A piece of wire used to construct a cuboid such that the base is 3 times its width. The condition is that the total S.A of the cuboid is 5096cm^2. Find the minimum length of wire required.

Question

S.A = surface area

Clueless on how to solve it. Any help would be appreciated. Thank you.

Answer 1

Greetings to you,

Your question deals with the subject of optimization and using the calculus techniques of the first derivative to obtain a "minimum or maximum," buzzwords for optimization, value. Thus you will need two equations, one that models the length of the wire and one that models the surface area of a cuboid figure. Note that the wire equation will be the one to undergo the first derivative with the keyword "minimum" associated with it.

The word cuboid is a fancy word for a rectangular prism that appears to be a cube, like a rubics cube, so by drawing and sketching a picture, one can develop the surface area equation model and this same model will be use to derive the length of the wire, which is used to construct the cuboid.

Step 1: Find Surface Area Equation Model

Surface Area = 2B + Ph → two times the area of the base plus the perimeter of the base times the height

x = width;

3x = length or "base" in the description above;

h = height

B = Area of the base = (x·3x) = 3x²; P = Perimeter of base = (x + x + 3x + 3x) = 8x

Surface Area = 2(3x²) + 8xh → 6x² + 8xh

Set equation equal to 5096 cm² as follows: 5096 = 6x² + 8xh

Use algebra to solve and isolate for variable h as follows: h = (5096 - 6x²) ÷ 8x

Step 2: Find Equation Model for length of wire strand

To do this, you will add up all the edges (12 edges in total to be exact) of your cuboid diagram/sketch to get the following organized and group according to their three dimensions.

L_w = (x + x + x + x) + (3x + 3x + 3x + 3x) + (h + h + h + h) which simplifies to the following below:

L_w = (4x) + (12x) + (4h)

L_w = 16x + 4h

Plugging in the surface area expression solved in terms of h yields the following:

L_w = 16x + 4(5096 - 6x²) ÷ 8x → L_w = 2548x^-1 + 13x

Step 3: Find minimized value via 1st Derivative and solve for x

L_w = 2548x^-1 + 13x

L' = -2548x^-2 + 13

0 = -2548x^-2 + 13

2548x^-2 = 13

x = ±14, so we choose the positive result for it does not makes logical sense to have a negative length in our model, thus x = 14 cm.

Step 4: Find h value

h = (5096 - 6x²) ÷ 8x

h = (5096 - 6(14)²) ÷ (8(14)) → 35 cm

Step 5: Find minimum wire length using L equation

L_w = 16x + 4h → 16(14cm) + 4( 35 cm) = 91 cm is the minimum length needed to build a cuboid that has a surface area 5096 cm²

Answer 2

Assume the shape is 6 rectangles. Label the dimensions L (length), W (width), and H (height).

When you say "base is 3x the width", I assume length is 3x the width......L = 3W

The total surface area is 2LW + 2LH + 2 WH = 5096

The length of wire required is 4L + 4W + 4H (Call this "X")

In each equation, replace L with 3W:

6W^2 + 8WH = 5096 I

X = 16W + 4H II

Solve (I) for H:

8WH = 5096 - 6W^2

H = 637/W - 3W/4

Substitute into (II):

X = 16W + 2548/W - 3W

X = 13W + 2548/W

To find the minimum or maximum X, take the 1st derivative and set = to 0:

X' = 13 - 2548 /W^2 = 0

W^2 = 2548/13 = 196, W = +/- 14 (Ignore the - answer)

SO:

W = 14

Total length = X = 13W + 2548/W

X = 182 + 182 = 364

Answer 3

I am assuming a cuboid is a fancy word for a box. So, the surface area will be the sum of the 6 rectangles. Top and bottom will be lw and 2 sides will be lh and two sides will be wh. Now, substitute l/3 for w to eliminate a variable so SA = 2L²/3 + 2Lh + 2Lh/3 . but SA is 5096.

so, 5096 = 2L² / 3 + 8Lh / 3. I will solve for h in this equation.

15288 - 2L² / 8L = h

also, the total length of wire will be 4L + 4h + 4w again, substituting to eliminate w we get

f(L) = 4L + 4h + 4L/3

lets substitute to eliminate h

f(L) = 4L + 4(15288 - 2L² / 8L) + 4L/3 is the function for the total length of wire. Find the minimum by a graph or probably first derivative test.