Question

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 amunt of cheese sold, in hundreds of pounds. The "profit" is -$80 when no cheese is sold.a. Find the profit functionb. Find the profit from selling 200lbs of Brie Cheese

Answer 1

Answer:

a) $P(x) = 15x^{4} + 10x^{3} - 80$

b) The profit from selling 200lbs of Brie Cheese is $240.

Step-by-step explanation:

a. Find the profit function

We have that:

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

The profit function is P(x), which is the integral of P'(x).

So

$P(x) = \int {P'(x)} \, dx = \int {60x^{3} + 30x^{2}} \, dx = \frac{60x^{4}}{4} + \frac{30x^{3}}{3} + K = 15x^{4} + 10x^{3} + K$

There, I applied the integral rules of sum and power.

Since P(0) = -80, K = -80

Then

$P(x) = 15x^{4} + 10x^{3} - 80$

b. Find the profit from selling 200lbs of Brie Cheese

200 lbs is 200/100 = 2 hundreds of pounds.

So this is P(2).

$P(x) = 15x^{4} + 10x^{3} - 80$

$P(2) = 15*2^{4} + 10*2^{3} - 80$

$P(2) = 240$

The profit from selling 200lbs of Brie Cheese is $240.