SOLUTION: Rectangle QRST between curve {{{y=-x^2+4}}} and {{{y=x^2-4}}}. Let P be the point of intersection of the side QT and the x-axis. Let α be the length of the perimeter of this r

Algebra ->  Points-lines-and-rays -> SOLUTION: Rectangle QRST between curve {{{y=-x^2+4}}} and {{{y=x^2-4}}}. Let P be the point of intersection of the side QT and the x-axis. Let α be the length of the perimeter of this r      Log On


   

Question 750024: Rectangle QRST between curve y=-x%5E2%2B4 and y=x%5E2-4. Let P be the point of intersection of the side QT and the x-axis. Let α be the length of the perimeter of this rectangle. We are to find the x-coordinate of the point P where α is maximized and also to find the maximum value of α. P(x,0), where 0 Therefore, when x=(A), α is maximized and its maximum value is (B)
solve for A and B
((this is the picture of the graph http://i44.tinypic.com/30sk9ir.jpg)
Answer by KMST(5398) About Me  (Show Source):
You can put this solution on YOUR website!
Both curves and the rectangle QRST are symmetrical with respect to the y-axis.
TI'll call the coordinates of P (a,0), to distinguish that x-coordinate value a from the variable x%7D%7D.%0D%0Ahe+x-coordinates+of+points+Q%2C+P+and+T+are+the+same+%7B%7B%7Bx=a.
The x-coordinates of points R and S are the same x=-a.
The y-coordinate of points Q and R, on curve y=-x%5E2%2B4 is y=-a%5E2%2B4.
The y-coordinate of points S and T, on curve y=x%5E2-4 is y=a%5E2-4.
The width ST (or QR) of the rectangle is w=2a.
The height RS (or QT) of the rectangle is h=-a%5E2%2B4-%28a%5E2-4%29=-a%5E2%2B4-a%5E2%2B4=-2a%5E2%2B8=2%28-a%5E2%2B4%29.
The perimeter of the rectangle is
alpha=2%28w%2Bh%29=2%282a%2B2%28-a%5E2%2B4%29%29=-4a%5E2%2B4a%2B16
alpha is a quadratic function in a
It's maximum is at a=-4%2F%282%28-4%29%29=highlight%281%2F2%29
because a parabola such as y=ax%5E2%2Bbx%2Bc has a vertex as x=-b%2F2a
The equation in vertex form would be
alpha=-4%28a-1%2F2%29%5E2%2Bhighlight%2817%29
-4%28a-1%2F2%29%5E2%2B17=-4%28a%5E2-a%2B1%2F4%29%2B17=-4a%5E2%2B4a-1%2B17=-4a%5E2%2B4a%2B16
So the maximum value of alpha happens at highlight%28A=1%2F2%29
and the maximum value of alpha is highlight%28B=17%29