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A business is currently selling 30 chairs per day for a price of $25 p
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22 Mar 2021, 10:16
22 Mar 2021, 10:16
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A business is currently selling 30 chairs per day for a price of $25 per chair. A worker predicts that if the business chooses to lower the cost of the chairs, then for every $1 the price is lowered, one more chair will be sold. If the prediction is accurate, what is the maximum revenue the business can earn in a day from selling chairs? (Revenue is the amount of money that the business takes in, without consideration of expenses or other costs.)
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Show: :: OA
$750
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
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A business is currently selling 30 chairs per day for a price of $25 p
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22 Mar 2021, 23:36
22 Mar 2021, 23:36
1
Carcass wrote:
A business is currently selling 30 chairs per day for a price of $25 per chair. A worker predicts that if the business chooses to lower the cost of the chairs, then for every $1 the price is lowered, one more chair will be sold. If the prediction is accurate, what is the maximum revenue the business can earn in a day from selling chairs? (Revenue is the amount of money that the business takes in, without consideration of expenses or other costs.)
Show: :: OA
$750
Revenue = (Number of chairs)(cost per chair)
Currently, \(R = (30)(25) = 750\)
As per the prediction, \(R = (30 + x)(25 - x)\), where \(x = 1, 2, 3, ......\)
Now, \(R_1 = (31)(24) = 744\)
\(R_2 = (32)(23) = 736\)
\(R_3 = (33)(22) = 726\)
We can see the values are decreasing, so the Maximum revenue could be \(750\) only
Alternative Approach:
Let's solve the quadratic, \(R = (30 + x)(25 - x)\)
\(R = -x^2 - 5x + 750\)
Parabola is a quadratic function with the maximum value at the vertex
x-coordinate of the vertex is given by \(\frac{-b}{2a} = \frac{-(-5)}{(2)(-1)} = \frac{-5}{2}\)
This means - Maximum revenue would be generated if the number of chairs are \(27.5\) and the cost per chair is \(27.5\) too, which is not possible!
Re: A business is currently selling 30 chairs per day for a price of $25 p
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24 Mar 2021, 02:58
24 Mar 2021, 02:58
2
Revenue = (Cost per chair)(Number of chairs sold)
Revenue = (30+x)(25-x)
To maximize revenue, you can take the derivative approach, but that would be slightly complicated for such a question
The best approach would be substituting the values of x since you will already be having the on-screen calculator with you.
At x=0, Revenue=750
At x=1, Revenue=744
At x=2, Revenue= 736
At x=3, Revenue= 726
.
.
Hence the Revenue keeps decreasing as we increase x. This shows that the maximum revenue will only be when 30 chairs are sold at the price of 25 each i.e 750 dollars
Revenue = (30+x)(25-x)
To maximize revenue, you can take the derivative approach, but that would be slightly complicated for such a question
The best approach would be substituting the values of x since you will already be having the on-screen calculator with you.
At x=0, Revenue=750
At x=1, Revenue=744
At x=2, Revenue= 736
At x=3, Revenue= 726
.
.
Hence the Revenue keeps decreasing as we increase x. This shows that the maximum revenue will only be when 30 chairs are sold at the price of 25 each i.e 750 dollars
