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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 \(2a\), with \(a>0\), then its base is the line segment connecting the points \((-a,0)\) and \((a,0)\).

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

The area of such a rectangle is then a function of \(a\):

\(A(a)=2a(4-a^2)=8a-2a^3\)

Differentiating with respect to \(a\) gives

\(A'(a)=8-6a^2\)

which has critical points at

\(8-6a^2=0\implies a=\pm\dfrac2{\sqrt3}\)

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

This means the largest area of this rectangle is

\(A\left(\dfrac2{\sqrt3}\right)=\dfrac{32}{3\sqrt3}\)

abidemiokin

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

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

\(A=xy\)

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:

\(A=2x(4-x^2)\)

If the area is at the maximum, then \(\frac{dA}{dx} = 0\)

\(A = 8x-2x^3\\\frac{dA}{dx}=8-6x^2\)

\(8-6x^2=0\\6x^2=8\\x^2=\frac{8}{6} \\x^2=\frac{4}{3}\\x=\sqrt{\frac{4}{3} } \\x=\frac{2}{\sqrt{3} }\)

To get the largest area, we will substitute \(x = \frac{2}{\sqrt{3} }\) into the area

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

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

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

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