Respuesta :
The function is used to model real life situation
- The value of a is -2000
- 150 computers must be produced to maximize the profit of $5,000,000
- A range of 140 to 160 computers would give the company at least $4,800,000
(a) The value of a
The function is given as:
[tex]\mathbf{y = a(x - 100)(x - 200) }[/tex]
The profit on 120 computers is 3200000.
So, we have:
[tex]\mathbf{3200000 = a(120 - 100)(120 - 200) }[/tex]
[tex]\mathbf{3200000 = -1600a}[/tex]
Divide both side by -1600
[tex]\mathbf{a = -2000}[/tex]
Hence, the value of (a) is -2000
(b) The maximum profit
The function becomes
[tex]\mathbf{y = -2000(x - 100)(x - 200)}[/tex]
Expand
[tex]\mathbf{y = -2000(x\² - 200x - 100x + 20000)}[/tex]
[tex]\mathbf{y = -2000(x\² - 300x + 20000)}[/tex]
Expand
[tex]\mathbf{y = -2000x\² + 600000x - 40000000}[/tex]
Differentiate
[tex]\mathbf{y' = -4000x + 600000}[/tex]
Set to 0
[tex]\mathbf{-4000x + 600000 = 0}[/tex]
Collect like terms
[tex]\mathbf{4000x = 600000}[/tex]
Divide both sides by 4000
[tex]\mathbf{x = 150}[/tex]
Substitute 150 for x in [tex]\mathbf{y = -2000(x - 100)(x - 200)}[/tex]
[tex]\mathbf{y = -2000(150 - 100)(150 - 200)}[/tex]
[tex]\mathbf{y = 5000000}[/tex]
150 computers must be produced to maximize the profit of $5,000,000
c) Range when the target is at least $4,800,000
This means that:
[tex]\mathbf{y \ge 4800000}[/tex]
So, we have:
[tex]\mathbf{-2000x\² + 600000x - 40000000 \ge 4800000}[/tex]
Collect like terms
[tex]\mathbf{-2000x\² + 600000x - 40000000 - 4800000 \ge 0}[/tex]
[tex]\mathbf{-2000x\² + 600000x - 44800000 \ge 0}[/tex]
Divide through by -2000
[tex]\mathbf{x\² - 300x +22400 \le 0}[/tex]
Factorize
[tex]\mathbf{(x - 140)(x - 160) \le 0}[/tex]
Express as inequality
[tex]\mathbf{140 \le x \le 160}[/tex]
A range of 140 to 160 computers would give the company at least $4,800,000
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