a tent on a square base of side 2a consists of four vertical sides of height b surmounted by a regular pyramid of height h if the volume enclosed is V show that the area of the canvass in the tent is 2V/a - 8/3ah + 4a√(h^2+a^2) show that the least area of the canvass corresponding to a given volume is given by a=√5/h and b=1/2h
Answers
Given : a tent on a square base of side 2a consists of four vertical sides of height b surmounted by a regular pyramid of height h
the volume enclosed is V
To Find : show that the area of the canvass in the tent is 2V/a - (8/3)ah + 4a√(h²+a²)
show that the least area of the canvass corresponding to a given volume is given by a=√5/h and b=1/2h
Solution:
Volume of cuboid shape = 2a*2a*b = 4a²b
Volume of pyramid shape = (2a)²h/3 = 4a²h/3
V = 4a²b + 4a²h/3
Lateral Surface area of Square pyramid = 2(2a)√(2a)²/4 + h²
= 4a√(a² + h²)
Surface area of cuboid shape = 4*2a*b = 8ab ( assuming nothing at base)
area of the canvass in the tent = 8ab + 4a√(a² + h²)
V = 4a²b + 4a²h/3
=> 4a²b = V - 4a²h/3
=> 4ab = V/a - 4ah/3
=> 8ab = 2V/a - 8ah/3
area of the canvass in the tent = 2V/a - 8ah/3 + 4a√(a² + h²)
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