Question

A box of volume 108 m3 with square bottom and no top is constructed out of two different materials. The cost of the bottom is $40/m2 and the cost of the sides is $30/m2 . Find the dimensions of the box that minimize total cost. (Let s denote the length of the side of the square bottom of the box and h denote the height of the box.)

(s, h) =

Answer 1

To minimize the total cost of the box, we need to find the dimensions that minimize the sum of the cost of the bottom and sides. By taking the derivative of the cost function and solving for the dimensions, we can find the optimal values. The dimensions that minimize the cost are approximately (s, h) = (3.78 m, 7.09 m).

Step-by-step explanation:

To minimize the total cost, we need to find the dimensions of the box that minimize the sum of the cost of the bottom and the sides.

The volume of the box is given as 108 m³, which can be expressed as s² * h, where s is the length of the side of the square bottom and h is the height of the box.

The cost of the bottom is $40/m² and the cost of the sides is $30/m². The cost can be expressed as 40 * s² + 30 * (s * 4 * h).

To minimize the cost, we need to find the values of s and h that minimize the cost function. We can do this by taking the derivative of the cost function with respect to s and h, setting them equal to zero, and solving for s and h. Once we have the values of s and h, we can calculate the cost.

Let's find the optimal dimensions of the box by using calculus:

Step 1: Write the cost function:

Cost = 40s² + 30s * 4h

Step 2: Write the volume constraint:

s² * h = 108

Step 3: Solve the volume constraint for h:

h = 108 / s²

Step 4: Substitute the value of h into the cost function:

Cost = 40s² + 30s * 4(108 / s²)

Step 5: Simplify the expression:

Cost = 40s² + 4320 / s

Step 6: Take the derivative of the cost function with respect to s:

d(Cost)/ds = 80s - 4320 / s²

Step 7: Set the derivative equal to zero and solve for s:

80s - 4320 / s² = 0

80s = 4320 / s²

80s³ = 4320

s³ = 54

s = ∛54

s ≈ 3.78 m

Step 8: Substitute the value of s into the volume constraint to find h:

h = 108 / s²

h = 108 / (3.78)²

h ≈ 7.09 m

So, the dimensions of the box that minimize the total cost are approximately (s, h) = (3.78 m, 7.09 m).

Answer 2

To minimize the total cost of constructing the box, we need to find the values of s and h that minimize the cost function. The cost function can be expressed as Cost = Cost of Bottom + Cost of Sides. By taking the partial derivatives of the cost function and setting them equal to zero, we find that the dimensions of the box that minimize total cost are s = 0 and h = 0.

Step-by-step explanation:

To minimize the total cost of constructing the box, we need to find the values of s and h that minimize the cost function. The cost function can be expressed as:

Cost = Cost of Bottom + Cost of Sides

Cost of Bottom = Cost per square meter * Area of the bottom = $40/s2 * s2 = $40s

Cost of Sides = Cost per square meter * Surface Area of the sides = $30/s * 4s * h = $120h

Total Cost = $40s + $120h

To minimize the total cost, we need to find the critical points of the cost function by taking partial derivatives with respect to s and h and setting them equal to zero.

Partial derivative with respect to s: dCost/ds = 40 - 120h/s

Partial derivative with respect to h: dCost/dh = 120

Setting both partial derivatives equal to zero:

40 - 120h/s = 0

120 = 0

Solving these two equations simultaneously:

40 - 120h/s = 0 => 40 = 120h/s => 3 = h/s

120 = 0 => h = 0

From the equation h/s = 3, we can see that the height of the box is 3 times the length of the side. Substituting this into the equation for the cost of the bottom, we have:

Cost of Bottom = $40s

Cost of Sides = $120(3s) = $360s

Total Cost = $40s + $360s = $400s

To minimize the total cost, we need to minimize the value of s. Therefore, the dimensions of the box that minimize total cost are s = 0 and h = 0.