IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

pottermania1990

find the area of the largest rectangle that can b inscribed in a semi circle of radius r .

uzairmehdi

is it r2/2 units

magiclko

see the attached figure,

here if area = a = 2xy

we have r² = x²+ y²

or we can have it as

a² = 4r²x²- 4x⁴

differentiatin wrt to x, nd then equating da/dx = 0, u'll get the value of x, nd then on substituting it for a, u'll get the value of a

its exactly the same as ur previous one... so m nt solving it whole, as its time taking, do knock again, if u find any prob

!!!

iitkgp_bipin

The largest rectangle should have one of the sides on the diameter. Let this side be x and the other side be y.

Draw a figure and you will find that :
r² = (x/2)² + y²

Now area is give by :
A² = x²y² = x²{r² - (x/2)²}

Differentiate it wrt x and put dA/dx = 0 to find the maxima.

This will give x = r

2

Hence A² = (r

2)²{r² - (r

2)²/2²} = r⁴

Hence maximum area = r²

BuddyGuy

U can also solve it using

as the variable.(see figure)

'r' being the radius of the given semicircle,

Let 'l' and 'b' be the length and breadth of the rectangle.

So,l=2rcos

b=rsin

so, area A=2r²sin

cos

=r²sin2

dA/d

= r²cos2

x2

for maximum area,

dA/d

=0

or,r²cos2

x2

i.e,cos2

=0

or,

=

/4 ... 0<

<

/2

putting the value of

in the length and breadth ,u get the dimensions of the rectangle

l=

2xr

b=r/

2

so,

A=r²

BuddyGuy

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