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): You can put this solution on YOUR website!


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



And the vertical dimension is:




John


Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!


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



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