# You are designing a rectangular poster to contain 50 in^2 of printing with a 4-in. margin at the top and bottom and a 2-in margin at each side. What overall dimensions will minimize the amount of paper used?

18" x 9"

Also from the same equation on simplifying, it is xy-8y-4x+32=50.

Since the area of the paper of size x inches by y inches is xy, let it be denoted as A. Thus A-8y-4x=18 Or A= 8y+4x+18

Dimension of minimum paper size would be 18 inches by 9 inches.

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To minimize the amount of paper used, the area of the printed portion should be maximized. Let the width of the printed portion be ( x ) inches and the height be ( y ) inches.

Given: Total area of the poster = 50 in^2 Margins:

- Top and bottom: 4 in each
- Sides: 2 in each

Area of printed portion = ( x \times y ) Total width = ( x + 4 + 4 ) = ( x + 8 ) Total height = ( y + 2 + 2 ) = ( y + 4 )

Total area of the poster in terms of ( x ) and ( y ): [ (x + 8) \times (y + 4) = 50 ]

To minimize the amount of paper used, maximize the area of the printed portion. We can rewrite the area of the printed portion in terms of one variable using the above equation:

[ y = \frac{50}{x + 8} - 4 ]

Area of printed portion: [ A = x \times \left( \frac{50}{x + 8} - 4 \right) ]

To find the minimum area, differentiate ( A ) with respect to ( x ), set it to zero, and solve for ( x ).

[ \frac{dA}{dx} = \frac{50}{(x + 8)^2} - \frac{50}{x + 8} ]

Setting ( \frac{dA}{dx} = 0 ): [ \frac{50}{(x + 8)^2} - \frac{50}{x + 8} = 0 ]

Solving the above equation will give the value of ( x ). Once ( x ) is found, substitute it back into the equation for ( y ) to get the corresponding ( y ) value.

Solving for ( x ) and ( y ) will give the dimensions that minimize the amount of paper used.

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