Question

A rectangle has its base on the x-axis and its upper two vertices on the parabola \(y = {\bf{4}} - {x^{\bf{2}}}\). What is the largest possible Area of the rectangle.

Short answer

The Area of the rectangle inside the parabola is \(6.16\) sq. units.

Step-by-step solution

Step 1-Find the function for Area of rectangle

The figure below represents the sketch of the parabola and rectangle inside it.

The Area of rectangle is:

\(A = 2xy\)

For \(y = 4 - {x^2}\),

\(\begin{aligned}{c}A = 2x\left( {4 - {x^2}} \right)\\ = 8x - 2{x^3}\end{aligned}\)

Step 2-Differentiate the function of Area

Differentiate the function \(A\left( x \right) = 8x - 2{x^3}\).

\(\begin{aligned}{c}A'\left( x \right) = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {8x - 2{x^3}} \right)\\ = 8 - 6{x^2}\\ = 2\left( {4 - 3{x^2}} \right)\end{aligned}\)

Step 3-Find the values of x at which Area is maximum

Find the roots of \(A'\left( x \right) = 0\).

\(\begin{aligned}{c}4 - 3{x^2} = 0\\{x^2} = \frac{4}{3}\\x = \frac{2}{{\sqrt 3 }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {0 < x < 2} \right)\end{aligned}\)

So, according to the first derivative test, the value of Area will be maximum when \(x = \frac{2}{{\sqrt 3 }}\).

Step 4-Find the maximum Area of the rectangle

Substitute \(\frac{2}{{\sqrt 3 }}\) for x in the equation \(A = 8x - 2{x^3}\).

\(\begin{aligned}{c}A = 8\left( {\frac{2}{{\sqrt 3 }}} \right) - 2{\left( {\frac{2}{{\sqrt 3 }}} \right)^3}\\ = \frac{{16}}{{\sqrt 3 }} - \frac{{16}}{{3\sqrt 3 }}\\ = \frac{{32}}{{3\sqrt 3 }}\\ \approx 6.16\end{aligned}\)

Thus, the Area of the rectangle inside the parabola is \(6.16\) sq. units.