Question

A record club has found that the marginal​ profit,

Upper P prime (x )​, in​ cents, is given by

Upper P prime (x )equals negative 0.0008 x cubed plus 0.20 x squared plus 46.8 x for x less than or equals 200​,

where x is the number of members currently enrolled in the club. Approximate the total profit when 120 members are enrolled by computing the sum

Summation from i equals 1 to 6 Upper P prime (x Subscript i Baseline )Upper Delta x with Upper Delta x equals 20.

Answer 1

Given :

`$P'(x) = -0.0008x^3+0.20x^2+46.8x,$` for x ≤ 200

Total profit when 120 members are enrolled is :

`$\sum_(i=1)^6P'(x_i) \Delta x$` with
`\Delta x = 20`

Using the left end points, we get,

The values of
`x_i` are : { 0, 20, 40, 60, 80, 100}

Therefore,

`$P'(x_1) = P'(0)=-(0.0008)(0)^3+(0.20)(0)^2+(46.8)(0)$`

= 0

`$P'(x_2) = P'(20)=-(0.0008)(20)^3+(0.20)(20)^2+(46.8)(20)$`

= 1009.6

`$P'(x_3) = P'(40)=-(0.0008)(40)^3+(0.20)(40)^2+(46.8)(40)$`

= 2140.8

`$P'(x_4) = P'(60)=-(0.0008)(60)^3+(0.20)(60)^2+(46.8)(60)$`

= 3355.2

`$P'(x_5) = P'(80)=-(0.0008)(80)^3+(0.20)(80)^2+(46.8)(80)$`

= 4614.4

`$P'(x_6) = P'(100)=-(0.0008)(100)^3+(0.20)(100)^2+(46.8)(100)$`

= 5880

`$\sum_(i=1)^6P'(x_i) \Delta x = P'(x_1)\Delta x + P'(x_2)\Delta x + P'(x_3)\Delta x + P'(x_4)\Delta x + P'(x_5)\Delta x + P'(x_6)\Delta x $`

= (0)(20) + (1009.6)(20) + (2140.8)(20) + (3355.2)(20) + (4614.4)(20) + (5880)(20)

= (20)( 0 + 1009.6 + 2140.8 + 3355.2 + 4614.4 + 5880)

= (20)(17,000)

= 340,000 cents

`$=(340000)/(100) \ \text{dollars}$`

= 3400 dollars

Hence, the required total profit is 3400 dollars.