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slader A rectangular poster is to contain 512 square inches of print. The margins at the top and bottom of the poster are to be 2 inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the poster be so that the least amount of poster is used

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Answer:

Dimensions of the poster are:

w  =  18 in

Y  =  36 in

Step-by-step explanation:

Print area of the rectangular poster is 512 in²

Let call "x"  and  "y"  dimensions for the print area of the poster then

A(p)  = Print area of the poster = x*y

512 = x*y     ⇒  y  = 512/x

Total area of the poster is:

A(t) = ( y + 4 ) * ( x + 2 )

A(t) = y*x +2*y +4*x + 8    And as  y = 512/x

Total area of the poster as a function of x is:

A(x)  =( 512/x)*x + 2* (512/x) + 4*x + 8

A(x)  = 512  + 1024/x + 4*x + 8    ⇒  A(x)  = 520  + 1024/x + 4*x

Taking derivatives on both sides of the equation we get:

A´(x)  =  - 1024/x² + 4

A´(x)  = 0         ⇒    - 1024 /x²  = -4    ⇒  4*x² = 1024

x² = 1024/4     ⇒  x²  = 256

x = 16 inches

And   y  =  512/x     ⇒  y = 512/16    ⇒ y = 32 inches

So we found x  and  y dimensions of the print area, then the dimensions of the poster are:

w = x + 2    ⇒  w  = 16  + 2   w  = 18 in

Y  = y + 4   ⇒  Y  = 32 + 4   Y  =  36 in

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