1- The owner of a garden supply store wants to construct a fence to enclose a re

Question

1- The owner of a garden supply store wants to construct a fence to enclose a rectangular outdoor storage area adjacent to the store, using part of the side of the store (which is 270 feet long) for part of one of the sides. (See the figure below.) There are 470 feet of fencing available to complete the job. Find the length of the sides parallel to the store and perpendicular that will maximize the total area of the outdoor enclosure.

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just the answer .. :)

The owner of a garden supply store wants to construct a fence to enclose a rectangular outdoor storage area adjacent to the store, using part of the side of the store (which is 270 feet long) for part of one of the sides. (See the figure below.) There are 470 feet of fencing available to complete the job. Find the length of the sides parallel to the store and perpendicular that will maximize the total area of the outdoor enclosure. A box is to be made out of a 12 by 20 piece of cardboard. Squares of equal size will be cut out of each corner, and then the ends and sides will be folded up to form a box with an open top. Find the length L, width W, and height H of the resulting box that maximizes the volume. (Assume that W L).

Explanation / Answer

1

Since one side is the wall, we let x be perpendicular and y be parallel:
2x+y = 480

y=480-2x

Area is the the product of the two sides, thus.
xy=480x - 2x^2

for maximum, d(xy)/dx must be 0. thus,

480-4x=0

x= 120. thus, y = 240.

Since 240<280, the side of the store is not a constraint.

thus, the perpendicular sides are 120 feet long and the parallel side is 240 feet long.

2

When squares wide side lengths of "s" are cut away from the cardboard, the following dimensions for the folded box are created:
length = 20 - 2s (two squares' lengths cut away)
width = 8 - 2s (same thing)
height = s (the flaps are as wide as the squares were)

V = lwh = (20 - 2s)(8 - 2s)(s); multiply...
V = 4s

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