Two parabolas y=16-3x^{2} and y=x^{2}-9 are drawn. A rectangle is contructed with sides parallel to the coordinate axes with the two upper vertices on the graph of y=16-3x^{2} and the two lower vertices on the graph of y=x^{2}-9. Determine the maximum possible area of this rectangle.

Steve S. | Tutoring in Precalculus, Trig, and Differential CalculusTutoring in Precalculus, Trig, and Diffe...

5.05.0(3 lesson ratings)(3)

0

Two parabolas y=16-3x^2 and y=x^2-9 are drawn. A rectangle is constructed with sides parallel to the coordinate axes with the two upper vertices on the graph of y=16-3x2 and the two lower vertices on the graph of y=x2-9. Determine the maximum possible area of this rectangle.

Both parabolas have x = 0 as their Axis of Symmetry, so we can find the half of the rectangle for x ≥ 0 and double it.

Upper right vertex is (x,16-3x^2) and

lower right vertex is (x,x^2-9).

Area of rectangle:

A = 2x(16-3x^2 – (x^2-9))

A = 2x(25 – 4x^2)

A = –2x(2x+5)(2x–5)

Zeros: x = –5/2, x = 0, x = 5/2

End behavior is down to the right and up to the left.

Domain of A is x ≥ 0, so local maximum will be between
x = 0 and x = 5/2. Use derivative to find it.

A’ = (–2x)’(2x+5)(2x–5)+(–2x)(2x+5)’(2x–5)+(–2x)(2x+5)(2x–5)’