Question

The area (in sq. units) of the largest rectangle \[ABCD\] whose vertices \[A\] and \[B\] lie on the \[x\] -axis and vertices \[\;C\] and \[\;D\] lie on the parabola \[y = {x^2} - 1\] below the x -axis, is
A.\[\dfrac{2}{{3\sqrt 3 }}\]
B.\[\dfrac{4}{3}\]
C.\[\dfrac{1}{{3\sqrt 3 }}\]
D.\[\dfrac{4}{{3\sqrt 3 }}\]

Answer 1

Hint: We will draw a diagram for the given problem. As the point \[A\] and \[B\]lie on the \[x\] -axis, then we can assume the coordinate of the points. Similarly, we will assume the coordinate of the points \[\;C\] and \[\;D\]. By using distance formula \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]we can find the length of the sides of the rectangle. We will find the area of the rectangle and from the derivative of the area we can find the maximum area.

Formula Used:
The distance formula is distance formula \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
The derivative formulas are
\[\dfrac{d}{{dx}}\left( {u \cdot v} \right) = u \cdot \dfrac{{dv}}{{dx}} + v \cdot \dfrac{{du}}{{dx}}\]
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
To find the maxima or minima of a function \[f\], we equate the first order derivative of \[f\]to \[0\].

Complete answer:
1.Points \[\;C\] and \[\;D\] lie on the parabola \[y = {x^2} - 1\]. So the line joining by the points \[\;C\] and \[\;D\] must be parallel to \[x\] -axis. The \[y\] -axis divides the parabola equally.
2. So the \[x\] coordinate of \[\;C\] is the same as the \[x\] coordinate of \[\;D\] with a negative sign. Similarly the \[x\] coordinate of \[\;B\] is the same as the \[x\] coordinate of \[\;A\] with a negative sign.
3. Also the the \[x\] coordinate of \[\;C\]is same as the \[x\] coordinate of \[\;B\] and the \[x\] coordinate of \[\;D\]is same as the \[x\] coordinate of \[\;A\].
4. We assume the coordinates of the points \[\;A\] and \[\;B\] as \[\left( {a,0} \right)\] and \[\left( { - a,0} \right)\] respectively.

We will put \[x = a\] in the parabola equation\[y = {x^2} - 1\].
\[y = {a^2} - 1\].
We get the \[y\]coordinate of the points \[\;C\] and \[\;D\] . Let us assume the coordinates of \[\;C\] and \[\;D\]such that \[\left( { - a,{a^2} - 1} \right)\] and \[\left( {a,{a^2} - 1} \right)\] respectively.
Now we will apply distance formula \[d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \] to calculate length of \[AB\] and \[BC\].
The length of \[AB\] is
\[AB = \sqrt {{{\left( { - a - a} \right)}^2} + {{\left( {0 - 0} \right)}^2}} \] [Here \[{x_1} = a,{y_1} = 0;{x_2} = - a,{y_2} = 0\]]
\[\begin{array}{l} = \sqrt {{{\left( {2a} \right)}^2}} \\ = 2a\,{\rm{units}}\end{array}\]
The length of \[BC\] is
\[BC = \sqrt {{{\left( { - a - \left( { - a} \right)} \right)}^2} + {{\left( {{a^2} - 1 - 0} \right)}^2}} \] [Here \[{x_1} = - a,{y_1} = 0;{x_2} = - a,{y_2} = {a^2} - 1\]]
\[\begin{array}{l} = \sqrt {{{\left( {{a^2} - 1} \right)}^2}} \\ = \left( {{a^2} - 1} \right)\,{\rm{units}}\end{array}\]
The area of the rectangle \[ABCD\] is \[A = {\rm{length}} \times {\rm{width}}\]
\[{\rm{ = 2a}}\left( {{a^2} - 1} \right)\] sq. units
The derivative formulas are
\[\dfrac{d}{{dx}}\left( {u \cdot v} \right) = u \cdot \dfrac{{dv}}{{dx}} + v \cdot \dfrac{{du}}{{dx}}\]
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
We will find the derivative of \[A = 2a\left( {{a^2} - 1} \right)\] with respect to \[a\].
\[\dfrac{{dA}}{{da}}{\rm{ = 2a}} \times {\rm{2a + 2}}\left( {{a^2} - 1} \right)\]
 \[\begin{array}{l}{\rm{ = }}4{a^2}{\rm{ + 2}}{a^2} - 2\\ = 6{a^2} - 2\end{array}\]
Again derivative with respect to \[a\]
\[\dfrac{{{d^2}A}}{{d{a^2}}}{\rm{ = 12a}}\]
We can get maxima or minima point of any function when \[\dfrac{{dA}}{{da}} = 0\].
So, \[6{a^2} - 2 = 0\]
\[a = \pm \dfrac{1}{{\sqrt 3 }}\]
Now we will put values of \[a\] in \[\dfrac{{{d^2}A}}{{d{a^2}}}{\rm{ = 12a}}\] to find maxima or minima point
Putting \[a = - \dfrac{1}{{\sqrt 3 }}\]
\[\dfrac{{{d^2}A}}{{d{a^2}}}{\rm{ = 12}}\left( { - \dfrac{1}{{\sqrt 3 }}} \right) < 0\]
Since \[\dfrac{{{d^2}A}}{{d{a^2}}} < 0\]at \[a = - \dfrac{1}{{\sqrt 3 }}\], so \[A\] has maxima value at \[a = - \dfrac{1}{{\sqrt 3 }}\] .
Putting \[a = \dfrac{1}{{\sqrt 3 }}\]
\[\dfrac{{{d^2}A}}{{d{a^2}}}{\rm{ = 12}}\left( {\dfrac{1}{{\sqrt 3 }}} \right) > 0\]
Since \[\dfrac{{{d^2}A}}{{d{a^2}}} > 0\]at \[a = \dfrac{1}{{\sqrt 3 }}\], so \[A\] has minimum value at \[a = \dfrac{1}{{\sqrt 3 }}\] .
Now we will put \[a = \dfrac{1}{{\sqrt 3 }}\] in \[A = 2a\left( {{a^2} - 1} \right)\]
\[A = 2 \cdot \left( { - \dfrac{1}{{\sqrt 3 }}} \right)\left[ {{{\left( {\dfrac{1}{{\sqrt 3 }}} \right)}^2} - 1} \right]\]
\[ = - \dfrac{2}{{\sqrt 3 }}\left[ {\dfrac{1}{3} - 1} \right]\]
  \[ = \dfrac{4}{{3\sqrt 3 }}\,\,{\rm{sq}}{\rm{.}}\,{\rm{units}}\]
The maximum area of the rectangle is\[\dfrac{4}{{3\sqrt 3 }}\,\,{\rm{sq}}{\rm{.}}\,{\rm{units}}\]. Hence option D is correct.

Note: Students are often confused with the maxima and minima points of a function. Remember that if \[\dfrac{{{d^2}F}}{{d{x^2}}} < 0\] at \[x = a\], then the value of the function at \[x = a\] will be maximum and if \[\dfrac{{{d^2}F}}{{d{x^2}}} > 0\] at \[x = a\], then the value of the function at \[x = a\] will be minimum.