Chapter 4, Problem 41RE

To determine

Tofind:thesmallest possible area of an isosceles that can be circumscribed in a circle.

Answer to Problem 41RE

The minimum area is $\text{3}\sqrt{\text{3}}{\text{r}}^{\text{2}}$ .

Explanation of Solution

Given:

Radius of circle is $\text{r}$ .

Concept used:

Area of triangle $\text{=}\frac{\text{1}}{\text{2}}\text{×base×height}$ .

Trigonometric formula:

  $\text{tan2θ=}\frac{\text{2tanθ}}{{\text{1-tan}}^{\text{2}}\text{θ}}$ .

If $\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}\text{>0}\text{.}$ the concave will be open upward and local minima can be found.

If $\frac{{\text{d}}^2\text{y}}{{\text{dx}}^2}\text{<0}\text{.}$ the concave will open downward and local maxima can be found.

Calculation:

By making a diagram of the circle inscribed in the isosceles triangle

The radius of circle $\text{r}$ will tends to perpendicular and base will be the half of base of the isosceles triangle let it be denoted by d.

Through this it can be solved as:

  $\text{tanθ=}\frac{\text{r}}{\text{d}}$ or $\text{tan2θ=}\frac{\text{h}}{\text{d}}$ .

  $\text{dtan2θ=h}$ .

Area of triangle $\text{=}\frac{\text{1}}{\text{2}}\text{×base×height}$ .

  $\Rightarrow \frac{\text{1}}{\text{2}}\text{×}\left(\text{2d}\right)\text{×h=d×h=d×dtan2θ}$ .

Using trigonometric formula:

  $\text{tan2θ=}\frac{\text{2tanθ}}{{\text{1-tan}}^{\text{2}}\text{θ}}$ .

Area= $\frac{{\text{2d}}^{\text{2}}\text{tanθ}}{{\text{1-tan}}^{\text{2}}\text{θ}}$ .

Substituting $\text{tanθ=}\frac{\text{r}}{\text{d}}$ .

  $\Rightarrow \frac{{\text{2d}}^{\text{2}}\left(\frac{r}{d}\right)}{\text{1-}{\left(\frac{r}{d}\right)}^2}$ .

  $\Rightarrow \frac{{\text{2rd}}^3}{d^2-r^2}$ .

  $A\left(d\right)=\frac{{\text{2rd}}^3}{d^2-r^2}$ .

Differentiating with respect to d to get:

  $\text{<msup><mo>A</mo><mo>′</mo></msup>}\left(\text{d}\right)\text{=}\frac{\left({\text{d}}^{\text{2}}{\text{-r}}^{\text{2}}\right)\left({\text{6rd}}^{\text{2}}\right)\text{-}\left({\text{2rd}}^{\text{3}}\right)\left(\text{2d}\right)}{\left({\text{d}}^{\text{2}}{\text{-r}}^{\text{2}}\right)}$ .

  $\text{<msup><mo>A</mo><mo>′</mo></msup>}\left(\text{d}\right)\text{=}\frac{{\text{2rd}}^{\text{2}}}{\left({\text{d}}^{\text{2}}{\text{-r}}^{\text{2}}\right)}\left({\text{d}}^{\text{2}}{\text{-3r}}^{\text{2}}\right)$

  $\text{<msup><mo>A</mo><mo>′</mo></msup>}\left(\text{d}\right)\text{=}\frac{{\text{2rd}}^{\text{2}}}{\left({\text{d}}^{\text{2}}{\text{-r}}^{\text{2}}\right)}\left(\text{d+}\sqrt{\text{3}}\text{r}\right)\left(\text{d-}\sqrt{\text{3}}\text{r}\right)$ .

When $\text{d<}\sqrt{\text{3}}\text{r,A}\left(\text{d}\right)$ is decreasing.

When $\text{d>}\sqrt{\text{3}}\text{r,A}\left(\text{d}\right)$ is increasing.

Therefor the $\text{d=}\sqrt{\text{3}}\text{r}$ is Minima.

  $\text{A}\left(\sqrt{\text{3}}\text{r}\right)\text{=}\frac{\text{2r}{\left(\sqrt{\text{3}}\text{r}\right)}^{\text{3}}}{{\left(\sqrt{\text{3}}\text{r}\right)}^{\text{2}}{\text{-r}}^{\text{2}}}$ .

  $\Rightarrow \frac{\text{6}\sqrt{\text{3}}{\text{r}}^{\text{4}}}{{\text{3r}}^{\text{2}}{\text{-r}}^{\text{2}}}$ = $\Rightarrow \frac{\text{6}\sqrt{\text{3}}{\text{r}}^{\text{4}}}{{\text{2r}}^{\text{2}}}$ = $\text{3}\sqrt{\text{3}}{\text{r}}^{\text{2}}$ .

Hence the minimum area is $\text{3}\sqrt{\text{3}}{\text{r}}^{\text{2}}$ .