Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

A rectangle is inscribed in a semi-circle of radius

r

with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area ?

Accepted Answer

Let the dimensions of the rectangle be $x$ and $y$ . Then,

$\frac{x^{2}}{4} + y^{2} = r^{2}$

$\Rightarrow x^{2} + 4 y^{2} = 4 r^{2}$

$\Rightarrow x^{2} = 4 (r^{2} - y^{2})$

Area of rectangle $= xy$

$\Rightarrow A = xy$

Squaring both sides, we get

$\Rightarrow A^{2} = x^{2} y^{2}$

$\Rightarrow Z = 4 y^{2} (r^{2} - y^{2})$ [From eq. $(1)$ ]

$\Rightarrow \frac{dZ}{dy} = 8 {yr}^{2} - 16 y^{3}$

For the maximum or minimum values of $Z$ , we must have

$\frac{dz}{dy} = 0$

$\Rightarrow 8 {yr}^{2} - 16 y^{3} = 0$

$\Rightarrow 8 r^{2} = 16 y^{2}$

$\Rightarrow y^{2} = \frac{r^{2}}{2}$

$\Rightarrow y = \frac{r}{\sqrt{2}}$

Substituting the value of $y$ in eq. $(1)$ , we get

$\Rightarrow x^{2} = 4 (r^{2} - {(\frac{r}{\sqrt{2}})}^{2})$

$\Rightarrow x^{2} = 4 (r^{2} - \frac{r^{2}}{2})$

$\Rightarrow x^{2} = 4 (\frac{r^{2}}{2})$

$\Rightarrow x^{2} = 2 r^{2}$

$\Rightarrow x = r \sqrt{2}$

Now,

$\frac{d^{2} Z}{{dy}^{2}} = 8 r^{2} - 48 y^{2}$

$\Rightarrow \frac{d^{2} Z}{{dy}^{2}} = 8 r^{2} - 48 (\frac{r^{2}}{2})$

$\Rightarrow \frac{d^{2} Z}{{dy}^{2}} = - 16 r^{2} < 0$

So, the area is maximum when $x = r \sqrt{2}$ and $y = \frac{r}{\sqrt{2}}$

Area $= xy$

$\Rightarrow A = r \sqrt{2} \times \frac{r}{\sqrt{2}}$

$\Rightarrow A = r^{2}$ .

A rectangle is inscribed in a semi-circle of radius $r$ with one of its sides on diameter of semi-circle. Find the dimension of the rectangle so that its area is maximum. Find also the area ?

Important Questions on Maxima and Minima

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