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A laptop company is choosing which components they want to put into th 28 Jan 2022, 08:45
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95% (hard)

Question Stats:

47% (02:54) correct 53% (03:00) wrong based on 102 sessions

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A laptop company is choosing which components they want to put into their next model. This model laptop will sell for $2000, and every additional component will bring the selling price up an additional $200. However, the company predicts that each time they increase the price, the number of sales will drop by around 5% from the base prediction. What is the best number of components to upgrade in order to maximize revenue?

(A) Six
(B) Five
(C) Four
(D) Three
(E) Two


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B
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A laptop company is choosing which components they want to put into th 31 Jan 2022, 17:57
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Bunuel
A laptop company is choosing which components they want to put into their next model. This model laptop will sell for $2000, and every additional component will bring the selling price up an additional $200. However, the company predicts that each time they increase the price, the number of sales will drop by around 5% from the base prediction. What is the best number of components to upgrade in order to maximize revenue?

(A) Six
(B) Five
(C) Four
(D) Three
(E) Two

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Breaking Down the Info:

Set the number of increases in price as x. Then the revenue would be \(R = (2000 + 200x)*(1 - 0.05x)\). We could multiply everything out and use the vertex of this parabola to find the maximum point (or use calculus if you prefer ...)

\(R = 2000 - 100x + 200x - 10x^2 = -10x^2 + 100x + 2000\). The vertex is \(-\frac{b}{2a} = \frac{100}{20} = 5\). Thus 5 increases is the optimal point.

Answer: B
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A laptop company is choosing which components they want to put into th 15 Jun 2023, 21:11
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Can you explain how it is 5?
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A laptop company is choosing which components they want to put into th 15 Jun 2023, 23:23
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Given: A laptop company is choosing which components they want to put into their next model. This model laptop will sell for $2000, and every additional component will bring the selling price up an additional $200. However, the company predicts that each time they increase the price, the number of sales will drop by around 5% from the base prediction.

Asked: What is the best number of components to upgrade in order to maximize revenue?

Let the number of components be x and base volume prediction be V.

R = Revenue = Price * Volume = ($2000 + $200x) (V - 5% xV) = $200V (10 + x)(1 - 5x/100) = $200V (10+x)(1-x/20) = $10V (10+x)(20-x) = $10V (200 +10x - xˆ2) = $10V (225 - 25 + 10x - xˆ2) = $10V (225 - (x-5)ˆ2)

To maximize revenues x = 5
Maximum Revenue = $10V *225 = $2250V

IMO B
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A laptop company is choosing which components they want to put into th 30 Aug 2024, 19:20
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KarishmaB MartyMurray Can you please explain this question ? gmatophobia
I think R = 200v *(10 +x) * .95^x should be used.

Revenue R = ( 2000 + 200x ) * ( 1 - .05x) looks too simplistic..CAn you kindly clarify ?­
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A laptop company is choosing which components they want to put into th 31 Aug 2024, 00:20
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Bunuel
A laptop company is choosing which components they want to put into their next model. This model laptop will sell for $2000, and every additional component will bring the selling price up an additional $200. However, the company predicts that each time they increase the price, the number of sales will drop by around 5% from the base prediction. What is the best number of components to upgrade in order to maximize revenue?

(A) Six
(B) Five
(C) Four
(D) Three
(E) Two


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­
Base Revenue = 2000 * No of people who buy (say P)

At every step, the per person revenue increases by 10% while the number of people reduce by only 5%. So obviously, it makes sense to carry out some of these steps to increase total revenue. The total revenue will be maximum for a certain number of steps, say n.

Revenue after n steps = 2000 * (1+.1n) * p * (1 - .05n)
To maximize revenue, we need (1+.1n) * (1 - .05n) to take its maximum value.

\((1+.1n) * (1 - .05n) = 1 + .05n - .005n^2 \)
This is a quadratic which is downward opening parabola (because a is negative) and its maximum value will be obtained at n = -b/2a

So n = .05/.01 = 5

Answer (B)

Note that even at n = 6 or 7, the revenue obtained will be higher than base revenue but it will be lower than the revenue obtained at n = 5.

