A company makes a profit of $y (in thousand dollars) when it produces x computers,
where y is given by the formula y = a(x - 100)(x - 200) for x 20 If 120
computers are produced, the profit will be $3,200,000.
a) Find the value of a.

b) What is the maximum profit the company can make? At this profit, how many
computers should be produced?

c) If the company targets to make at least $4,800,000, what is the range of the
number of computers to be produced?


Answers

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:

[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


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Answers

Answer:

this is the answer....

Answer:

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Step-by-step explanation:

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Answer:

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Answer:

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-TheUnknownScientist 72

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Answers

Answer:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

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15d = -105

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____

RainbowSalt2222 ☔

Maria is recording the percentages she earned on each quiz in her math class. Here are her results for the last 7 quizzes.
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Answers

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Answer:

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Answer:

Step-by-step explanation:

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>> Answer

_________

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Answer:

Not quite in proof format but here's a short explaination. You should be able to fill out the table yourself.

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Answer:

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Answers

Answer:

90

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Answers

3/8 -> 3/8 (it is fully simplified)

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Have a nice day!

     If this is not what you are looking for - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

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Step-by-step explanation:

hope I helped BESTIEE

Answer:

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