SOLUTION: The base of a rectangle is on the x-axis and its two upper vertices are on the parabola y=16-x^2. of all such rectangles, what are the dimensions of the one with greatest area?
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-> SOLUTION: The base of a rectangle is on the x-axis and its two upper vertices are on the parabola y=16-x^2. of all such rectangles, what are the dimensions of the one with greatest area?
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Question 310349: The base of a rectangle is on the x-axis and its two upper vertices are on the parabola y=16-x^2. of all such rectangles, what are the dimensions of the one with greatest area? Found 2 solutions by solver91311, Edwin McCravy:Answer by solver91311(24713) (Show Source):
Because of the fact that the parabola is symmetric to the y-axis, the rectangle must also be symmetric to the y-axis. That means that the two lower vertices are and . The measure of the base of the rectangle is therefore . The upper vertices, being points on the parabola are: and . Therefore, the measure of the height of the rectangle is simply . Therefore the area of the rectangle is:
Such a function has local extremes at the points where the first derivative is zero:
Discard the negative root since we need a positive measure of length
We are looking for a maximum, so we want to see if the value of the second derivative is negative at the extreme point.
Which it is...
So, the horizontal dimension of the largest area rectangle is
I think you need to draw a graph to understand what is going on:
The base of the green rectangle is 2x, because
the base extends from -x to +x on the x axis,
and that is a distance of 2x.
The height of the green rectangle is y, so
The area of the green rectangle is
Area = base * height or
Since y and x are related by the equation of the parabola,
we can substitute for
Set
Use the positive value
Using the second derivative test,
, which is negative when ,
which proves that A is a maximum when .
Therefore, the base of the largest rectangle,
and its height is.
Edwin
