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It’s well known that a company's profit=20,000x-25x^2, where x is the

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New post Updated on: 12 Oct 2016, 03:58
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It’s well known that a company's profit=20,000x-25x^2, where x is the number of workers. If the profit of the company is the greatest, what is the value of x?

A. 50
B. 100
C. 150
D. 250
E. 400

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Originally posted by MathRevolution on 12 Oct 2016, 03:31.
Last edited by Bunuel on 12 Oct 2016, 03:58, edited 1 time in total.
Renamed the topic and edited the question.
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Re: ) It’s well known that a company's profit=20,000x-25x^2, where  [#permalink]

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New post 12 Oct 2016, 03:59
MathRevolution wrote:
It’s well known that a company's profit=20,000x-25x^2, where x is the number of workers. If the profit of the company is the greatest, what is the value of x?

A. 50 B. 100 C. 150 D. 250 E. 400


E - Used plugin after plotting a graph of this equation.
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Re: It’s well known that a company's profit=20,000x-25x^2, where x is the  [#permalink]

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New post 12 Oct 2016, 11:16
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MathRevolution wrote:
It’s well known that a company's profit=20,000x-25x^2, where x is the number of workers. If the profit of the company is the greatest, what is the value of x?

A. 50
B. 100
C. 150
D. 250
E. 400


So, \(Profit = 20000x - 25x^2\)

Or, \(Profit = x ( 20000 - 25x )\)

Quote:
Where x is the number of workers. If the profit of the company is the greatest, what is the value of x?


Now, to maximize profit we must minimize : ( 20000 - 25x )

Among the given options only (E) is the highest value for x , that can maximize the Final Value (Profit )

So, Answer will be (E)

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It’s well known that a company's profit=20,000x-25x^2, where x is the  [#permalink]

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New post 12 Oct 2016, 16:40
2
2
Alternate approach:

To find Max or Min of a polynomial f(x), Equate f'(x)=0

f(x) = 20,000x-25x^2

Differentiate f(x)

f'(x) = 20,000 - 25*2*x = 0

50x = 20,000

x=400

D


Note: To confirm whether x is maximum or minimum, substitute x in f''(x)

If f''(x) <0, then f(x) is maximum

If f''(x) >0, then f(x) is minimum

In this case, f''(x)= -50 < 0
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It’s well known that a company's profit=20,000x-25x^2, where x is the  [#permalink]

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New post 13 Oct 2016, 07:33
It’s well known that a company's profit=20,000x-25x^2, where x is the number of workers. If the profit of the company is the greatest, what is the value of x?

A. 50 B. 100 C. 150 D. 250 E. 400
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Re: It’s well known that a company's profit=20,000x-25x^2, where x is the  [#permalink]

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New post 17 Oct 2016, 00:33
==>In general, from y=ax2+bx+c, the value of y from symmetric axis, x=-b/2a is the least or the greatest. For the profit has to be the greatest, x=-20,000/2(-25) = 400, hence the greatest profit. The answer is E.
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Re: It’s well known that a company's profit=20,000x-25x^2, where x is the  [#permalink]

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New post 11 Nov 2016, 03:06
Abhishek009 wrote:
MathRevolution wrote:
It’s well known that a company's profit=20,000x-25x^2, where x is the number of workers. If the profit of the company is the greatest, what is the value of x?

A. 50
B. 100
C. 150
D. 250
E. 400


So, \(Profit = 20000x - 25x^2\)

Or, \(Profit = x ( 20000 - 25x )\)

Quote:
Where x is the number of workers. If the profit of the company is the greatest, what is the value of x?


Now, to maximize profit we must minimize : ( 20000 - 25x )

Among the given options only (E) is the highest value for x , that can maximize the Final Value (Profit )

So, Answer will be (E)


Hi,
Sorry, slightly unclear on the highlivhted portion above, but should we not try to maximise the highlighted portion as bigger the multiple of x, higher the value of the profit. Please can anyone let me know what am i missing here.

Thanks
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Re: It’s well known that a company's profit=20,000x-25x^2, where x is the  [#permalink]

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New post 12 Nov 2016, 06:34
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MathRevolution wrote:
It’s well known that a company's profit=20,000x-25x^2, where x is the number of workers. If the profit of the company is the greatest, what is the value of x?

A. 50
B. 100
C. 150
D. 250
E. 400


Notice the given formula P = 20,000x - 25x^2 is a quadratic function. Furthermore, the graph of this function will be a parabola opening downward, and the vertex of this parabola will be the maximum point of the graph. Since we need to determine a value of x in which the company makes the greatest profit, we need to determine the value of x at the vertex of the parabola.

Recall that x = -b/(2a) is the formula to find the x-value of the vertex; therefore, the number of workers that maximizes the profit is:

x = -20,000/[2(-25)] = -20,000/-50 = 400

Answer: E
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Re: It’s well known that a company's profit=20,000x-25x^2, where x is the  [#permalink]

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New post 12 Nov 2016, 06:52
Noted, thanks for the clarification!
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Re: It’s well known that a company's profit=20,000x-25x^2, where x is the  [#permalink]

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