SOLUTION: Two verticies of a rectangle are on the x-axis and the other two vertices on the curve y=8/(x^2+4). Find the maximum area of the rectangle.
I did find y'=-16x/(x^2+4)^2. I could
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-> SOLUTION: Two verticies of a rectangle are on the x-axis and the other two vertices on the curve y=8/(x^2+4). Find the maximum area of the rectangle.
I did find y'=-16x/(x^2+4)^2. I could
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Question 1192564: Two verticies of a rectangle are on the x-axis and the other two vertices on the curve y=8/(x^2+4). Find the maximum area of the rectangle.
I did find y'=-16x/(x^2+4)^2. I couldn't figure out what do afterwards, I did try setting this to zero, but that isn't correct. Answer by ikleyn(52790) (Show Source):
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Two verticies of a rectangle are on the x-axis and the other two vertices on the curve y=8/(x^2+4).
Find the maximum area of the rectangle.
I did find y'=-16x/(x^2+4)^2. I couldn't figure out what do afterwards,
I did try setting this to zero, but that isn't correct.
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The rectangle has the horizontal side 2x and vertical side .
So the area of the rectangle is
A(x) = = . (1)
To find the maximum of the area, we should calculate the derivative A'(x) and equate it to zero.
It is enough to calculate the NUMERATOR of the derivative: it is
16*(x^2+1) - 16x*(2x) = 16x^2 + 16 - 32x^2 = 16 - 16x^2.
After that, we equate the derivative to zero - - - again, it is enough to equate the NUMERATOR to zero, which gives
16 = 16x^2, or x^2 = 1, or x = +/-1, of which we select x= 1.
So the maximum area of the rectangle is (use formula (1) at x= 1 )
= = = 8. ANSWER
