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# A batch of widgets costs p + 15 dollars for a company to pro

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A batch of widgets costs p + 15 dollars for a company to pro [#permalink]  13 Sep 2017, 12:48
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A batch of widgets costs p + 15 dollars for a company to produce and each batch sells for p(9 – p) dollars. For which of the following values of p does the company make a profit?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 7
[Reveal] Spoiler: OA

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Re: A batch of widgets costs p + 15 dollars for a company to pro [#permalink]  12 Oct 2017, 18:27

Instead of solving the equation for "p" which take more time and have chance of error too..its better to put values and find out the answer.
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Re: A batch of widgets costs p + 15 dollars for a company to pro [#permalink]  15 Oct 2017, 07:04
Rajneesh Garg wrote:

Instead of solving the equation for "p" which take more time and have chance of error too..its better to put values and find out the answer.

That's could be an idea. Anyway, if we write the equation for positive profits p^2-8p+15>0 it is easy to notice that it can be rewritten as (x-5)(x-3)>0.

Thus, the interval of solutions is 3<x<5. Then, the only choice between 3 and 5 excluded is choice B, 4.
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Re: A batch of widgets costs p + 15 dollars for a company to pro [#permalink]  12 Dec 2017, 16:54
Expert's post
Carcass wrote:

A batch of widgets costs p + 15 dollars for a company to produce and each batch sells for p(9 – p) dollars. For which of the following values of p does the company make a profit?

(A) 3

(B) 4

(C) 5

(D) 6

(E) 7

Since profit = revenue - cost, or P = R - C, we need to determine a value of p such that revenue > cost. We have that p + 15 = cost and p(9 - p) = revenue. Therefore, the profit is P = p(9 - p) - (p + 15) = 9p - p^2 - p - 15 = -p^2 + 8p - 15.

Notice that the graph of P = -p^2 + 8p - 15 is a parabola that opens downward, so the maximum value is at its vertex. Recall that we can use the formula -b/2a to find the x-value (in this case, the p-value) of the vertex:

p = -b/2a = -8/(2(-1)) = -8/-2 = 4

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Jeffrey Miller
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Re: A batch of widgets costs p + 15 dollars for a company to pro   [#permalink] 12 Dec 2017, 16:54
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