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?
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.
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 ?
To find the dimensions of the tent that minimize the cost of the material used, we need to set up expressions for the cost in terms of the dimensions.
Let's denote the side length of the equilateral triangle (the cross-section of the prism) as x and the height of the prism as h.
The volume of a right prism is given by the formula: V = base area * height. In this case, the base area is the area of an equilateral triangle, which can be calculated using the formula: base area = (sqrt(3) / 4) * side length^2.
Given that the volume of the tent is 2.2 m^3, we have the equation: 2.2 = (sqrt(3) / 4) * x^2 * h.
To determine the cost of the material, we need to consider the different costs for the floor and the rest of the material. Let's denote the cost per square meter of the floor material as c, and the cost per square meter of the remaining material as 1.4c (since it is 1.4 times as expensive).
The cost of the floor material can be calculated as the area of the triangular floor multiplied by the cost per square meter: Cost_floor = c * base area. Substituting the formula for the base area, we have: Cost_floor = c * (sqrt(3) / 4) * x^2.
The cost of the rest of the material (the ends and the top) can be calculated as the combined area of the four triangular faces multiplied by the cost per square meter: Cost_material = 1.4c * 4 * base area. Substituting the formula for the base area, we have: Cost_material = 1.4c * 4 * (sqrt(3) / 4) * x^2.
Now, let's express the total cost in terms of x and h:
Cost_total = Cost_floor + Cost_material
Cost_total = c * (sqrt(3) / 4) * x^2 + 1.4c * 4 * (sqrt(3) / 4) * x^2
Cost_total = c * (sqrt(3) / 4) * x^2 + 1.4c * (sqrt(3) / 4) * x^2
To find the dimensions that minimize the cost of the material used, we need to minimize the cost function Cost_total with respect to x and h. We can do this by taking partial derivatives with respect to x and h and setting them equal to zero.
I will explain further in the next message.