Rectangle ABCD has 2 vertices on y=-(1/4)x^2 + 9 and 2 vertices on y=(1/2)x^2 -18. find the maximum possible perimeter for rectangle ABCD.

napessos Dec 6, 2023

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Let P be the point of intersection of the parabolae y=−(1/4)x2+9 and y=(1/2)x2−18. Let A,B,C,D be the vertices of the rectangle such that A∈y=−(1/4)x2+9 and D∈y=(1/2)x2−18. Since the slope of y=−(1/4)x2+9 is negative, we know that A and D are consecutive vertices.

Since the slope of y=(1/2)x2−18 is positive, we know that B and C are consecutive vertices. To maximize the perimeter, we want to maximize the length of AD.

Let A=(−a,−0.25a2+9) and D=(a,0.5a2−18). Then AD=2a and the distance between the parabolae at x=a and x=−a is 0.75a2+9. Therefore, the area of rectangle ABCD is 2a⋅(0.75a2+9)=1.5a3+18a.

To maximize the area, we need to maximize 1.5a3+18a. Taking the derivative of this expression, we get 4.5a2+18=0. Solving this quadratic equation, we get a=−22 and a=22.

Since the area of rectangle ABCD is negative when a=−22, the maximum area of rectangle ABCD occurs when a=22. Therefore, the maximum perimeter of rectangle ABCD is 2⋅2*sqrt(2)⋅(0.75⋅4+9)=48*sqrt(2) square units.

BuiIderBoi Dec 6, 2023