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

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- Apr 8th 2007, 08:53 PMcamherokidOptimization Problem4
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 - Apr 8th 2007, 10:59 PMearboth
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.