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LammettHash
The region bounded by the parabola and the x-axis is symmetric, so any inscribed rectangle whose base lies in the x-axis will have its base extend symmetrically around the origin, i.e. if the base has length [tex]2a[/tex], with [tex]a>0[/tex], then its base is the line segment connecting the points [tex](-a,0)[/tex] and [tex](a,0)[/tex].

The height of such a rectangle will then by [tex]4-(\pm a)^2=4-a^2[/tex].

The area of such a rectangle is then a function of [tex]a[/tex]:

[tex]A(a)=2a(4-a^2)=8a-2a^3[/tex]

Differentiating with respect to [tex]a[/tex] gives

[tex]A'(a)=8-6a^2[/tex]

which has critical points at

[tex]8-6a^2=0\implies a=\pm\dfrac2{\sqrt3}[/tex]

We omit the negative root. Checking the sign of the second derivative at the positive critical point (it's negative) confirms that [tex]a=\dfrac2{\sqrt3}[/tex] is the site of a local maximum.

This means the largest area of this rectangle is

[tex]A\left(\dfrac2{\sqrt3}\right)=\dfrac{32}{3\sqrt3}[/tex]
abidemiokin

The largest possible area of a rectangle with a base on the x-axis and vertices on curve y = 4 - x² is [tex]\frac{32}{3\sqrt{3} }[/tex]

The formula for finding the area of a rectangle is expressed as:

[tex]A=xy[/tex]

x is the length of the base

y is the vertices on the curve (width)

Since the area of a rectangle with its base on the x-axis, then the base we will use is 2x

The area of the rectangle will become:

[tex]A=2x(4-x^2)[/tex]

If the area is at the maximum, then [tex]\frac{dA}{dx} = 0[/tex]

[tex]A = 8x-2x^3\\\frac{dA}{dx}=8-6x^2[/tex]

[tex]8-6x^2=0\\6x^2=8\\x^2=\frac{8}{6} \\x^2=\frac{4}{3}\\x=\sqrt{\frac{4}{3} } \\x=\frac{2}{\sqrt{3} }[/tex]

To get the largest area, we will substitute [tex]x = \frac{2}{\sqrt{3} }[/tex] into the area

[tex]A=8x-2x^3\\A=8(\frac{2}{\sqrt{3}} )-2(\frac{2}{\sqrt{3} } )^3\\A=\frac{16}{\sqrt{3} } -\frac{16}{3\sqrt{3} } \\A=\frac{48-16}{3\sqrt{3} } \\A=\frac{32}{3\sqrt{3} }[/tex]

Hence the largest possible area of a rectangle with a base on the x-axis and vertices on curve y = 4 - x² is [tex]\frac{32}{3\sqrt{3} }[/tex]

Learn more here: https://brainly.com/question/2272944

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