please help me please

Question

Answer 1

10.

To reach break even, cost = revenue ---> 50x^2 + 1000 = 500x - x^2

51x^2 - 500x + 1000 = 0

Find the roots using quadratic formula: for ax^2 + bx + c = 0, x = (- b ± sqrt(b^2 - 4ac))/2a

Here a = 51, b = -500 and c = 1000

x = (500 ± sqrt (500^2 - 4 * 51 * 1000))/2 * 51

x = (500 ± sqrt (46000))/102

x = (500 ± 214.4761)/102

x = 7.004, 2.799254

Check whether the least value satisfying the both equations.

Check the cost and revenue with x = 2.799254

C(x) = 50x^2 + 1000 = 50 * 2.799254^2 + 1000 = 1391.791

R(x) = 500x - x^2 = 500*2.799254 - 2.799254^2 = 1391.791

So, the least number of equipments to be sold to reach breakeven is 1392

Answer 2

11.

a) Cost of producing 50 items, C(50) = 100 * 50 + 2000 / 50 = 5000 + 40 = 5040

c) Average cost per item of producing 50 items = 5040/50 = 100.8

e) Mimimum cost for producing 1 item = C(1) = 100 * 1 + 2000 /1 = 2100

Answer 3

11 d)

Find the first derivate of the function and make it zero

C'(x) = 100 - 2000/x^2

0 = 100 - 2000/x^2

100 = 2000/x^2

x^2 = 20

x = ± 4.472136

Find the second deriavative

C"(x)= 2000 * 2/x^3

Substitute x = 4.472136

C"(x) = 2000 * 2/ 4.472136^3 = 44.72136 > 0

production = 4.472136 items