Question

A record club has found that the marginal​ profit,

Answer 1

Solution :

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

$=\frac{340000}{100} \ \text{dollars}$

= 3400 dollars

Hence, the required total profit is 3400 dollars.