A trash company is designing an open-top, rectangular container that will have a volume of 135 ft^3.The cost of making the bottom of the container is $5 per square foot, and the cost of the sides is $4 per square foot. Find the dimensions of the container that will minimize total cost.
If the base has dimensions x and y, and the height is z, you want to
minimize 5xy + 4*2z(x+y)
subject to
xyz = 135
Now use your favorite linear programming technique. See whether you can arrive at the solution of
540 at (6,6,15/4)
These things usually work out so that two of the dimensions are equal, thus reducing the variables from three to two.
To minimize the total cost, we need to find the dimensions of the container that will minimize the sum of the cost of the bottom and the cost of the sides.
Let's assume the length of the container is L ft, the width is W ft, and the height is H ft.
The volume of a rectangular container is given by the formula:
Volume = Length × Width × Height
Given that the volume is 135 ft^3, we can write the equation:
L × W × H = 135
The formula for the cost of the bottom is:
Cost of the bottom = Area of the bottom × Cost per square foot
The area of the bottom is given by the formula:
Area of the bottom = Length × Width
The formula for the cost of the sides is:
Cost of the sides = 2 × (Area of side 1 + Area of side 2 + Area of side 3) × Cost per square foot
The area of each side is given by the formulas:
Area of side 1 = Length × Height
Area of side 2 = Width × Height
Area of side 3 = Length × Width
So, the total cost is:
Total cost = Cost of the bottom + Cost of the sides
Total cost = (Length × Width) × 5 + 2 × ((Length × Height) + (Width × Height) + (Length × Width)) × 4
To find the dimensions that will minimize the total cost, we will use optimization techniques by finding the critical points. To do this, we will differentiate the total cost equation with respect to one variable, set it equal to zero, and solve for that variable.
Let's differentiate the total cost equation with respect to Length:
d(Total cost)/dLength = 5W + 2(2H + W) × 4 = 0
5W + 8H + 8W = 0
Next, let's differentiate the total cost equation with respect to Width:
d(Total cost)/dWidth = 5L + 2(2H + L) × 4 = 0
5L + 8H + 8L = 0
Now, let's solve these two equations to find the values of Length (L), Width (W), and Height (H) that minimize the total cost.
To find the dimensions of the container that will minimize the total cost, we need to formulate an equation for the cost of the container in terms of its dimensions.
Let's assume the length of the container is L, the width is W, and the height is H.
The volume of a rectangular container is given by V = L * W * H, and in this case, the volume is given as 135 ft^3. So we have the equation:
L * W * H = 135
Now, let's express the cost of the container in terms of its dimensions. The cost of the bottom is $5 per square foot, and the area of the bottom is L * W. So the cost of the bottom is:
Cost of the bottom = $5 * (L * W)
Similarly, the cost of the sides is $4 per square foot, and the area of the sides is 2LH + 2WH (since there are two lengths and two widths). So the cost of the sides is:
Cost of the sides = $4 * (2LH + 2WH)
Therefore, the total cost of the container is:
Total cost = Cost of the bottom + Cost of the sides
= $5 * (L * W) + $4 * (2LH + 2WH)
= 5LW + 8LH + 8WH
Now, we want to minimize the total cost. To do that, we need to find the values of L, W, and H that will make the total cost as small as possible.
We can use calculus to find the critical points of the function Total cost = 5LW + 8LH + 8WH, and then check for the minimum value.
First, let's express one of the variables (either L, W, or H) in terms of the other two variables using the volume equation L * W * H = 135.
For example, we can express L in terms of W and H by dividing the volume equation by WH:
L = 135 / (W * H)
Substituting this expression for L into the total cost equation, we get:
Total cost = 5(W * H) * W + 8(135 / (W * H)) * H + 8W * H
Simplifying, we obtain:
Total cost = 5W^2 + (1080 / W) + 8W^2
To find the critical points, we differentiate the total cost function with respect to W and set it to zero:
dTotal cost / dW = 10W - (1080 / W^2) = 0
Multiplying both sides by W^2, we have:
10W^3 - 1080 = 0
Simplifying, we find:
W^3 - 108 = 0
Solving this equation, we get:
W = cbrt(108) ≈ 4.843
Now, substitute this value of W back into the volume equation to find the corresponding values of L and H:
L = 135 / (W * H)
≈ 135 / (4.843 * H)
Substitute W = 4.843 and solve for L:
L ≈ 135 / (4.843 * H)
Hence, the dimensions of the container that will minimize the total cost are approximately:
Length (L) ≈ 135 / (4.843 * H)
Width (W) ≈ 4.843
Height (H) ≈ H
Therefore, to minimize the total cost, set the width to approximately 4.843 ft and calculate the length and height of the container using the equations above.