Question

A company makes a profit of $y (in thousand dollars) when it produces x computers,

Answer 1

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:

$y = a(x - 100)(x - 200)$

Item a:

If 120  computers are produced, the profit will be $3,200,000, hence when $x = 120, y = 3200000$, and this is used to find a.

$y = a(x - 100)(x - 200)$

$3200000 = a(120 - 100)(120 - 200)$

$-1600a = 3200000$

$a = -\frac{3200000}{1600}$

$a = -2000$

The value of a is -2000.

Item b:

We first place the quadratic function into standard form, thus:

$y = -2000(x - 100)(x - 200)$

$y = -2000(x^2 - 300x + 20000)$

$y = -2000x^2 + 600000x - 40000000$

Which has coefficients $a = -2000, b = 600000, c = -40000000$.

Then, we have to find the vertex:

$x_V = -\frac{b}{2a} = -\frac{600000}{2(-2000)} = 150$

$\Delta = b^2 - 4ac = (600000)^2 - 4(-2000)(-40000000) = 40000000000$

$y_V = -\frac{\Delta}{4a} = -\frac{40000000000 }{4(-2000)} = 5000000$

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:

$y = -200x^2 + 600000x - 40000000$

$4800000 = -200x^2 + 600000x - 40000000$

$-200x^2 + 600000x - 44800000 = 0$

The coefficients are $a = -200, b = 600000, c = -44800000$.

Then:

$\Delta = b^2 - 4ac = (600000)^2 - 4(-2000)(-44800000) = 1600000000$

$x_1 = \frac{-b - \sqrt{Delta}}{2a} = \frac{-600000 - \sqrt{1600000000}}{2(-2000)} = 160$

$x_2 = \frac{-b + \sqrt{Delta}}{2a} = \frac{-600000 + \sqrt{1600000000}}{2(-2000)} = 140$

Between 140 and 160 computers need to be produced.

A similar problem is given at brainly.com/question/24705734