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# Optimization Problems

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

### 1 Answer by Expert Tutors

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

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

A’ = –2(2x+5)(2x–5)–4x(4x)

A’ = –2(4x^2–25)–16x^2

A’ = –8x^2 + 50 – 16x^2

A’ = 50 – 24x^2

Solve for x_max:

50 – 24(x_max)^2 = 0

24(x_max)^2 = 50

(x_max)^2 = 50/24 = 25*3/(4*3*3) = 25*3/36

x_max = 5√(3)/6 ≈ 1.44337567297407

A_max = 2(5√(3)/6)(25 – 4(5√(3)/6)^2)

A_max = 2(5√(3)/6)(25 – 4(25*3/36))

A_max = 2(5√(3)/6)(25 – (25/3))

A_max = 2(5√(3)/6)(50/3) = 500√(3)/18

A_max = 250√(3)/9 ≈ 48.11252243247