Find the area of the largest rectangle with one corner at the origin, the opposite corner in the first quadrant on the graph of the parabola f(x)=540−5x^2, and sides parallel to the axes

Question

The maximum possible area is?

Answer 1

Area of the rectangle will be base×height. The base will be x and the height will be f(x)

A(x) = xf(x) = 540x-5x³

The domain of A(x) is [0, √108]. √108 is where the parabola crosses the x-axis.

A'(x) = 540 - 15x²

Setting A'(x) to zero and solving for x:

x²=540/15 = 36

x = ±6, Use the positive value to be in the first quadrant. 6 is in the domain of A(x)

A(6) = 2160

Answer 2

Area S = xy; y = 540 - 5x²; S = 540x - 5x³; S' = 540 - 15x² = 0; x² = 36; x = 6 (because in 1st quadrant)

S'' = - 30x = - 180 < 0, so, we have maximum. S_max = 540·6 - 5·6³ = 2160