A lidless box is to be made using 2 m^2 of cardboard. Find the dimensions of the box with the largest possible volume.

Question

A lidless box is to be made using 2 m² of cardboard. Find the dimensions of the box with the largest possible volume.

A lidless cardboard box is to be made with a volume of 4 m³. Find the dimensions of the box that requires the least amount of cardboard.

Answer 1

Unfortunately, this problem requires a bit of algebra, but you can use Wolfram Alpha to do some simplification for you.

Since we are dealing with a box, the volume is V = LWH, and the total surface area is A = 2LW + 2LH + 2HW

a) First, we set our total surface area to equal 2, so 2 = 2LW + 2LH + 2HW. The 2's cancel out, and we can solve for one of the variables (I chose L in this case):

LH + LW + HW = 1

L (H+W) = 1 - HW

L = (1 - HW) / (H + W)

This means our volume is V = LWH = WH (1 - HW) / (H + W)

We are trying to minimize our volume, so we are looking for solutions where dV/dH = 0 and dV/dW = 0. They are equivalent just with the terms swapped.

dV/dH = W (1 - HW) / (H +W) + WH (-W) / (H + W) + WW(1-HW) * -1 / (H+W)² = 0

W is common to each term, so that cancels out. 1/(H+W) is also common, so that cancels too

1 - HW - HW - H(1-HW)/(H+W) = 0

1 - 2HW - H(1-HW)/(H+W) = 0

H + W - 2HW - H + H²W = 0

H² - 2HW + W = 0, so (H - W)² = 0 implying that H = W

Inserting that into our equation for L, we have:

L = (1 - HW) / (H + W) = (1 - W²) / (2W)

Inserting that into our Volume equation, we have:

V = LWH = (1 - W²) / (2W) · W² = (W/2) · (1 - W²) = W/2 - W³/2

Taking that derivative and setting it to zero, we have

dV/dW = 0 = 1/2 - 3W²/2

1 = 3W², so W = 1/sqrt(3). This also means that H = 1/sqrt(3), and

L = (1 - W²) / (2W) = (1 - 1/3) / (2 · sqrt(3)) = (2/3) / (2 ·sqrt(3)) = 1/(2·sqrt(3))

Putting it together, H = W = 1/sqrt(3), and L = 1/(2⋅sqrt(3))

b) We're going in reverse this time. V=LWH = 4, so L = 4/(HW)

A = 2LH + 2LW + 2HW = 2 · (4/(HW) · H + 4/(HW) · W + HW)

A = 2· ( 4/W + 4/H + HW)

Now, we're going to minimize the area, so dA/dW = 0 and dA/dH = 0

dA/dW = 2 · (-4/W² + 0 + W) = 0

4/W² = W

4 = W³, so W = 4^1/3, likewise, the other derivative produces H = 4^1/3

Our earlier equation for L gives:

L = 4/(HW) = 4/(4^2/3) = 4^1/3

Therefore, L = W = H = 4^1/3

This should make sense since the shape that produces the greatest volume with the least surface area is a sphere

Answer 2

box is made from 2 m by 2m with corners cut out, x by x, then the rest folded up

that's one version of this problem, commonly used way to make a lidless box

then V = (2-2x)(2-2x)x=4x-8x^2+4x^3

take the derivative, set =0 and solve for x

V'= 4-16x+12x^2=0

3x^2-4x+1 = 0

(3x-1)(x-1)=0

x=1,1/3

V(1) =0 that's minimum Volume

V(1/3) = (2-2/3)(2-2/3)(1/3) = max V =16/27 m^3

That means you have 4/9 m^2 material left over

If you use all the material to make the lidless box, that's a different solution

but on the same version of the problem to get a 4 m^3 lidless box, the square material required is

found with the reverse of the above

V=4=(x)(s-2x)

=about .63(3.8-1.26)= about 4 m^3

material is 3.8 by 3.8 m

surface area = base+4 sides = (3.8-1.26)^2 + 4(.63)(3.8-1.26)

min surface area=about 8.84 m^2