Find the dimension of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola
y=8-x^2
Hello,
I've attached an image of the parabola and the rectangle.
One point on the parabola has the coordinates P(s, 8-s²). The the area of the rectangle is:
a = 2s * (8-s²) = 16s - 2s³
Calculate the first derivative:
a'(s) = 16 - 6s²
You'll get an extremum (maximum or minimum) if a'(s) = 0. Thus s = √(8/3)
Therefore the dimensions of the rectangle are: l = 2√(8/3), w = 16/3, a_(max) = 32/3*√(8/3) ≈ 17.42
With this solution I assumed that the rectangle is above the x-axis. Otherwise the area of the rectangle is unlimited.