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icang [17]
3 years ago
14

A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

The base needs to be stronger than the sides, so it costs more. The base costs $9 per f t 2 , while the sides only cost $6 per f t 2 . We wish to find the dimensions of the box that minimize the cost of material used.
Mathematics
1 answer:
noname [10]3 years ago
8 0

Answer:

Width = 4ft

Height = 4ft

Length = 8ft

Step-by-step explanation:

Given

Volume = 128ft^3

L = 2W

Base\ Cost = \$9/ft^2

Sides\ Cost = \$6/ft^2

Required

The dimension that minimizes the cost

The volume is:

Volume = LWH

This gives:

128 = LWH

Substitute L = 2W

128 = 2W * WH

128 = 2W^2H

Make H the subject

H = \frac{128}{2W^2}

H = \frac{64}{W^2}

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

A = LW + 2(WH + LH)

The cost is:

Cost = 9 * LW + 6 * 2(WH + LH)

Cost = 9 * LW + 12(WH + LH)

Cost = 9 * LW + 12H(W + L)

Substitute: H = \frac{64}{W^2} and L = 2W

Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)

Cost =18W^2 +  \frac{768}{W^2}*3W

Cost =18W^2 +  \frac{2304}{W}

To minimize the cost, we differentiate

C' =2*18W +  -1 * 2304W^{-2}

Then set to 0

2*18W +  -1 * 2304W^{-2} =0

36W - 2304W^{-2} =0

Rewrite as:

36W = 2304W^{-2}

Divide both sides by W

36 = 2304W^{-3}

Rewrite as:

36 = \frac{2304}{W^3}

Solve for W^3

W^3 = \frac{2304}{36}

W^3 = 64

Take cube roots

W = 4

Recall that:

L = 2W

L = 2 * 4

L = 8

H = \frac{64}{W^2}

H = \frac{64}{4^2}

H = \frac{64}{16}

H = 4

Hence, the dimension that minimizes the cost is:

Width = 4ft

Height = 4ft

Length = 8ft

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