0 votes
320 views
in Derivatives by (37.2k points)
closed by

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

1 Answer

+1 vote
by (34.6k points)
selected by
 
Best answer

Let ABCD be the rectangle which is inscribed in a semi - circle with centre O and radius r.

Assume that the length of rectangle is 2x and breadth is 2y.

⇒ CD = AB = 2x

In right angle ΔACD,We get 

x2 + 4y2 = r...(i)

(By Pythagoras theorem)

Now area of rectangle ABCD is given by,

A = length x breadth

⇒  \(A=2x\times2y=4xy\) 

Let Z = A2 = 4(r2x2 - x4) ...(ii)

Then Z is maximum or minimum accordings as A is maximum or minimum.

Differentiating equation (ii) with respect to x, we get

\(\frac{dZ}{dx}=\) 4[r2 . 2x - 4x3]

For maximum or minimum value of Z, we have

\(\frac{dZ}{dx}=0\) 

⇒  r2 - 2x2 = 0

( ∵ x cannot be zero)

⇒  \(x^2=\frac{r^2}{2}\)

Now,

Thus area will be maximum when,

\(x=\frac{r}{\sqrt 2}\)

⇒  \(2x =\sqrt 2r,\)

Putting \(x=\frac{r}{\sqrt 2}\) in equation (i), we obtain,

\(y=\frac{r}{2\sqrt 2}\) 

⇒ 2y = \(\frac{r}{\sqrt 2}\) 

Now,

A = \(2x \times 2y\)

= \(\sqrt2r\,\times\,\frac{r}{\sqrt2}\) 

= r

Hence, the dimensions are \(\sqrt2r\) and \(\frac{r}{\sqrt2}\) and area = r2 sq units.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...