A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. Building the tank costs $10 per square meter for the base and $5 per square meter for the sides.
a) Write a function C, the cost of constructing the described tank as a function of l(the length) and h(the height).
b) Write a function C, the cost of constructing the described tank as a function of a single variable.
c) What is the cost of the least expensive tank? (show all work)
let its length be l m
and its height by h m
volume = 4lh
but 4lh = 36
h = 9/l
a) cost = 10(4l) + 5(2lh) + 5(8h)
= 40l + 10lh + 40h
b) cost = 10(4l) + 5(2l)(9/l) + 40(9/l
= 40l + 90 + 360/l ,where l ≠ 0
c) d(cost)/dl = 40 + 0 - 360/l^2 = 0 for a min of cost
40 = 360/l^2
l^2 = 9
l = √9 = 3
when l = 3, h = 9/3 = 3
dimensions for min cost = 3by4 for the base and a height of 3
cost = 330.00
test: take a value slightly higher and lower than l = 3
l = 3.1, cost = 330.13 , higher
l = 2.9 , cost = 330.14 , higher
answer looks good
a) To write the function C, we need to calculate the areas of the base and sides, and then multiply them by their respective costs.
The area of the base is given by length * width = l * 4.
The area of each side is given by length * height = l * h.
The total cost is the sum of the cost of the base and four sides:
C(l, h) = (l * 4 * 10) + (l * h * 5 * 4)
C(l, h) = 40l + 20lh
b) To write the function C as a single variable, we can express the height in terms of the length. Since we know the volume, we can write h in terms of l:
Volume = length * width * height
36 = l * 4 * h
h = 36 / (4l)
h = 9 / l
Now we substitute this back into the function C:
C(l) = 40l + 20l(9 / l)
C(l) = 40l + 180
c) To find the cost of the least expensive tank, we need to minimize the function C(l). Since the cost per square meter is the same for the base and sides, the cost is proportional to the perimeter of the base. Therefore, we want to minimize the perimeter.
The perimeter of the base is given by P = 2l + 2w = 2l + 2(4) = 2l + 8.
Since we want to minimize the cost, we want to minimize the perimeter. The minimum value of the perimeter occurs when l has the smallest possible value. In this case, l = 1.
Therefore, the cost of the least expensive tank is:
C(1) = 40(1) + 180 = 220.
a) To find the cost, we need to calculate the area of the base and the area of the sides separately, and then multiply by their respective costs per square meter.
The area of the base is given by: A = l * w, where l is the length and w is the width.
The area of the sides is given by: A = 2 * (l * h + w * h), where h is the height, and we multiply by 2 because there are two identical sides.
Therefore, the cost function C(l, h) is:
C(l, h) = (10 * (l * w)) + (5 * 2 * (l * h + w * h))
b) Since the width is fixed at 4 meters, we can express the cost function C(h) as a function of the height only, by substituting w = 4 into the formula from part a:
C(h) = (10 * (l * 4)) + (5 * 2 * (l * h + 4 * h))
c) To find the cost of the least expensive tank, we need to find the minimum value of the cost function C(h).
To solve for the minimum, we can take the derivative of C(h) with respect to h, set it equal to zero, and solve for h.
First, let's simplify the cost function C(h):
C(h) = 40l + 10lh + 40h + 8lh
Now, let's take the derivative:
C'(h) = 10l + 10l + 40
Setting the derivative equal to zero and solving for h:
10l + 10l + 40 = 0
20l = -40
l = -2
Since l cannot be negative in this context, we discard this solution.
Therefore, there is no minimum cost, as the least expensive tank cannot be constructed given the given specifications and cost constraints.
a) To write the cost function C as a function of l (length) and h (height), we need to calculate the surface area of each component (base and sides) and multiply it by the respective cost per square meter.
The base of the tank has a length of l and a width of 4 meters, so the base area is given by A_base = l * 4.
The sides of the tank have a length of l, a height of h, and there are two sides. So, the total combined area of the sides is A_sides = 2 * l * h.
Now, the cost function C can be written as:
C(l, h) = cost of base + cost of sides
C(l, h) = (A_base * cost per square meter for base) + (A_sides * cost per square meter for sides)
C(l, h) = (l * 4 * 10) + (2 * l * h * 5)
C(l, h) = 40l + 10lh
b) To write the cost function C as a function of a single variable, we can eliminate one of the variables (l or h). Since we already know the volume of the tank is 36 cubic meters, we can express one variable in terms of the other.
The volume of a rectangular prism is given by length * width * height. In this case, we have:
l * 4 * h = 36
Simplifying the equation:
l * h = 9
We can express l in terms of h as:
l = 9/h
Now we substitute this value of l in the cost function:
C(h) = 40(9/h) + 10h(9/h)
C(h) = 360/h + 90
c) To find the cost of the least expensive tank, we need to minimize the cost function C(h). To do this, we can find the critical points by taking the derivative of C(h) with respect to h and setting it equal to zero.
C'(h) = -360/h^2 + 10
Setting C'(h) = 0:
-360/h^2 + 10 = 0
360/h^2 = 10
360 = 10h^2
36 = h^2
h = √36
h = 6
Substituting h = 6 into the cost function:
C(6) = 360/6 + 90
C(6) = 60 + 90
C(6) = 150
Therefore, the cost of the least expensive tank is $150.