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December 11th, 2012, 04:46 PM   #1
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Dimensions of max area and area itself?

so far i got: y = 49-x^2
Area = 2x(49-x^2)
Area = 98x - 2x^3
6x^2 = -98
x^2 = sqrt(98/6) === 2sqrt(7)/(sqrt(6)





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December 11th, 2012, 06:11 PM   #2
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Re: Dimensions of max area and area itself?

You have the correct area function:



Differentiating and equating to zero, we find:



So, our critical number is found from:



Taking the positive root, we find:



and so:



To ensure we have a maximum, observe that:

, and since our critical number is positive, the second derivative is negative, showing we do have a maximum. Hence:

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December 11th, 2012, 06:13 PM   #3
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Re: Dimensions of max area and area itself?

thanks mark!
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