Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
267 views
in Perimeter and Area of Plane Figures by (25 points)
edited by

A piece of wire of length \( 20 cm \) is formed into the shade of sector of a circle of radius \( r cm \) and angle \( \theta \) radians.

a. show that \( \theta=\frac{20-2 r}{r} \) and that the area of the sector is \( r(10-r) cm ^{2} \). 

b. Hence find the values of \( r \) and \( \theta \) to give the maximum area 

Please log in or register to answer this question.

1 Answer

0 votes
by (44.1k points)
edited by

(a) Length of the wire is l = 20 cm

Given that wire is formed in to the shade of the sector of circle or radius r cm and angle θ.

l = 2r + \(\frac{2\pi r}{2\pi}\times\theta\)

l = 2r + rθ

2r + rθ = 20 (\(\because\) l = 20 cm)

rθ = \(\frac{20-2r}r\)---(1)

Area of sector = \(\frac{\pi r^2}{2\pi}\times\theta\) = \(\frac{r^2}2\times\frac{20-2r}r\) (From (1))

 =  \(\frac r2\times2(10-r)\) 

 = r (10 - r) cm2

(b) Area of sector is A = r (10 - r) = 10r - r2

\(\therefore\) \(\frac{dA}{dr}\) = 10 - 2r

\(\frac{dA}{dr}\) = 0 gives r = 5 \(\therefore\) Critical point is r = 5

\(\frac{d^2A}{dr^2} = -2 < 0\) 

\(\therefore\) Critical point r = 5 is local maximum of function A.

\(\therefore\) Maximum area = 5(10 -5)  = 5 x 5 = 25 cm2

Related questions

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

...