Question

A rod of 108 meters long is bent to form a rectangle. Find its dimensions if the area is maximum.

Answer 1

lenght multi breadth multi jeight or thickness

Answer 2

Answer:

Area is Maximum when Length = 27 m  and Breadth = 27 m.

Step-by-step explanation:

Given: Length of rod = 108 m

           Rod bent into a Rectangle.

To find: Dimensions of rectangle of maximum area

As Rod bent into a RectabngleThe Length of Rod becomes Perimeter of Rod.

Perimeter of Rectangle = 108 m

2 × ( Length + Breadth ) = 108

Length + Breath = 54 m

l + b = 54

b = 54 - l ......... (1)

Area of Rectangle = Length × Breadth

A = l × b

Now, we Derivate the area,

$^{\frac{\mathrm{d}A}{\mathrm{d} l}}= ^{\frac{\mathrm{d}(l\times b)}{\mathrm{d} l}}\\\\=\:^{\frac{\mathrm{d}(l\times (54-l))}{\mathrm{d} l}}$   ( From 1)

$=\:^{\frac{\mathrm{d}(54l-l^2)}{\mathrm{d} l}}\\\\=\:^{\frac{\mathrm{d}(54l)}{\mathrm{d} l}}-^{\frac{\mathrm{d}(l^2)}{\mathrm{d} l}}\\=\:54-2l$

For area to be maximum , we put $^{\frac{\mathrm{d}A}{\mathrm{d} l}}=0$

54 - 2l = 0

54 = 2l

l = $\frac{54}{2}$

l = 27 m

b = 54 - 27  ( from 1 )

  = 27 m

$\frac{\mathrm{d}^2 A}{\mathrm{d} l^2}=\:\frac{\mathrm{d}(54-2l)}{\mathrm{d} l}\\\\\frac{\mathrm{d}^2 A}{\mathrm{d} l^2}=\:\frac{\mathrm{d}(54)}{\mathrm{d} l}-\frac{\mathrm{d}(2l)}{\mathrm{d} l}\\\\\frac{\mathrm{d}^2 A}{\mathrm{d} l^2}=0-2\\\\\frac{\mathrm{d}^2 A}{\mathrm{d} l^2}=-2 \leq0$

Therefore, Area is Maximum when Length = 27 m  and Breadth = 27 m.