beach shelter canvas image
sdt3
The canvas wind shelter is to be constructed for use on Padre Island beaches. It is to have a back, 2 square ft of canvas are available, find the length of the shelter for which the space inside is maximized assuming all the canvas is used.
A) length=15 ft
B) length=17ft
C) length=18ft
D) length=16ft
E) none of these
Answer
I think you need to supply more information.
Is the shelter a triangle (like an upside-down V) with one end solid?
Also, 2 square feet total fabric doesn't make sense. 200 maybe?
The very first thing you need to do is compute the area of the cross-section as a function of the length of the side and the height. The largest cross-section comes from a tent with a side angle of 45 degrees (optimise area as a function of angle by taking the derivative).
At this point, you will develop a formula for the volume of the shelter as a function of length, given the constant amount of fabric available using the two formulas:
V = area*length (area is the cross section)
fabric amount = side length*2*length + end panel
I think you need to supply more information.
Is the shelter a triangle (like an upside-down V) with one end solid?
Also, 2 square feet total fabric doesn't make sense. 200 maybe?
The very first thing you need to do is compute the area of the cross-section as a function of the length of the side and the height. The largest cross-section comes from a tent with a side angle of 45 degrees (optimise area as a function of angle by taking the derivative).
At this point, you will develop a formula for the volume of the shelter as a function of length, given the constant amount of fabric available using the two formulas:
V = area*length (area is the cross section)
fabric amount = side length*2*length + end panel
Geometry/Calculus problem?
Gary
A canvas wind shelter is to be constructed for use on the beach
of an island. It is to have a back, two square
sides, and a top. If 150 square feet of canvas
are available, find the depth of the shelter for
which the space inside is maximized assuming
all the canvas is used.
Answer
Say: depth = P and width = L
surface of left side = right side = P²
surface of top = back = PL
total surface = 2P² + 2PL = 150 â L = (150 - 2P²) / 2P = (75 - P²) / P
volume = V = P²L= P²(75 - P²) / P = 75P - P³
V' = 75 - 3P² = 0
âD = â(0² - 4â(-3)â75) = â900 = 30
P = (0 - 30) / (2â(-3)) = 5
P = (0 + 30) / (2â(-3)) = -5
Negative depth makes no sense, so the only solution is P = 5
Say: depth = P and width = L
surface of left side = right side = P²
surface of top = back = PL
total surface = 2P² + 2PL = 150 â L = (150 - 2P²) / 2P = (75 - P²) / P
volume = V = P²L= P²(75 - P²) / P = 75P - P³
V' = 75 - 3P² = 0
âD = â(0² - 4â(-3)â75) = â900 = 30
P = (0 - 30) / (2â(-3)) = 5
P = (0 + 30) / (2â(-3)) = -5
Negative depth makes no sense, so the only solution is P = 5
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