Arectangular poster is to contain 81 square inches of print. the margins at the top and bottom and on each side are to be 5 inches. find the dimensions of the page which will minimize the amount of paper used.

lenght   19 in

width     19 in

Step-by-step explanation:

Let x  and y be dimensions of print area, that is

x*y  =  81 in²          then   y = 81/x

then total area of the poster is (including top, bottom and sides margins)

( x  +  10  )*( y  +  10 ) = A(t)

A(x)  =  ( x  +  10  )*( 81/x  +  10 )   solving

A(x)  =   81  + 10x + 810/x + 100

Taking derivatives both sides of the equation

A´(x)  =  10  - 810/x²

A´(x)  = 0                10  - 810/x²  =  0       (10x²-810)/x²  =  0

10x² -  810  = 0

x²  = 81

x  = 9 in      then   y  = 81/x      y = 81/9     y  = 9 in

Then remember, these are values for the print area, now we have to add 10 inches according to margins

The dimensions of the poster will be

lenght   19 in

width     19 in

i dont know sorry

step-by-step explanation:

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.