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Maximize the weekly profit

by gmattesttaker2 » Wed Dec 25, 2013 12:12 am
Hello,

Can you please assist with this:

A chair manufacturer can manufacture a maximum of 800 chairs per week. Her profit is given by the equation P = 1/10(600 - x)(x - 10), where P is her profit and x is the number of chairs manufactured during one week. Which of the following represents the number of chairs that will maximize her weekly profit?

A) 300
B) 305
C) 600
D) 610
E) 620

OA: B

I was able to eliminate C, D and E. I was wondering if there is a way to decide between A and B without plugging in the values. I was able to arrive at this parabola equation:

P = -x^2/10 + 61x - 600

I was wondering if there a way from here to find the correct answer.

Thanks a lot,
Sri
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by Uva@90 » Wed Dec 25, 2013 2:12 am
Hi Sri,
As you eliminate C,D,E. We will look into A and B alone,
Since P = 1/10(600 - x)(x - 10)
but, If X=300
P=(300)*(290)------>1
if X=305
P=(295)*(295)------>2

Equation 1 can be written as (295+5)(295-5) => 295^2-5^2
Equation 2 is 295^2
Hence you can eliminate option A
Ans is B

Regards,
Uva.
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by gmattesttaker2 » Wed Dec 25, 2013 6:35 pm
Uva@90 wrote:Hi Sri,
As you eliminate C,D,E. We will look into A and B alone,
Since P = 1/10(600 - x)(x - 10)
but, If X=300
P=(300)*(290)------>1
if X=305
P=(295)*(295)------>2

Equation 1 can be written as (295+5)(295-5) => 295^2-5^2
Equation 2 is 295^2
Hence you can eliminate option A
Ans is B

Regards,
Uva.

Hello Uva,

Thanks for this awesome approach. This is really cool. Thanks again for all your help.

Best Regards,
Sri
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