a builder intends to construct a storage shed having a volume of 900ft^3, a flat roof and a rectangular base whose width is three-fourths the length. the cost per square foot of the materials is 4000.00 for the floor,6000.00 for the sides and 3000.00 for the roof. what dimension will minimize the cost?
I like to avoid fractions if possible, so let the
width be 3x
and the length be 4x , (notice that 3x/4x) = 3/4)
let the height be h
V= (3x)(4x)(h)
h = 900/(12x^2) = 75/x^2
Cost = C = 4000(top) + 6000(sides) + 3000(roof)
= 4000(12x^2) + 6000(8xh+6xh) + 3000(12x^2)
= 84000x^2 + 84000xh
= 84000x^2 + 84000x(75/x^2
= 84000x^2 + 84000(75)/x
= 84000(x^2 + 75/x)
d(Cost)/dx = 84000(2x - 75/x^2) = 0 for a min of C
2x = 75/x^2
x^3 = 75/2 = 37.5
x = (37.5)^(1/3) = 3.347
width = 3(3.347) = 10.04
length = 4(3.347) = 13.39
height = 75/(3.347)^2 = 6.69 (all in feet)
Well, it seems like we have a task to clown around with some math! To minimize the cost, we need to find the dimensions of the storage shed that will result in the lowest overall cost. Let's get started!
Let's assume the length of the rectangular base is "x" feet. According to the problem, the width is three-fourths the length, so the width would be 3x/4.
Now, let's find the height of the storage shed. We know the volume is 900ft³, so we can express the height as 900 / (length × width).
Using our previous values for length and width, the height would be 900 / (x × 3x/4).
To calculate the cost, we need the total area of each part of the shed, which will be the sum of the area of the floor, sides, and roof.
The floor area is length × width.
The area of each side is height × width.
The roof area is length × width.
Now, let's use these formulas to calculate the cost. The total cost will be the sum of the cost of each part multiplied by its respective area.
Total Cost = (4000 × floor area) + (6000 × 2 × side area) + (3000 × roof area)
Plugging in the formulas for each area and simplifying the equation will give us a function in terms of "x" only.
Once we have that equation, we can find the value of "x" that minimizes the cost by either graphing the equation or by using calculus techniques like taking the derivative and setting it equal to zero.
Voila! We'll have the dimension of the shed that results in the minimum cost. Happy cost-minimizing construction, my friend!
To minimize the cost of constructing the storage shed, we need to find the dimensions that will minimize the surface area. Since the cost per square foot is different for each part, we need to consider the floor, the sides, and the roof separately.
Let's start by expressing the dimensions in terms of the length (L) of the base.
Width of the base = 3/4 * L
Height of the base = L (since it's a rectangular base)
Height of the shed = H
The volume of the shed is given as 900 ft³, so we can write:
L * (3/4 * L) * H = 900
Next, let's isolate H in terms of L:
H = 900 / (L * (3/4 * L))
H = 1200 / (L²)
Now, let's find the total surface area of the shed (including the floor, sides, and roof):
Floor area = L * (3/4 * L) = 3/4 * L²
Side area = 2 * H * (L + 3/4 * L) = 2 * H * (7/4 * L)
Roof area = (3/4 * L) * L = 3/4 * L²
Total surface area = Floor area + Side area + Roof area
Total surface area = (3/4 * L²) + (2 * H * (7/4 * L)) + (3/4 * L²)
Total surface area = 3/2 * L² + (7/2 * H * L)
Now, let's multiply the surface area by the respective cost per square foot for each part and express the total cost:
Cost = (4000 * (3/4 * L²)) + (6000 * (7/2 * H * L)) + (3000 * (3/4 * L²))
Cost = 3000 * (3/4 * L²) + 6000 * (7/2 * H * L) + 4000 * (3/4 * L²)
Cost = 2250 * L² + 10500 * H * L + 3000 * L²
To minimize the cost, we need to find the critical points by taking the partial derivatives with respect to L and H and setting them equal to zero:
dCost/dL = 4500 * L + 10500 * H = 0
dCost/dH = 10500 * L = 0
From the second derivative test, when the second derivative is positive, it is a minimum.
The second derivative of Cost with respect to L is given by:
d²Cost/dL² = 4500
Since d²Cost/dL² is positive (4500 > 0), we can conclude that L has a minimum value.
Therefore, to minimize the cost, we need to find the value of L that satisfies the equation:
4500 * L + 10500 * H = 0
Unfortunately, we cannot determine the exact value of L without additional information.
To minimize the cost of constructing the storage shed, we need to find the dimensions that result in the minimum surface area, as the cost is determined by the surface area of each component (floor, sides, roof) multiplied by the respective cost per square foot of materials.
Let's solve step by step:
Step 1: Define the variables.
Let:
- Length of the base be L (in feet)
- Width of the base be W (in feet)
Step 2: Express the volume in terms of the variables.
The volume of a rectangular shed is given as 900 ft^3. The volume formula is V = L * W * H, where H is the height of the shed. Since the shed has a flat roof, the height will be determined later. However, we can express H in terms of L and W for now.
Given that the width is three-fourths (3/4) the length, we have W = (3/4)L.
So, the volume equation becomes: 900 = L * (3/4)L * H
Step 3: Express H in terms of L.
We can solve the volume equation for H by dividing both sides by L * (3/4)L, giving us:
900 / (L * (3/4)L) = H
1200 / (L^2) = H
Step 4: Express the surface area in terms of the variables.
The surface area of the shed consists of the floor, two sides, and the roof.
- The floor area is L * W = L * (3/4)L = (3/4)L^2.
- The area of each side is H * L, but since there are two sides, the total side area is 2HL.
- The roof area is L * W = L * (3/4)L = (3/4)L^2.
So, the total surface area equation becomes:
A = (3/4)L^2 + 2HL + (3/4)L^2
A = (3/2)L^2 + 2HL
Step 5: Substitute the expression for H from Step 3 into the surface area equation.
A = (3/2)L^2 + 2(L * 1200 / (L^2))
A = (3/2)L^2 + 2400 / L
Step 6: Minimize the cost by finding the critical points.
To minimize the cost, we need to find the critical points of the surface area equation. We can do this by taking the derivative of A with respect to L, setting it to zero, and solving for L.
dA/dL = (3/2)(2L) - 2400 / L^2 = 0
3L - 2400 / L^2 = 0
3L^3 - 2400 = 0
L^3 = 800
L = ∛800 = 9.236
Step 7: Determine the dimensions that minimize the cost.
Since the length cannot be negative, the approximate length that minimizes the cost is L ≈ 9.236 ft.
Substituting this value of L into the equation for W, we get:
W = (3/4) * 9.236 ≈ 6.927 ft.
So, the approximate dimensions that minimize the cost of construction are:
Length ≈ 9.236 ft
Width ≈ 6.927 ft