A rectangular box with a volume of 64 ftcubed is to be constructed with a square base and top. The cost per square foot for the bottom is 15cents, for the top is 10cents, and for the sides is 2.5cents. What dimensions will minimize the cost?
To minimize the cost, we need to find the dimensions that result in the least total cost for constructing the box.
Let's assume the side length of the square base and top is x ft. Then the height of the box will be 64/x ft, since the volume of the box is 64 ft³.
The cost of the bottom and top can be calculated as the area multiplied by the cost per square foot. The cost of the sides can be calculated as the perimeter multiplied by the cost per foot.
Cost of the bottom = (x²) * 0.15
Cost of the top = (x²) * 0.10
Cost of the sides = 4 * (2x + 64/x) * 0.025
Total cost = Cost of the bottom + Cost of the top + Cost of the sides
Now, let's find the derivative of the total cost with respect to x, and set it equal to zero to find the minimum:
d(Total cost)/dx = d(Cost of the bottom)/dx + d(Cost of the top)/dx + d(Cost of the sides)/dx
We can simplify the equation by expanding and rearranging it to solve for x.
Let's calculate it step-by-step:
1. Calculate the derivative of the cost of the bottom:
d(Cost of the bottom)/dx = 2 * x * 0.15 = 0.30x
2. Calculate the derivative of the cost of the top:
d(Cost of the top)/dx = 2 * x * 0.10 = 0.20x
3. Calculate the derivative of the cost of the sides:
d(Cost of the sides)/dx = 4 * (2 - 64/x²) * 0.025
4. Simplify the derivative of the sides:
d(Cost of the sides)/dx = (8 - 256/x²) * 0.025
= (8 - 256/x²) * 0.025
= 0.2 - 6.4/x²
5. Calculate the derivative of the total cost:
d(Total cost)/dx = 0.30x + 0.20x + (0.2 - 6.4/x²)
= 0.50x + 0.2 - 6.4/x²
Now, let's set the derivative equal to zero and solve for x:
0.50x + 0.2 - 6.4/x² = 0
0.50x² + 0.2x - 6.4 = 0
Now we can solve this quadratic equation to find the value of x that minimizes the cost.
To find the dimensions that will minimize the cost, we need to determine the dimensions of the square base and the height of the box.
Let's assume the side length of the square base is x. Since the base is square, the area of the base will be x^2.
The height of the box will be h.
The volume of the rectangular box is given as 64 ft^3, so we can write the equation for the volume as:
Volume = x^2 * h = 64
Now, let's express the cost in terms of x and h.
The cost for the bottom is given as 15 cents per square foot, so the cost for the bottom will be:
Cost_bottom = 0.15 * x^2
The cost for the top is given as 10 cents per square foot, so the cost for the top will be:
Cost_top = 0.10 * x^2
Since there are two sides, the cost for the sides will be:
Cost_sides = 2 * 0.025 * (x * h + x * h) = 0.05 * 2 * x * h = 0.10 * x * h
Now, let's express the total cost in terms of x and h:
Total_cost = Cost_bottom + Cost_top + Cost_sides
= 0.15 * x^2 + 0.10 * x^2 + 0.10 * x * h
To minimize the cost, we can take the derivative of the total cost with respect to x, set it equal to zero, and solve for x.
d(Total_cost)/dx = 0.30 * x + 0.10 * h = 0
0.30 * x = -0.10 * h
x = -0.10 * h / 0.30
x = -(1/3) * h
Now, substitute this value of x in the equation for the volume:
x^2 * h = 64
(-(1/3) * h)^2 * h = 64
(1/9) * h^3 = 64
h^3 = 576
h = ∛576
h ≈ 8
Substitute the value of h back into the equation for x:
x = -(1/3) * h
x = -(1/3) * 8
x ≈ -8/3
Since the dimensions cannot be negative, we take the positive values:
x ≈ 8/3
h ≈ 8
Therefore, the dimensions that will minimize the cost are approximately:
Square base side length: 8/3 feet (or approximately 2.7 feet)
Box height: 8 feet
If the base has side x, and the height is y, then
x^2 y = 64
The cost is
c(x,y) = 15x^2 + 10x^2 + 2.5 * 4xy
But, y = 64/x^2, so
c(x) = 25x^2 + 640/x
Now just find where c'(x) = 0