Here is a video discussing the parabola form of quadratics: https://youtu.be/QOSVZ7JLuH0
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A laptop company is choosing which components they want to put into th 10 Sep 2024, 05:13
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KarishmaB Can you please elaborate a bit more on the highlighted part for better understanding ?
KarishmaB

Bunuel
A laptop company is choosing which components they want to put into their next model. This model laptop will sell for $2000, and every additional component will bring the selling price up an additional $200. However, the company predicts that each time they increase the price, the number of sales will drop by around 5% from the base prediction. What is the best number of components to upgrade in order to maximize revenue?

(A) Six
(B) Five
(C) Four
(D) Three
(E) Two


Are You Up For the Challenge: 700 Level Questions
­
Base Revenue = 2000 * No of people who buy (say P)

At every step, the per person revenue increases by 10% while the number of people reduce by only 5%. So obviously, it makes sense to carry out some of these steps to increase total revenue. The total revenue will be maximum for a certain number of steps, say n.

Revenue after n steps = 2000 * (1+.1n) * p * (1 - .05n)
To maximize revenue, we need (1+.1n) * (1 - .05n) to take its maximum value.

\((1+.1n) * (1 - .05n) = 1 + .05n - .005n^2 \)
This is a quadratic which is downward opening parabola (because a is negative) and its maximum value will be obtained at n = -b/2a

So n = .05/.01 = 5

Answer (B)

Note that even at n = 6 or 7, the revenue obtained will be higher than base revenue but it will be lower than the revenue obtained at n = 5.

Here is a video discussing the parabola form of quadratics: https://youtu.be/QOSVZ7JLuH0
Show more
­
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A laptop company is choosing which components they want to put into th 11 Sep 2024, 05:28
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sayan640
KarishmaB Can you please elaborate a bit more on the highlighted part for better understanding ?
KarishmaB

Bunuel
A laptop company is choosing which components they want to put into their next model. This model laptop will sell for $2000, and every additional component will bring the selling price up an additional $200. However, the company predicts that each time they increase the price, the number of sales will drop by around 5% from the base prediction. What is the best number of components to upgrade in order to maximize revenue?

(A) Six
(B) Five
(C) Four
(D) Three
(E) Two


Are You Up For the Challenge: 700 Level Questions
­
Base Revenue = 2000 * No of people who buy (say P)

At every step, the per person revenue increases by 10% while the number of people reduce by only 5%. So obviously, it makes sense to carry out some of these steps to increase total revenue. The total revenue will be maximum for a certain number of steps, say n.

Revenue after n steps = 2000 * (1+.1n) * p * (1 - .05n)
To maximize revenue, we need (1+.1n) * (1 - .05n) to take its maximum value.

\((1+.1n) * (1 - .05n) = 1 + .05n - .005n^2 \)
This is a quadratic which is downward opening parabola (because a is negative) and its maximum value will be obtained at n = -b/2a

So n = .05/.01 = 5

Answer (B)

Note that even at n = 6 or 7, the revenue obtained will be higher than base revenue but it will be lower than the revenue obtained at n = 5.

Here is a video discussing the parabola form of quadratics: https://youtu.be/QOSVZ7JLuH0
­
Show more
­Revenue increases by 10% (from base) at every step because price of the laptop will increase by 10% ($200 of $2000) but the number of sales will reduce so if in the base case, p people were buying, at every step, p will reduce by 5% of p (from base)

So if 3 steps are done say,

Revenue = (2000 + 30% of 2000)(p - 15% of p)
Revenue = 2000(1 + 3*.1)*p(1 - 3*.05)

In general terms,
Revenue = 2000*(1 + .1n) * p(1 - .05n)­
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A laptop company is choosing which components they want to put into th 13 Sep 2024, 13:35
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Easier method,
since on every increase of $200 there is a sales drop of 5%
i.e. on one component addition $2200 -->> .95 sales of base prediction
2-->>$2400--> .90 of base prediction
3-->>$2600-->> .85
4-->>$2800-->> .80
5-->>$3000-->> .75
6-->>$3200-->> .70
by simple multiplication was able to get result just under the time. however use of quadratic is time savior.
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