you provide no data on the cost of the door material, relative to the sides and floor.
However, so far we know
if the side of the triangle is x
the length of the tent is y,
the area of the triangles is 1/2 * x/2 * √3x^2/2 = √3x/8
v = √3x^2/8 y = 2.2
so,
y = 17.6/(√3x^2)
Now, if
cost of door = a
cost of floor = b
cost of sides/end = 1.4b
total cost = a(√3/8 x^2) + 2bxy + b√3/8 x^2 + 1.4bxy
= a(√3/8 x^2) + 2bx(17.6/(√3x^2)) + b√3/8 x^2 + 1.4bx(17.6/(√3x^2))
= a√3/8 x^2 + 35.2b/(x√3) + b√3/8 x^2 + 24.64b/(x√3)
= (a+b)√3/8 x^2 + 59.84b/(x√3)
dcost/dx = (a+b)√3/4 x - 59.84b/(√3 x^2)
cost is minimum when
(a+b)√3/4 x - 59.84b/(√3 x^2) = 0
3/4 (a+b) x^3 = 59.84b
x = cuberoot(79.79b/(a+b))
supply the value for a, and you have your answer for x. Then you can figure cost and area.
Imagine making a tent in the shape of a right prism whose cross-section is an
equilateral triangle (the door is on one of the triangular ends). Assume we want the volume
to be 2.2 m3, to sleep two or three people. The
oor of the tent is cheaper material than
the rest: assume that the material making up the ends and the top of the tent is 1.4 times
as expensive per square meter as the material touching the ground.
(a) What should the dimensions of the tent be so that the cost of the material used is a
minimum?
(b) What is the total area of the material used?
2 answers
the area of the triangles is 1/2 * x/2 * √3x^2/2 = √3x/8 ? isn't it 2*(area of one triangle) = 2*(sqrt(3)/4)x^2= (sqrt(3)/2)x^2 ?