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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?
Q. 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.
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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