Smallest Area Poster Dimensions: Solve the Puzzle!

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SUMMARY

The discussion focuses on determining the dimensions of a poster that minimizes area while maintaining a fixed printed area of 382 square centimeters. The correct approach involves recognizing that the actual dimensions of the poster are l + 4 cm and w + 8 cm, where l and w are the dimensions of the printed area. The formula for the total area is expressed as (l + 4)(w + 8), leading to the equation lw + 4w + 8l + 32. The goal is to minimize this expression by substituting w with 382/l, resulting in a function that can be minimized.

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the top and bottom margins of a poster are 2cm and the side margins are each 4cm. If the area of printed material on the poster is fixed at 382 square centimeters, find the dimensions of the poster with the smallest area.

so the area l*w = 382

and the perimiter = 2(l-4)+2(w-8)

but if i try to solve for l i get a negative number. can someone help me out?
 
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You are setting it up wrong. You say " l*w = 382" but 382 is the area of the printed material so the actual dimensions are l+4 and w+8. Also the perimeter is irrelevant. You are asked to find l and w such that the total area, (l+ 4)(w+ 8)= lw+ 4w+ 8l+ 32, is a minimum.

From lw= 382, w= 382/l and that becomes 382+ 4(382)/l+ 8l+ 32= 6112/l+ 8l+ 412. Minimize that.

Finally, remember that the answer, the dimensions of the poster, are l+ 4 and w+ 8, not l and w.
 

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