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It’s well known that a company's profit=20,000x25x^2, where x is the [#permalink]
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12 Oct 2016, 03:31
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It’s well known that a company's profit=20,000x25x^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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Last edited by Bunuel on 12 Oct 2016, 03:58, edited 1 time in total.
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Re: ) It’s well known that a company's profit=20,000x25x^2, where [#permalink]
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12 Oct 2016, 03:59
MathRevolution wrote: It’s well known that a company's profit=20,000x25x^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=20,000x25x^2, where x is the [#permalink]
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12 Oct 2016, 11:16
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MathRevolution wrote: It’s well known that a company's profit=20,000x25x^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=20,000x25x^2, where x is the [#permalink]
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12 Oct 2016, 16:40
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Alternate approach:
To find Max or Min of a polynomial f(x), Equate f'(x)=0
f(x) = 20,000x25x^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=20,000x25x^2, where x is the [#permalink]
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13 Oct 2016, 07:33
It’s well known that a company's profit=20,000x25x^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=20,000x25x^2, where x is the [#permalink]
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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=20,000x25x^2, where x is the [#permalink]
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11 Nov 2016, 03:06
Abhishek009 wrote: MathRevolution wrote: It’s well known that a company's profit=20,000x25x^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=20,000x25x^2, where x is the [#permalink]
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12 Nov 2016, 06:34
MathRevolution wrote: It’s well known that a company's profit=20,000x25x^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 xvalue 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=20,000x25x^2, where x is the [#permalink]
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12 Nov 2016, 06:52
Noted, thanks for the clarification!



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