Length of the side of a square base to provide a volume of 4000 in^3 with the minimum material to construct the entire box. is 6 in. The dimensions for this box would be (6 in) by (6 in) by 111.1 in.

Step-by-step explanation:

The volume of a square box is given by

Volume(V) = L*W*H,

where L is length, W is width, and H is height.

Since the base is square, W = L, so we can rewrite the volume equation as:

V = L*W*H

V = L*L*H

Volume = L^2H

The area of the box surface is the sum of all the sides:

Top and Bottom: 2*(L*W), or 2 (W^2)

2 Sides: 2*(W*H), or 2(W^2)

2 other sides: 2*(L*H)

The total surface area is

Area = 2*(L*W) + 2*(W*H) + 2*(L*H)

Since W=L, we can rewrite this equation as:

Area = 2*(L*L) + 2*(L*H) + 2*(L*H)

Total Surface Area = 2L^2 + 4LH

The target volume is 4000 in^3

We know that Volume = L^2H from above.

4000 in^3 = L^2H

we can rewrite this as:

H = 4000/L^2

Substitute this value of H into Total Surface Area = 2L^2 + 4LH

Total Surface Area = 2L^2 + 4L(4000/L^2)

Total Surface Area = 2L^2 + (1000/L)

We want to minimize total surface are, so we can either plot this function and look for the minimum, or we can take the first derivative and set it equal to zero (the first derivative tells us the slope of the line at any point. We want the point where the slope is zero, where it changes direction). I will plot the function.

Plot: See attached graph.

I set the above equation to the format: y = 2x^2 + (1000/x), where y is the total surface area and x is the box length.

I see a minimum at (6.2,230). This corresponds to a box length of 6.2 in and a surface area of 230 in^2.

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2 years ago