Respuesta :
Solution :
Given :
[tex]$P'(x) = -0.0008x^3+0.20x^2+46.8x,$[/tex] for x ≤ 200
Total profit when 120 members are enrolled is :
[tex]$\sum_{i=1}^6P'(x_i) \Delta x$[/tex] with [tex]\Delta x = 20[/tex]
Using the left end points, we get,
The values of [tex]x_i[/tex] are : { 0, 20, 40, 60, 80, 100}
Therefore,
[tex]$P'(x_1) = P'(0)=-(0.0008)(0)^3+(0.20)(0)^2+(46.8)(0)$[/tex]
= 0
[tex]$P'(x_2) = P'(20)=-(0.0008)(20)^3+(0.20)(20)^2+(46.8)(20)$[/tex]
= 1009.6
[tex]$P'(x_3) = P'(40)=-(0.0008)(40)^3+(0.20)(40)^2+(46.8)(40)$[/tex]
= 2140.8
[tex]$P'(x_4) = P'(60)=-(0.0008)(60)^3+(0.20)(60)^2+(46.8)(60)$[/tex]
= 3355.2
[tex]$P'(x_5) = P'(80)=-(0.0008)(80)^3+(0.20)(80)^2+(46.8)(80)$[/tex]
= 4614.4
[tex]$P'(x_6) = P'(100)=-(0.0008)(100)^3+(0.20)(100)^2+(46.8)(100)$[/tex]
= 5880
[tex]$\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 $[/tex]
= (0)(20) + (1009.6)(20) + (2140.8)(20) + (3355.2)(20) + (4614.4)(20) + (5880)(20)
= (20)( 0 + 1009.6 + 2140.8 + 3355.2 + 4614.4 + 5880)
= (20)(17,000)
= 340,000 cents
[tex]$=\frac{340000}{100} \ \text{dollars}$[/tex]
= 3400 dollars
Hence, the required total profit is 3400 dollars.
