Question

The maximum area (in sq. units) of a rectangle having its base on the x-axis and its two other vertices on the parabola, $y=12-{{x}^{2}}$ such that the rectangle lies inside the parabola is
$\begin{align}
  & \left( A \right)20\sqrt{2} \\
 & \left( B \right)18\sqrt{3} \\
 & \left( C \right)32 \\
 & \left( D \right)36 \\
\end{align}$

Answer 1

Hint: We start solving this question by assuming the length and height of the rectangle as a and h. then we find the coordinates of the vertices. As two of the vertices are on the parabola, we substitute the point on the parabola in the equation of the parabola. Then we get a relation between a and h. Then we find the area of the rectangle using the formula $Area=\left( length \right)\times \left( breadth \right)$. We substitute the relation we got between a and h and differentiate the area with respect to a, to obtain the condition for maximum area. We substitute the obtained condition in the area formula to find the maximum area.

Complete step-by-step solution:
The equation of the parabola we were given is $y=12-{{x}^{2}}$.
As we see, the parabola is symmetric to the y-axis. So, to form a rectangle it also should be symmetric about the y-axis.
Let us assume that the length and height of the rectangle are a and b respectively.
As the rectangle is symmetric about x-axis, we can write the coordinates of the vertices on the base as $\left( -\dfrac{a}{2},0 \right)$ and $\left( \dfrac{a}{2},0 \right)$. As the height of the rectangle is h, the coordinates of vertices of rectangle on the parabola are $\left( -\dfrac{a}{2},h \right)$ and $\left( \dfrac{a}{2},h \right)$.
The graph of those looks like

As the points $\left( -\dfrac{a}{2},h \right)$ and $\left( \dfrac{a}{2},h \right)$ are on the parabola, they must satisfy the equation of the parabola. So, by substituting the point in the equation of the parabola we get,
$\begin{align}
  & \Rightarrow y=12-{{x}^{2}} \\
 & \Rightarrow h=12-{{\left( \dfrac{a}{2} \right)}^{2}} \\
 & \Rightarrow h=12-\dfrac{{{a}^{2}}}{4} \\
\end{align}$
So, now we got a relation between the length and height of the rectangle.
Now let us consider the formula for the area of the rectangle.
$Area=\left( length \right)\times \left( breadth \right)$
Using the above formula, we get area of our rectangle as,
$\begin{align}
  & \Rightarrow A=a \times h \\
 & \Rightarrow A=a\left( 12-\dfrac{{{a}^{2}}}{4} \right) \\
 & \Rightarrow A=12a-\dfrac{{{a}^{3}}}{4} \\
\end{align}$
As we need to find the maximum area of the rectangle, let us use optimization technique, a function $f\left( x \right)$ is maximum or minimum, if ${f}'\left( x \right)=0$.
So, let us differentiate the area of rectangle.
$\begin{align}
  & \Rightarrow \dfrac{dA}{da}=\dfrac{d}{da}\left( 12a-\dfrac{{{a}^{3}}}{4} \right)=0 \\
 & \Rightarrow 12-\dfrac{3{{a}^{2}}}{4}=0 \\
 & \Rightarrow \dfrac{3{{a}^{2}}}{4}=12 \\
 & \Rightarrow 3{{a}^{2}}=48 \\
 & \Rightarrow {{a}^{2}}=16 \\
 & \Rightarrow a=4 \\
\end{align}$
Now we need to find if the condition is for maximum.
The condition for maximum is a double derivative of the function should be negative.
As $\dfrac{dA}{da}=12-\dfrac{3{{a}^{2}}}{4}$, let us differentiate it with a to find the double derivative.
\[\begin{align}
  & \Rightarrow \dfrac{{{d}^{2}}A}{d{{a}^{2}}}=\dfrac{d}{da}\left( 12-\dfrac{3{{a}^{2}}}{4} \right) \\
 & \Rightarrow \dfrac{{{d}^{2}}A}{d{{a}^{2}}}=-\dfrac{6a}{4}=-\dfrac{3a}{2} \\
\end{align}\]
Now let us substitute the values obtained in the double derivative of volume.
 \[\Rightarrow {{\left. \dfrac{{{d}^{2}}A}{d{{a}^{2}}} \right]}_{h=\sqrt{3}}}=-\dfrac{3\left( 4 \right)}{2}=-6<0\]
As the double derivative is negative when \[a=4\]. It is the value for which the volume is maximum.
Now let us substitute the value of a in the formula of area of rectangle.
$\begin{align}
  & \Rightarrow A=12\left( 4 \right)-\dfrac{{{4}^{3}}}{4} \\
 & \Rightarrow A=48-\dfrac{64}{4} \\
 & \Rightarrow A=48-16 \\
 & \Rightarrow A=32 \\
\end{align}$
Hence, maximum area of rectangle is 32 square units.
Hence answer is Option C.

Note: The major point one needs to remember while solving this problem is that rectangle is also symmetrical with respect to the y-axis. Otherwise one will consider the co-ordinates of vertices on the x-axis as (-l, 0) and (a-l, 0). Then the process of solving the question becomes lengthy and unsolvable.