A rectangular page is to contain 36 square inches of print. The margins on each side are 1 inch. Find the dimensions of the page such that the least amount of paper is used.

The length are breadth is x = 6 and y = 6 inch respectively.

Given;

Each rectangular page is to contain 36 square inches of print.

The margins on each side are 1 inch.

We have to find ;

The dimensions of the page such that the least amount of paper is used.

 Therefore,

Let, length and breadth of paper are x and y respectively.

Area is given by :

Area = xy    

xy = 36 inch²

Now, it is given than the margins on each side is 1 inch.

So, Area of paper is :  A = ( x + 2 )( y + 2 )

We want area to be minimum.

Putting value of x from 1 , we get :

A = ( + 2 )( y + 2 )

Now, Differentiating above equation and equating with 0 :

then;  xy = 36

        6x = 36

         x =  

         x = 6inches

Now, we know dimension cannot be negative.

So, y = 6 and x = 6 .

Therefore, length and breadth are 6 and 6 inch respectively.

For more information about the Area of dimension click the link given below.

Given :

A rectangular page is to contain 36 square inches of print.

The margins on each side are 1 inch.

To Find :

The dimensions of the page such that the least amount of paper is used.

Solution :

Let, length and breadth of paper are x and y respectively.

Area is given by :

Area = xy    ....1)

xy = 36 inch²

Now, it is given than the margins on each side is 1 inch.

So, area of paper is :

A = ( x + 2 )( y + 2 )

We want area to be minimum.

Putting value of x from 1 , we get :

A = ( 36/y + 2 )( y + 2 )

Now, differentiating above equation and equating with 0 :

Now, we know dimension cannot be negative.

So, y = 6 and x = 6 .

Therefore, length and breadth are 6 and 6 inch respectively.

just needed some points but try socratic

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.