Question

Find the dimensions of the rectangle of parameter 36 cm which is sweep out a volume a large as possible when revolved about one of its sides also find the maximum volume

Answer 1

Answer:

hey dear ,

Step-by-step explanation:

h/

To obtain

the absolute maxima or minima for the function

f(x)f(x)

(i) Find f′(x)f′(x) and put f′(x)=0f′(x)=0

(ii) Obtain the points from f′(x)=0f′(x)=0

(iii) By phthogoras theorem, find f′′(x)f″(x) and

check the value of f′′(x)f″(x) for each of the points obtained if f′′(x)>0f″(x)>0

Step 1

Perimeter of the rectangle is

p=36p=36

Let the length of the rectangle be = xx

Let the breadth of the rectangle be = yy

p=2(x+y)=36p=2(x+y)=36

⇒x+y=18⇒y=18−x⇒x+y=18⇒y=18−x

As the rectangle revolves, the

length = xx

width = 18−x18−x

height = xx

∴∴ volume of the cubid is

v=x×(18−x)×xv=x×(18−x)×x

=x2(18−x)=x2(18−x)

=18x2−x3=18x2−x3

Step 2

Differentiating w.r.t xx we get

dvdx=36x−3x2dvdx=36x−3x2

differentiating w.r.t xx we get,

d2vdx2=36−6xd2vdx2=36−6x

Step 3

For maximum or minimum we must have

dvdx=0dvdx=0

⇒36x−3x2=0⇒36x−3x2=0

⇒3x=36⇒3x=36

x=12x=12

∴y=18−12∴y=18−12

=6=6

∴(d2vdx2)(12,6)=36−6×12∴(d2vdx2)(12,6)=36−6×12

=36−72=36−72

−36<0−36<0

Hence vv is maximum when x=12x=12

Step 4

∴∴ The dimensions of the rectangle is

x=12andy=6x=12andy=6

The maximum volume of the box is given by

v=12×12×6v=12×12×6

=864=864 cubic units.

hope it's helpful for you .

Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer}}}}}}$

__________________________

$\huge{\fbox{\fbox{\bigstar{\underline{\red{Step\:1}}}}}}$

Perimeter of the rectangle is

p=36

Let the length of the rectangle be = x

Let the breadth of the rectangle be = y

p=2(x+y)=36

⇒x+y=18⇒y=18−x

As the rectangle revolves, the

length = x

width = 18−x

height = x

∴ volume of the cubid is

v=x×(18−x)×x

=x2(18−x)

=18x2−x3

$\huge{\fbox{\fbox{\bigstar{\underline{\red{Step\:2}}}}}}$

Differentiating w.r.t x we get

dvdx=36x−3x2

differentiating w.r.t x we get,

d2vdx2=36−6x

$\huge{\fbox{\fbox{\bigstar{\underline{\red{Step\:3}}}}}}$

For maximum or minimum we must have

dvdx=0

⇒36x−3x2=0

⇒3x=36

x=12

∴y=18−12

=6

∴(d2vdx2)(12,6)=36−6×12

=36−72

−36<0

Hence v is maximum when x=12

$\huge{\fbox{\fbox{\bigstar{\underline{\red{Step\:4}}}}}}$

∴ The dimensions of the rectangle is

x=12andy=6

The maximum volume of the box is given by

v=12×12×6

=864 cubic units

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Answer\:864\:cubic\:units}}}}}}$

__________________________

#ANSWERWITHQUALITY #BAL