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 floor 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?
Now change the problem so that the floor is more expensive material than the rest; assume that the material is 1.4 times as expensive per square meter as the material making up the top of the tent.
(c)What should the dimensions of the ten be so that the cost of the material used is a minimum?
(d) what is the total area of the material used?
the area of a triangle of side s is √3/4 s^2
So, if the triangles have side x and the tent has length y, we have
(√3/4)x^2 * y = 2.2
the area of cloth used is
a = 2(√3/4)x^2 + 3xy
and the cost c is
c = 1.4(2(√3/4)x^2 + xy)+xy
Now just find dc/dx, after substituting in for y.
I assume that then you can do (b)-(d)
To find the dimensions of the tent that minimize the cost of the material used, we need to consider the surface area of the tent. The surface area is divided into two parts - the floor and the top/ends. Let's solve parts (a) and (b) first, where the floor is cheaper than the rest of the tent.
(a) To minimize the cost of the material used:
Step 1: Determine the dimensions of the tent.
Let's assume the side length of the equilateral triangle cross-section is "x", and the height of the prism is "h".
Step 2: Find the surface area of the floor.
The floor is an equilateral triangle, so its area is given by: Floor A = (√3 / 4) * x^2.
Step 3: Find the surface area of the top + ends.
The top and ends of the tent form 3 congruent rectangles. The width of each rectangle is "x", and the length is "h". Therefore, the total surface area of the top and ends is given by: Top + Ends A = 3 * x * h.
Step 4: Find the total surface area of the tent.
Total A = Floor A + Top + Ends A = (√3 / 4) * x^2 + 3 * x * h.
Step 5: Express the cost of the material in terms of surface area.
The cost of the floor material is "Cf" per square meter, while the cost of the top/ends material is 1.4 times more expensive, so it is 1.4 * "Cf" per square meter.
Step 6: Express the cost function.
The cost function "C" in terms of the cost of materials used can be written as:
C = Cf * Floor A + (1.4 * Cf) * Top + Ends A = Cf * (√3 / 4) * x^2 + (1.4 * Cf) * 3 * x * h.
Step 7: Express total volume and height.
The volume of the tent is given as 2.2 m^3. Since the tent is a right prism, the volume is given by: Volume = Floor A * h = (√3 / 4) * x^2 * h.
Hence, h = 2.2 / (√3 / 4) / x^2.
Step 8: Substitute the expression for "h" into the cost function.
C = Cf * (√3 / 4) * x^2 + (1.4 * Cf) * 3 * x * (2.2 / (√3 / 4) / x^2).
Simplifying the expression, we get:
C = Cf * (√3 / 4) * x^2 + (1.4 * Cf) * 6.6 / (√3 / 4).
Step 9: Differentiate C with respect to "x" to find the critical points.
Differentiating C with respect to "x", we get:
dC/dx = Cf * (√3 / 2) * x - (1.4 * Cf) * 2.2 * (√3 / 4) / x^3.
Setting this to zero, we have:
Cf * (√3 / 2) * x = (1.4 * Cf) * 2.2 * (√3 / 4) / x^3.
Simplifying, we get:
x^4 = (2.2 * 4) / (1.4 * 3) = 8 / 3.
So, x = (8 / 3)^(1/4).
(b) To find the total area of the material used:
Using the expression for the floor area and the top + ends area, we can substitute the value of "x" from part (a) into the formulas and calculate the total area.
Total A = Floor A + Top + Ends A.
Now that we've solved parts (a) and (b), let's move on to parts (c) and (d) where the floor is more expensive than the rest of the tent.
(c) To minimize the cost of the material used:
We follow a similar approach as in part (a), but this time the cost function will have different coefficients.
Step 5: The cost of the floor material is now 1.4 * "Cf" per square meter, while the cost of the top/ends material is "Cf" per square meter.
Step 6: Express the cost function.
The cost function "C" in terms of the cost of materials used can be written as:
C = (1.4 * Cf) * Floor A + Cf * Top + Ends A = (1.4 * Cf) * (√3 / 4) * x^2 + Cf * 3 * x * h.
Step 7: Express the height "h".
The volume is still 2.2 m^3, and since the height is now expressed in terms of the floor area, we have:
h = 2.2 / (√3 / 4) / x^2.
Step 8: Substitute the expression for "h" into the cost function.
C = (1.4 * Cf) * (√3 / 4) * x^2 + Cf * 3 * x * (2.2 / (√3 / 4) / x^2).
Step 9: Differentiate C with respect to "x" to find the critical points and solve for "x".
(d) To find the total area of the material used:
Using the expression for the floor area and the top + ends area, substitute the value of "x" from part (c) into the formulas and calculate the total area.
To solve this problem, we need to find the dimensions that minimize the cost of the materials used for the tent.
(a) Let's find the dimensions of the tent when the floor is cheaper material than the rest.
First, let's define the variables:
- Let the height of the prism be h
- Let the side length of the equilateral triangle cross-section be s
The volume of a right prism is given by the formula: V = base area × height.
In this case, the base area is an equilateral triangle, so we need to find the formula for the area of an equilateral triangle.
The area of an equilateral triangle can be found using the formula: A = (s^2√3)/4.
Given that the volume is 2.2 m^3, we have:
2.2 = (A × h)
2.2 = ((s^2√3)/4 × h)
Now, let's solve for s in terms of h and substitute its value into the equation to get rid of s:
s = (2.4 × 4√3)∛((2.2 × 4)/((√3 × h)))
s = 0.67∛(8.8/h)
Now, let's find the cost of the material used.
The cost of the material is cheaper for the floor and more expensive for the top and ends. Given that the floor material is cheaper by a factor of 1.4, we can set the cost of the floor material as 1 unit and the cost of the top and end material as 1.4 units.
The cost of the floor material is proportional to its area, while the cost of the top and end materials is 1.4 times the area of each.
The surface area of the floor is given by: Afloor = s^2
The surface area of each top and end is given by: Atopend = (s × h)/2
The total cost is given by: C = Afloor (1 unit) + 2Atopend (1.4 units)
Now, substitute the expressions for the area of the floor and top/end into the cost equation:
C = (s^2) + 2((s × h)/2)(1.4)
C = s^2 + 1.4 × s × h
To minimize the cost, we need to find the values of s and h that give the minimum value for C. Let's differentiate C with respect to s and h and set these derivatives equal to zero to get the critical points:
dC/ds = 2s + 1.4h = 0
dC/dh = 1.4s = 0
Solve these equations simultaneously to find the values of s and h that yield the minimum cost.
(b) To find the total area of the material used, we need to find the area of the floor, top, and ends.
Total Area = Afloor + 2Atopend
Total Area = s^2 + 2((s × h)/2)
Now, substitute the values of s and h that we obtained from solving part (a) into this equation to get the answer.
Now, let's move on to part (c) and (d) where the floor is more expensive than the rest.
(c) The approach to find the dimensions that minimize the cost of the materials used will be the same as in part (a), with the exception that the cost of the floor material will be 1.4 times the cost of the top and end materials.
(d) The calculation of the total area of the material used will also be the same as in part (b), except the cost of the floor material will be 1.4 times the cost of the top material.
Using these steps, you should be able to find the dimensions, minimum cost, and total area of the material used for both scenarios.