Consider a rectangle QRST such that the side QR is parallel to the c-axis, the two vertices Q and R are on the y > 0 portion of the graph of the quadratic function y = -x^2 + 4, and the two...

Consider a rectangle QRST such that the side QR is parallel to the c-axis, the two vertices Q and R are on the y > 0 portion of the graph of the quadratic function y = -x^2 + 4, and the two other vertices S and T are on the y < 0 portion of the graph of the quadratic function y = 1/2x^2 - 2 as shown in the picture attached. Let P be the point of intersection of the side QT and the x-axis. Let 'ℓ' be the length of the perimeter of this rectangle. We are to find the x-coordinate of the point P where ℓ Is maximized and also to find the maximum value of ℓ.

Let P be (α,0), where 0 < α < 2. Since

PQ = a - α^2 , QR = b * α , PT= c - 1/d * α^2

we have

ℓ = e + f * α - g * α^2

Therefore, when α = h, ℓ is maximized and its value is i.

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Let P be at the point `(alpha,0)` such that `bar(PQ),bar(PT)` are perpendicular to the x-axis.

Since Q lies on `y=-x^2+4` and the x-coordinate of Q is `alpha` then the coordinates of Q are `(alpha,4-alpha^2)`

By symmetry the coordinates for R are `(-alpha,4-alpha^2)`

Since T lies on `y=1/2 x^2-2` the coordinates for T are `(alpha,1/2alpha^2-2)`

By symmetry the coordinates for S are `(-alpha,1/2 alpha^2-2)`

Then QR=ST=`2alpha`.

Also `PQ=4-alpha^2,PT=|1/2alpha^2-2|`

Now since the x-intercept of `y=-x^2+4` is 2 we know that `alpha <=2` so `|1/2alpha^2-2|=-1/2alpha^2+2`

so `QT=RS=6-3/2 alpha^2`

If `l` is the perimeter, the expression for `l` is:

`l=2(2alpha)+2(6-3/2alpha^2)=-3alpha^2+4alpha+12`

In order to maximize `l` , note that the expression for the perimeter is a quadratic. The graph would be a parabola opening down, so the maximum occurs at the vertex. The x-coordinate of the vertex of a parabola in the form `y=ax^2+bx+c` is `x=-b/(2a)` .

Thus `l` is maximized when `alpha=-4/(2(-3))=2/3`

With `alpha=2/3` then `l=-3(2/3)^2+4(2/3)+12=40/3`

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The maximal perimeter of a rectangle under the given restrictions is `40/3` units and occurs when the point P is at `(2/3,0)`

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