I guess we are not allowed to cut this canvas up so the tent will be ten feet long and the opening at the end will be an equilateral triangle with slopes(hypotenuse of right triangle bounded by center pole h, slope, and ground g, of 5 ft.
Maximize the area of that triangle
A = .5 g h
dA/dh= .5 g dh/dh + .5 h dg/dh
= .5 g + .5 h dg/dh
= 0 for max or min
g^2 + h^2 = 5^2 = 25
g^2 = 25 - h^2
g = sqrt (25-h^2) = (25-h^2)^.5
dg/dh = .5(25-h^2)^-.5 (-2h)
= - h (25-h^2)^-.5
so
0 = .5(25-h^2)^.5 - .5h^2 (25-h^2)^-.5
(25-h^2)^.5 = h^2 /(25-h^2)
25 - h^2 = h^2
2 h^2 = 25
h^2 = 5^2 /2
h = 5/sqrt 2 = 5 sqrt 2/2 = 3.54 ft
A square piece of canvas measuring 10 ft. by 10 ft. is to be used to make a pup tent with open ends as shown. How long should the center pole be if the volume covered by the tent is to be maximum?
1 answer