The marginal profit in dollars on Brie cheese sold at a cheese store is given by P'(x)=x(60x^2+30x), where x is the amount of cheese sold in hundreds of pounds. The "profit" is -$50 when cheese is sold.

a. Find the profit function. P(x)=
b. Find the profit from selling 300 pounds of Brie cheese.
Please help, I do not know where to begin

P'(x) = 60x^3 + 30x^2

So,
P(x) = 15x^4 + 10x^3 + c

If you meant the profit is -50 when no cheese is sold, then P(0) = -50, and c=50. So,

P(x) = 15x^4 + 10x^3 - 50

Now plug in x=3 to find P when 300 lbs are sold.

To find the profit function, you need to integrate the marginal profit function P'(x) with respect to x.

a. Find the profit function, P(x):
P'(x) = x(60x^2 + 30x)

To integrate this function, you need to use the power rule for integration. The power rule states that integrating x^n with respect to x results in (1/(n+1)) * x^(n+1).

Applying the power rule to each term in P'(x), we get:
P(x) = ∫[x(60x^2 + 30x)]dx
= ∫[60x^3 + 30x^2]dx

Integrating each term individually:
∫[60x^3]dx = (60/4) * x^4 + C1 (where C1 is the constant of integration)
∫[30x^2]dx = (30/3) * x^3 + C2 (where C2 is the constant of integration)

Combining the two results:
P(x) = (60/4) * x^4 + (30/3) * x^3 + C
= 15x^4 + 10x^3 + C (where C is the constant of integration)

To find the value of C, we use the given information that the profit is -$50 when cheese is sold. This means that when x = 0 (representing no cheese sold), P(x) should equal -50.

Setting P(x) = -50:
-50 = 15(0)^4 + 10(0)^3 + C
-50 = 0 + 0 + C
C = -50

Therefore, the profit function is:
P(x) = 15x^4 + 10x^3 - 50

b. Find the profit from selling 300 pounds of Brie cheese:
To find the profit from selling 300 pounds of cheese, substitute x = 3 (since x represents hundreds of pounds) into the profit function:

P(3) = 15(3)^4 + 10(3)^3 - 50
= 15(81) + 10(27) - 50
= 1215 + 270 - 50
= 1435

Therefore, the profit from selling 300 pounds of Brie cheese is $1435.

To find the profit function, we need to integrate the given marginal profit function with respect to x.

a. Finding the profit function:
Given: P'(x) = x(60x^2 + 30x)

Integrating both sides of the equation, we get:
∫ P'(x) dx = ∫ (x(60x^2 + 30x)) dx

Using the power rule of integration and distributing, we have:
P(x) = ∫ (60x^3 + 30x^2) dx

Integrating each term separately, we get:
P(x) = 60 * ∫ (x^3) dx + 30 * ∫ (x^2) dx

Using the power rule of integration again, we have:
P(x) = 60 * (x^4/4) + 30 * (x^3/3) + C

Where C is the constant of integration. Since we know that the profit is -$50 when cheese is sold, we can substitute this value into the equation to find the value of C.

P(-1) = 60 * ((-1)^4/4) + 30 * ((-1)^3/3) + C = -50

Simplifying, we get:
15 + 10 + C = -50
25 + C = -50
C = -75

Therefore, the profit function is given by:
P(x) = 60 * (x^4/4) + 30 * (x^3/3) - 75

b. Finding the profit from selling 300 pounds of Brie cheese:
To find the profit from selling 300 pounds of Brie cheese, we substitute x = 3 (since x represents the amount in hundreds of pounds) into the profit function.

P(3) = 60 * (3^4/4) + 30 * (3^3/3) - 75
P(3) = 60 * (81/4) + 30 * (27/3) - 75
P(3) = 60 * 20.25 + 30 * 9 - 75
P(3) = 1215 + 270 - 75
P(3) = 1410

Therefore, the profit from selling 300 pounds of Brie cheese is $1410.

kjb