Question

Find the maximum volume of a cylinder generated by rotating a rectangle of perimeter 48 cm about one of its sides

Answer 1

Solution:

Let L be length of rectangle and B be the breadth of rectangle.

Perimeter= 48 cm

2 (L +B)= 48

L +B= 24

L= 24 - B or B = 24 -L------(1)

1. If L> B

And rotate the rectangle along Length, we are getting the cylinder having Radius r, and height equal to Breadth B.

Then, 2 π r = L

r = $\frac{L}{2\pi }$

Volume of Cylinder = π r² h

$V_{1}$= Volume of Cylinder=$\pi r^2 B\\\\ \pi \times \frac{L^2}{4(\pi)^2}\times B\\\\ \frac{BL^2}{4\pi}   \\\\ V_{1}= \frac{(24-L)L^2}{2\pi } \\\\ V_{1}=(24 L^2 - L^3)\times \frac{1}{2 \pi }\\\\V'_{1}=(48L - 3L^2)\times\frac{1}{2\pi }$------Using (1)

For Maxima or Minima

 $V'_{1}$=0

48 L- 3 L²=0

48 = 3 L

Dividing both sides by 3, we get

L = 16

Putting the value of L in (1), we get

B = 24 - 16

B= 8

To check whether it is either maximum volume or minimum Volume

It must be $V"_{1}$<0

$[tex]V"_{1}=\frac{1}{2\pi }(48 -6L)=\frac{1}{2\pi}\times (-24)=-12 \pi$

Hence , at L= 16 and B= 8 , we are getting maximum volume of cylinder.Or,  at B= 16 and L= 8 , we are getting maximum volume of cylinder

So,Maximum volume of cylinder when Length= 16 cm, Breadth=8 cm is given by

=$V_{1}= \frac{BL^2}{4\pi  }\\\\= \frac{8 \times 16\times 16}{4 \times 3.14}\\\\ V_{1}=163.057$  square cm

If you will replace L that is length by Breadth in the above equation you will get the same result.