Question

The marginal profit in dollars on Brie cheese sold at a cheese store is given by Upper P prime (x )equals x (60 x squared plus 90 x )comma where x is the amount of cheese​ sold, in hundreds of pounds. The​ "profit" is minus​$50 when no cheese is sold.

a. Find the profit function.
b. Find the profit from selling 400 pounds of Brie cheese.

Answer 1

After integrating the marginal profit function and applying the initial condition, the profit function is P(x) = 15x^4 + 45x^3 - 50. To find the profit from selling 400 pounds of Brie cheese, plug x = 4 into the profit function to get $6670.

Step-by-step explanation:

To solve for the profit function, we need to integrate the given marginal profit function P'(x) = x(60x^2 + 90x), given that the profit is -$50 when no cheese is sold. By integrating the given function concerning x and using the initial condition, we find the indefinite integral:

∫ P'(x) dx = ∫ x(60x^2 + 90x) dx

P(x) = 15x^4 + 45x^3 + C

We apply the initial condition P(0) = -50 to find the constant of integration C:

-50 = 15(0)^4 + 45(0)^3 + C ⇒ C = -50

Now the profit function is:

P(x) = 15x^4 + 45x^3 - 50

For part (b), to find the profit from selling 400 pounds of cheese, we use the profit function with x = 4 (since x is measured in hundreds of pounds):

P(4) = 15(4)^4 + 45(4)^3 - 50

P(4) = 15(256) + 45(64) - 50

P(4) = 3840 + 2880 - 50

P(4) = 6670

Thus, the profit from selling 400 pounds of Brie cheese is $6670.

Answer 2

Part A:

P(x)=15x^4+30x^3-50

Part B:

P(4)=$4270

Step-by-step explanation:

Part A:

In order to find the profit function P(x) we have to integrate the P'(x)

P'(x)=x(60x^2+90x)

P'(x)=60x^3+90x^2

`\int\limits {p'(x)} \, dx =\int\limits {60x^3+90x^2} \, dx`

P(x)=15x^4+30x^3+C

when x=0, C=-50

P(x)=15x^4+30x^3-50

Part B:

x=4

P(x)=15x^4+30x^3-50

P(4)=15*4^4+30*4^3-50

P(4)=$4270

Profit from selling 400 pounds is $4270