Question

A small tie shop finds that at a sales level of x ties per day its marginal profit is MP (x )dollars per​ tie, where MP (x )equals 1.85 plus 0.12 x minus 0.0024 x squared. ​Also, the shop will lose ​$65 per day at a sales level of x equals 0. Find the profit from operating the shop at a sales level of x ties per day.

Answer 1

Correct question:

A small tie shop finds that at a sales level of x ties per day its marginal profit is MP(x )dollars per​ tie, where

MP(x) =1.85 + 0.12x - 0.0024x². ​Also, the shop will lose ​$65 per day at a sales level of x= 0. Find the profit from operating the shop at a sales level of x ties per day.

Answer:

P(x) =1.85x +0.06x² - 0.0008x³ - 65

Explanation:

Given the marginal profit function:

MP(x) =1.85 +0.12x - 0.0024x², P(0)= -65

We are to find P(x).

P(x) = ∫MP(x) dx

P(x) = ∫(1.85 + 0.12x - 0.0024x²) dx

= ∫1.85 dx+∫0.12x dx+∫(-0.0024x²)dx + C

= 1.85x + 0.06x² - 0.0008x³ + C

Initial condition at P(0) = - 65

where x(0), P(x) = -65

we have:

-65 = 1.85(0)+0.06(0)² - 0.0008x(0)³ + C

-65 = 0 + 0 - 0 + C

-65 = C

C = -65

P(x) =1.85x + 0.06x² - 0.0008x³ - 65

Answer 2

Question:

A small tie shop finds that at a sales level of x ties per day its marginal profit is MP(x )dollars per​ tie, where

MP(x) =1.85 + 0.12x - 0.0024x². ​Also, the shop will lose ​$65 per day at a sales level of x= 0. Find the profit from operating the shop at a sales level of x ties per day.

Answer:

P(x) = 1.85x + 0.06x^2 - 0.0008x^3 - 65

Explanation:

MP(x) = 1.85 + 0.12(x) - 0.0024(x^2)

At sales level, x=0, loss = $65

P(0) = - $65

Integrating MP(x) with respect to x

P(x) = ∫MP(x) dx = ∫ 1.85 + 0.12(x) - 0.0024(x^2)

P(x) = 1.85x + 0.06x^2 - 0.0008x^3 + C

Therefore :

P(x) = 1.85x + 0.06x^2 - 0.0008x^3 - 65