Application of Extremum in Plane Geometry and Solid geometry - Practice Questions & MCQ

Application of Extremum in Plane Geometry and Solid Geometry

There are some interesting applications of maxima and minima. Let’s go through some illustrations to understand more:

Illustration 1

Find the rectangle having a maximum area such that it is inscribed in the semicircle of radius $r$.
Let the centre of the semicircle be at the origin $(0,0)$
Area $($ Rectangle $)=\mathrm{A}=\mathrm{xy}$

From the figure

$
\begin{aligned}
& A=(2 r \cos \theta)(r \sin \theta) \\
& A=r^2 \sin 2 \theta
\end{aligned}
$
Differentiating it w.r.t. $\Theta$, and putting $\mathrm{f}^{\prime}(\mathrm{x})=0$, we can get the point of maxima

$
\theta=\pi / 4
$
Hence, $A=r^2($ when $\theta=\pi / 4)$
$\Rightarrow$ sides of rectangle $=2 r \cos \theta=2 r \cos (\pi / 4)=\sqrt{2} r$ and $r \sin \theta=r / \sqrt{2}$

Illustration 2

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is:

Radius of cylinder $=\sqrt{\mathrm{R}^2-\left(\frac{\mathrm{h}}{2}\right)^2}$
volume of cylinder $=\pi \mathrm{r}^2 \mathrm{~h}=\pi\left(\mathrm{R}^2-\frac{\mathrm{h}^2}{4}\right) \times \mathrm{h}$
for maximum volume

$
\begin{aligned}
& \frac{\mathrm{dv}}{\mathrm{dh}}=\pi\left(\mathrm{R}^2-\frac{3 \mathrm{~h}^2}{4}\right)=0 \\
& \Rightarrow \mathrm{~h}^2=\frac{4 \mathrm{R}^2}{3} \\
& \Rightarrow \mathrm{~h}=2 \sqrt{3} \quad[\text { as } \mathrm{R}=3]
\end{aligned}
$