Arectangular page is to contain 24 square inches of print. the margin at the top and at the bottom of the page are 1.5 inches, and the margins on the left and on the right are 1 inch wide. what should be the dimensions of the paper so that the least amount of paper is used.

Dimensions of the paper

height   h  = 9 in

length    L  = 6 in

A(min)   =  54 in²

Step-by-step explanation:

Let   x  and  y  be dimensions of the print area then

x*y =  24  in²    then   y  =  24/x

The total area is:

Heigth (h)     h  =  y  + 2*1.5      h  =  y  + 3  

lenght  (L)      L  =  x + 2      

A(page)  =  h*L      then    

A  =  ( y  +  3 ) * ( x  +  2  )

A(x)  = ( 24/x  + 3 ) * ( x  + 2 )

A(x)  = 24  +  48/x  +  3x  +  6

A(x)  =  30  +  48 /x  +  3x

Taking derivatives on both sides of the equation

A´(x)  =  -48/x²  + 3

A´(x)  = 0           -48/x²  + 3  = 0       -48/x²  =  -3

x²  =  48/3     x²  = 16    x  = 4  in 

and  y   =  24 / 4    =   6  in

Then  dimensions of paper  

h =  y  +  3  

h  =  9 in

L  =  x  +  2

L  =  4  +  2

L  = 6 in

A(min)  =  9*6

A (min)  = 54  in²

4

step-by-step explanation:

6 (donuts) divided by 3 is 2.

so 2/3 means 2 times 2, which is 4/6.

you get two and your friend gets four.

the function makes a strait line horizontally with no x intercept but a y intercept at 9.