2}$ .

$(10, 100 x - 5 x^{2} + 200)$ is a local maxima

Step 12