##### exam calculus question

label Calculus
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Find the maximum possible area of a rectangle that could fit in the space between the x-axis and the parabola -2x4 + 8.

Jun 3rd, 2015

Thank you for the opportunity to help you with your question!

The area of the rectangle = 2X * Y.
Y = -2X^4+8
by substitution, then the rectangle
Area = 2X * - (-2X^4+8)

set the equation to 0

0 =  2X +2x^4+8

taking the derivative of the equation:

0 = 2+8x^3

x^3=-2/8

and solve for X;

x=(-2)^1/3/2

please clear the equation i am not understanding...............................

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 3rd, 2015

I don't really understand this, can you please explain ?

"by substitution, then the rectangle
Area = 2X * - (-2X^4+8)"

Jun 3rd, 2015
The space in question is bounded by the parabola and the X axis. If the rectangle has its base on the x-axis, then the upper corners are touching the parabola. Also, note that if we solve for X and find Y, then X * Y = ½ the rectangle (on the right side of the Y-axis. The other half is on the left side of the Y-axis. Therefore, the total area = 2X * Y.

Jun 3rd, 2015

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Jun 3rd, 2015
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Jun 3rd, 2015
Oct 21st, 2017
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