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Find possible dimensions for a closed box with volume 1176 cubic​ inches, surface area 1036 square​ inches, and length that is twice the width.

User Jaquan
by
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1 Answer

3 votes

Answer:

Length (L) ≈ 37.17 inches

Width (W) ≈ 18.58 inches

Height (H) ≈ 0.475 inches

Explanation:

Let's denote the dimensions of the closed box as follows:

Length: L

Width: W

Height: H

We are given the following information:

Volume of the box = 1176 cubic inches

Surface area of the box = 1036 square inches

Length is twice the width (L = 2W)

To find the dimensions, we can set up a system of equations based on the given information and the formulas for volume and surface area of a rectangular box:

Volume (V) of a rectangular box = Length (L) × Width (W) × Height (H)

V = L × W × H

Surface Area (SA) of a rectangular box = 2lw + 2lh + 2wh

SA = 2(LW) + 2(LH) + 2(WH)

Now, let's proceed with the calculations:

Given:

V = 1176 cubic inches

SA = 1036 square inches

L = 2W

Volume equation:

1176 = L × W × H

Surface Area equation:

1036 = 2(LW) + 2(LH) + 2(WH)

Now, substitute L = 2W into the surface area equation:

1036 = 2(2W)(W) + 2(2W)H + 2(W)H

1036 = 4W^2 + 4WH + 2WH

Now, we have two equations:

1176 = 2W^2H

1036 = 6W^2H

Now, we can solve this system of equations to find the values of W and H.

Divide equation 2 by 6:

W^2H = 1036/6

W^2H = 172.67

Now, substitute this value of W^2H into equation 1:

1176 = 2(172.67)

1176 = 345.33

Now, solve for W:

W = √(345.33)

W ≈ 18.58 inches

Now, substitute the value of W into equation 2 to find H:

1036 = 6(18.58)^2H

1036 = 2180.69H

Now, solve for H:

H = 1036 / 2180.69

H ≈ 0.475 inches

Finally, we can find the value of L using the given relationship L = 2W:

L = 2 × 18.58

L ≈ 37.17 inches

So, the possible dimensions of the closed box are approximately:

Length (L) ≈ 37.17 inches

Width (W) ≈ 18.58 inches

Height (H) ≈ 0.475 inches

User Ntwrkguru
by
8.2k points
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