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A rectangular display in a furniture store needs walls on three sides only, with 6-foot openings on the two facing sides. If the display needs to have 450 ft of floor space, then what is the least number of linear feet of wall material that can be used to create the display? r

A rectangular display in a furniture store needs walls on three sides only, with 6-foot openings on the two facing sides. If the display needs to have 450 ft of floor space, then what is the least number of linear feet of wall material that can be used to create the display? r

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Question:** A rectangular display in a furniture store needs walls on three sides only, with 6-foot openings on the two facing sides. If the display needs to have 450 square feet of floor space, then what is the least number of linear feet of wall material that can be used to create the display? **Diagram Explanation:** The diagram represents a rectangular area with three sides having walls. It shows: - Two vertical walls labeled "L" on the left and right, each with a 6-foot gap at the entrance. - A horizontal wall labeled "W" at the top connecting the two vertical sides. - The floor space is required to be 450 square feet. To solve, let L be the length of the vertical sides and W be the width. The equation for area becomes W * L = 450. The total wall length is W + 2(L - 6), which is minimized given the constraints.
Transcribed Image Text:**Question:** A rectangular display in a furniture store needs walls on three sides only, with 6-foot openings on the two facing sides. If the display needs to have 450 square feet of floor space, then what is the least number of linear feet of wall material that can be used to create the display? **Diagram Explanation:** The diagram represents a rectangular area with three sides having walls. It shows: - Two vertical walls labeled "L" on the left and right, each with a 6-foot gap at the entrance. - A horizontal wall labeled "W" at the top connecting the two vertical sides. - The floor space is required to be 450 square feet. To solve, let L be the length of the vertical sides and W be the width. The equation for area becomes W * L = 450. The total wall length is W + 2(L - 6), which is minimized given the constraints.
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