Respuesta :
Using quadratic function concepts, it is found that:
a) The value of a is -2000.
b) The maximum profit the company can make is of $5,000,000, when 150 computers should be produced.
c) Between 140 and 160 computers need to be produced.
The profit is modeled by:
[tex]y = a(x - 100)(x - 200)[/tex]
Item a:
If 120 computers are produced, the profit will be $3,200,000, hence when [tex]x = 120, y = 3200000[/tex], and this is used to find a.
[tex]y = a(x - 100)(x - 200)[/tex]
[tex]3200000 = a(120 - 100)(120 - 200)[/tex]
[tex]-1600a = 3200000[/tex]
[tex]a = -\frac{3200000}{1600}[/tex]
[tex]a = -2000[/tex]
The value of a is -2000.
Item b:
We first place the quadratic function into standard form, thus:
[tex]y = -2000(x - 100)(x - 200)[/tex]
[tex]y = -2000(x^2 - 300x + 20000)[/tex]
[tex]y = -2000x^2 + 600000x - 40000000[/tex]
Which has coefficients [tex]a = -2000, b = 600000, c = -40000000[/tex].
Then, we have to find the vertex:
[tex]x_V = -\frac{b}{2a} = -\frac{600000}{2(-2000)} = 150[/tex]
[tex]\Delta = b^2 - 4ac = (600000)^2 - 4(-2000)(-40000000) = 40000000000 [/tex]
[tex]y_V = -\frac{\Delta}{4a} = -\frac{40000000000 }{4(-2000)} = 5000000[/tex]
The maximum profit the company can make is of $5,000,000, when 150 computers should be produced.
Item c:
We are working with a concave down parabola, hence the range is between the roots of:
[tex]y = -200x^2 + 600000x - 40000000[/tex]
[tex]4800000 = -200x^2 + 600000x - 40000000[/tex]
[tex]-200x^2 + 600000x - 44800000 = 0[/tex]
The coefficients are [tex]a = -200, b = 600000, c = -44800000[/tex].
Then:
[tex]\Delta = b^2 - 4ac = (600000)^2 - 4(-2000)(-44800000) = 1600000000[/tex]
[tex]x_1 = \frac{-b - \sqrt{Delta}}{2a} = \frac{-600000 - \sqrt{1600000000}}{2(-2000)} = 160[/tex]
[tex]x_2 = \frac{-b + \sqrt{Delta}}{2a} = \frac{-600000 + \sqrt{1600000000}}{2(-2000)} = 140[/tex]
Between 140 and 160 computers need to be produced.
A similar problem is given at https://brainly.com/question/24705734
