Math / Calculus

Question[-/1.38 Points] DETAILS MY NOTES LARAPCALC10 5.1.
Find the profit function for the given marginal profit and initial condition. \begin{tabular}{|c|c|} \hline Marginal Profit & Initial Condition \\ \hline $\frac{d P}{d x}=-30 x+270$ & $P(5)=\$ 630$ \\ \hline \end{tabular} $P(x)=$ $\square$ Need Help? Read it Watch it Master It

Studdy Solution

STEP 1

What is this asking? We need to find a profit formula, $P(x)$ , and we're given how quickly the profit changes ( $\frac{dP}{dx}$ ) and a known profit at a specific point in time. Watch out! Don't forget to use the initial condition to find the constant of integration!
This is super important to get the right formula.

STEP 2

1. Integrate the Marginal Profit
2. Determine the Constant of Integration
3. State the Profit Function

STEP 3

Alright, so we're given the marginal profit, which is how much the profit changes as $x$ changes.
It's like the speed of profit growth!
Mathematically, it's the derivative of the profit function, $\frac{dP}{dx} = -30x + 270$ .
To find the actual profit function, $P(x)$ , we need to integrate the marginal profit!

STEP 4

Let's do this!
The integral of $-30x + 270$ with respect to $x$ is: $\int (-30x + 270) \, dx$

STEP 5

We can break this down: $\int (-30x) \, dx + \int 270 \, dx$ $-30 \int x \, dx + 270 \int 1 \, dx$ $-30 \cdot \frac{x^2}{2} + 270x + C$ $-15x^2 + 270x + C$ Remember that $C$ is our constant of integration – it's super important!

STEP 6

We know that $P(5) = \$630$ .
This means when $x = \textbf{5}$ , the profit $P(x)$ is $\textbf{\$630}$ .
Let's plug these values into our integrated formula: $630 = -15(\textbf{5})^2 + 270(\textbf{5}) + C$

STEP 7

Now, let's solve for $C$ : $630 = -15(25) + 1350 + C$ $630 = -375 + 1350 + C$ $630 = 975 + C$ $C = 630 - 975$ $C = \textbf{-345}$ Great! We found our constant of integration!

STEP 8

Now we just substitute the value of $C$ back into our integrated formula.
This gives us the profit function: $P(x) = -15x^2 + 270x - 345$

STEP 9

The profit function is $P(x) = -15x^2 + 270x - 345$ .

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Studdy Solution

STEP 1

STEP 2

1. Integrate the Marginal Profit
2. Determine the Constant of Integration
3. State the Profit Function

STEP 3

STEP 4

Let's do this!
The integral of $-30x + 270$ with respect to $x$ is: $\int (-30x + 270) \, dx$

STEP 5

We can break this down: $\int (-30x) \, dx + \int 270 \, dx$ $-30 \int x \, dx + 270 \int 1 \, dx$ $-30 \cdot \frac{x^2}{2} + 270x + C$ $-15x^2 + 270x + C$ Remember that $C$ is our constant of integration – it's super important!

STEP 6

We know that $P(5) = \$630$ .
This means when $x = \textbf{5}$ , the profit $P(x)$ is $\textbf{\$630}$ .
Let's plug these values into our integrated formula: $630 = -15(\textbf{5})^2 + 270(\textbf{5}) + C$

STEP 7

Now, let's solve for $C$ : $630 = -15(25) + 1350 + C$ $630 = -375 + 1350 + C$ $630 = 975 + C$ $C = 630 - 975$ $C = \textbf{-345}$ Great! We found our constant of integration!

STEP 8

Now we just substitute the value of $C$ back into our integrated formula.
This gives us the profit function: $P(x) = -15x^2 + 270x - 345$

STEP 9

The profit function is $P(x) = -15x^2 + 270x - 345$ .

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Studdy Solution

STEP 1

STEP 2

1. Integrate the Marginal Profit2. Determine the Constant of Integration3. State the Profit Function

STEP 3

STEP 4

Let's do this!The integral of −30x+270-30x + 270−30x+270 with respect to xxx is: ∫(−30x+270) dx \int (-30x + 270) \, dx ∫(−30x+270)dx

STEP 5

STEP 6

We know that P(5)=$630P(5) = \$630P(5)=$630.This means when x=5x = \textbf{5}x=5, the profit P(x)P(x)P(x) is $630\textbf{\$630}$630.Let's plug these values into our integrated formula: 630=−15(5)2+270(5)+C 630 = -15(\textbf{5})^2 + 270(\textbf{5}) + C 630=−15(5)2+270(5)+C

STEP 7

Now, let's solve for CCC: 630=−15(25)+1350+C 630 = -15(25) + 1350 + C 630=−15(25)+1350+C 630=−375+1350+C 630 = -375 + 1350 + C 630=−375+1350+C630=975+C 630 = 975 + C 630=975+CC=630−975 C = 630 - 975 C=630−975C=-345 C = \textbf{-345} C=-345Great! We found our constant of integration!

STEP 8

Now we just substitute the value of CCC back into our integrated formula.This gives us the *profit function*: P(x)=−15x2+270x−345 P(x) = -15x^2 + 270x - 345 P(x)=−15x2+270x−345

STEP 9

The profit function is P(x)=−15x2+270x−345P(x) = -15x^2 + 270x - 345P(x)=−15x2+270x−345.

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1. Integrate the Marginal Profit
2. Determine the Constant of Integration
3. State the Profit Function

Let's do this!
The integral of $-30x + 270$ with respect to $x$ is: $\int (-30x + 270) \, dx$

We know that $P(5) = \$630$ .
This means when $x = \textbf{5}$ , the profit $P(x)$ is $\textbf{\$630}$ .
Let's plug these values into our integrated formula: $630 = -15(\textbf{5})^2 + 270(\textbf{5}) + C$

Now, let's solve for $C$ : $630 = -15(25) + 1350 + C$ $630 = -375 + 1350 + C$ $630 = 975 + C$ $C = 630 - 975$ $C = \textbf{-345}$ Great! We found our constant of integration!

Now we just substitute the value of $C$ back into our integrated formula.
This gives us the profit function: $P(x) = -15x^2 + 270x - 345$

The profit function is $P(x) = -15x^2 + 270x - 345$ .